PHY101 Physics Handouts PDF

PHYSICS –PHY101 VU PHYSICS 101 AN INTRODUCTION TO PHYSICS This course of 45 video lectures, as well as accompanying notes, have been developed and presented by Dr. P...

PHYSICS –PHY101 VU PHYSICS 101 AN INTRODUCTION TO PHYSICS This course of 45 video lectures, as well as accompanying notes, have been developed and presented by Dr. Pervez Amirali Hoodbhoy, professor of physics at Quaid-e-Azam University, Islamabad, for the Virtual University of Pakistan, Lahore. © Copyright Virtual University of Pakistan 1 PHYSICS –PHY101 VU TABLE OF CONTENTS I. GENERAL INFORMATION II. LECTURE SUMMARIES Page # Lecture 1 Introduction to physics and this course 4 Lecture 2 Kinematics – I 6 Lecture 3 Kinematics – II 8 Lecture 4 Force and Newton’s Laws 10 Lecture 5 Applications of Newton’s Laws – I 12 Lecture 6 Applications of Newton’s Laws – II 14 Lecture 7 Work and Energy 16 Lecture 8 Conservation of Energy 19 Lecture 9 Momentum 22 Lecture 10 Collisions 25 Lecture 11 Rotational Kinematics 27 Lecture 12 Physics of Many Particles 30 Lecture 13 Angular Momentum 35 Lecture 14 Equilibrium of Rigid Bodies 39 Lecture 15 Oscillations - I 42 Lecture 16 Oscillations – II 45 Lecture 17 Physics of Materials 48 Lecture 18 Physics of Fluids 51 Lecture 19 Physics of Sound 54 Lecture 20 Wave Motion 56 Lecture 21 Gravitation 59 Lecture 22 Electrostatics – I 62 Lecture 23 Electrostatics – II 65 Lecture 24 Electric Potential 68 Lecture 25 Capacitors and Currents 71 Lecture 26 Currents and Circuits 74 Lecture 27 The Magnetic Field 78 Lecture 28 Electromagnetic Induction 82 Lecture 29 Alternating Current 85 Lecture 30 Electromagnetic Waves 90 Lecture 31 Physics of Light 94 Lecture 32 Interaction of Light with Matter 98 Lecture 33 Interference and Diffraction 102 Lecture 34 The Particle Nature of Light 106 Lecture 35 Geometrical Optics 110 Lecture 36 Heat – I 115 Lecture 37 Heat – II 121 Lecture 38 Heat – III 125 Lecture 39 Special Relativity – I 129 Lecture 40 Special Relativity – II 135 Lecture 41 Matter as Waves 140 Lecture 42 Quantum Mechanics 147 Lecture 43 Introduction to Atomic Physics 153 Lecture 44 Introduction to Nuclear Physics 160 Lecture 45 Physics of the Sun 168 © Copyright Virtual University of Pakistan 2 PHYSICS –PHY101 VU GENERAL INFORMATION Purpose: This course aims at providing the student a good understanding of physics at the elementary level. Physics is essential for understanding the modern world, and is a definite part of its culture. Background: It will be assumed that the student has taken physics and mathematics at the F.Sc level, i.e. the 12th year of schooling. However, B.Sc students are also likely to find the course useful. Calculus is not assumed and some essential concepts will be developed as the course progresses. Algebra and trigonometry are essential. However, for physics, the more mathematics one knows the better. Scope and Duration: The course has 45 lectures, each of somewhat less than one hour duration. All main fields of physics will be covered, together with several applications in each. Language: For ease of communication, all lectures are in Urdu. However, English or Latin technical terms have been used where necessary. The student must remember that further study and research in science is possible only if he or she has an adequate grasp of English. Textbook: There is no prescribed textbook. However, you are strongly recommended to read a book at the level of “College Physics” by Halliday and Resnick (any edition). There are many other such books too, such as “University Physics” by Young and Freedman. Study any book that you are comfortable with, preferably by a well-established foreign author. Avoid local authors because they usually copy. After listening to a lecture, go read the relevant chapter. Please remember that these notes cover only some things that you should know and are not meant to be complete. Assignments: There will be total Eight Assignment in this course and its schedules will be announced from time to time. The book you choose to consult will have many more. Those students who are seriously interested in the subject are advised to work out several of the questions posed there. In physics you cannot hope to gain mastery of the subject without extensive problem solving. Examinations: Their schedules will be announced from time to time. Tutors: Their duty is to help you, and they will respond to all genuine questions. However, please do not overload them as they have to deal with a large number of students. Happy studying! Acknowledgements: I thank the Virtual University team and administration for excellent cooperation, as well as Mansoor Noori and Naeem Shahid, for valuable help. © Copyright Virtual University of Pakistan 3 PHYSICS –PHY101 VU Summary of Lecture 1 – INTRODUCTION TO PHYSICS 1. Physics is a science. Science works according to the scientific method. The scientific method accepts only reason, logic, and experimental evidence to tell between what is scientifically correct and what is not. Scientists do not simply believe – they test, and keep testing until satisfied. Just because some “big scientist” says something is right, that thing does not become a fact of science. Unless a discovery is repeatedly established in different laboratories at different times by different people, or the same theoretical result is derived by clear use of established rules, we do not accept it as a scientific discovery. The real strength of science lies in the fact that it continually keeps challenging itself. 2. It is thought that the laws of physics do not change from place to place. This is why experiments carried out in different countries by different scientists – of any religion or race – have always led to the same results if the experiments have been done honestly and correctly. We also think that the laws of physics today are the same as they were in the past. Evidence, contained in the light that left distant stars billions of years ago, strongly indicates that the laws operating at that time were no different than those today. The spectra of different elements then and now are impossible to tell apart, even though physicists have looked very carefully. 3. This course will cover the following broad categories: a) Classical Mechanics, which deals with the motion of bodies under the action of forces. This is often called Newtonian mechanics as well. b) Electromagnetism, whose objective is to study how charges behave under the influence of electric and magnetic fields as well as understand how charges can create these fields. c) Thermal Physics, in which one studies the nature of heat and the changes that the addition of heat brings about in matter. d) Quantum Mechanics, which primarily deals with the physics of small objects such as atoms, nuclei, quarks, etc. However, Quantum Mechanics will be treated only briefly for lack of time. 4. Every physical quantity can be expressed in terms of three fundamental dimensions: Mass (M), Length (L), Time (T). Some examples: Speed LT −1 Acceleration LT −2 Force MLT −2 Energy ML2T −2 Pressure ML−1T −2 You cannot add quantities that have different dimensions. So force can be added to force, but force can never be added to energy, etc. A formula is definitely wrong if the dimensions on the left and right sides of the equal sign are different. © Copyright Virtual University of Pakistan 4 PHYSICS –PHY101 VU 5. Remember that any function f ( x) takes as input a dimensionless number x and outputs a quantity f (which may, or may not have a dimension). Take, for example, the function θ3 θ5 f (θ ) = sin θ. You know the expansion: sin θ = θ − − ⋅⋅⋅ If θ had a dimension + 3! 5! then you would be adding up quantities of different dimensions, and that is not allowed. 6. Do not confuse units and dimensions. We can use different units to measure the same physical quantity. So, for example, you can measure the mass in units of kilograms, pounds, or even in sair and chatak! In this course we shall always use the MKS or Metre- Kilogram-Second system. When you want to convert from one hsystem to another, be methodical as in the example below: mi mi ft 1 m 1 hr m 1 = 1 × 5280 × × = 0.447 hr hr mi 3.28 ft 3600 s s When you write it out in this manner, note that various quantities cancel out cleanly in the numerator and denominator. So you never make a mistake! 7. A good scientist first thinks of the larger picture and then of the finer details. So, estimating orders of magnitude is extremely important. Students often make the mistake of trying to get the decimal points right instead of the first digit – which obviously matters the most! So if you are asked to calculate the height of some building using some data and you come up with 0.301219 metres or 4.01219 × 106 metres, then the answer is plain nonsense even though you may have miraculously got the last six digits right. Physics is commonsense first, so use your intelligence before submitting any answer. 8. Always check your equations to see if they have the same dimensions on the left side as on the right. So, for example, from this principle we can see the equation v 2 = u 2 + 2at is clearly wrong, whereas v 2 = u 2 + 13a 2t 2 could possibly be a correct relation. (Here v and u are velocities, a is acceleration, and t is time.) Note here that I use the word possibly because the dimensions on both sides match up in this case. 9. Whenever you derive an equation that is a little complicated, see if you can find a special limit where it becomes simple and transparent. So, sometimes it is helpful to imagine that some quantity in it is very large or very small. Where possible, make a “mental graph” so that you can picture an equation. So, for example, a formula for the distribution of molecular speeds in a 2 2 gas could look like f (v) = ve-(v-v0 ) / a. Even without knowing the value of a you can immediately see that a) f (v) goes to zero for large values of v, and v = 0. b) The maximum value of f (v) occurs at v0 and the function decreases on both side of this value. v0 © Copyright Virtual University of Pakistan 5 PHYSICS –PHY101 VU Summary of Lecture 2 – KINEMATICS I 1. x(t) is called displacement and it denotes the position of a body at time. If the displacement is positive then that body is to the right of the chosen origin and if negative, then it is to the left. 2. If a body is moving with average speed v then in time t it will cover a distance d=vt. But, in fact, the speed of a car changes from time to time and so one should limit the use of this formula to small time differences only. So, more accurately, one defines an average speed over the small time interval Δt as: distance travelled in time Δt average speed = Δt 3. We define instantaneous velocity at any time t as: x(t ) − x(t1 ) Δx v= 2 ≡. t2 − t1 Δt Here Δx and Δt are both very small quantities that tend to zero but their ratio v does not. x2 Δx x1 t1 Δt t2 4. Just as we have defined velocity as the rate of change of distance, similarly we can define instantaneous acceleration at any time t as: v(t ) − v(t1 ) Δv a= 2 ≡. t2 − t1 Δt Here Δv and Δt are both very small quantities that tend to zero but their ratio a is not zero, in general. Negative acceleration is called deceleration. The speed of a decelerating body decreases with time. 5. Some students are puzzled by the fact that a body can have a very large acceleration but can be standing still at a given time. In fact, it can be moving in the opposite direction to its acceleration. There is actually nothing strange here because position, velocity, and acceleration are independent quantities. This means that specifying one does not specify the other. © Copyright Virtual University of Pakistan 6 PHYSICS –PHY101 VU 6. For constant speed and a body that is at x=0 at time t=0, x increases linearly with time, x ∝ t (or x = vt ). If the body is at position x0 at time t = 0, then at time t, x = x0 + vt. 7. For constant acceleration and a body that starts from rest at t = 0, v increases linearly with time, v ∝ t (or v = at ). If the body has speed v0 at t = 0, then at time t , v = at + v0. 8. We know in (6) above how far a body moving at constant speed moves in time t. But what if the body is changing its speed? If the speed is increasing linearly (i.e. constant acceleration), then the answer is particularly simple: just use the same formula as in (6) but use the average speed: (v0 + v0 + at ) / 2. So we get x = x0 + (v0 + v0 + at )t / 2 = x0 + v0t + 12 at 2. This formula tells you how far a body moves in time t if it moves with constant acceleration a, and if started at position x0 at t=0 with speed v0. 9. We can eliminate the time using (7), and arrive at another useful formula that tells us what the final speed will be after the body has traveled a distance equal to x − x0 after time t , v 2 = v02 + 2a ( x − x0 ). 10. Vectors: a quantity that has a size as well as direction is called a vector. So, for example, the wind blows with some speed and in some direction. So the wind velocity is a vector. 11. If we choose axes, then a vector is fixed by its components along those axes. In one dimension, a vector has only one component (call it the x-component). In two dimensions, a vector has both x and y components. In three dimensions, the components are along the x,y,z axes. G 12. If we denote a vector r = ( x, y ) then, rx = x = r cos θ , and ry = y = r sin θ. Note that x 2 + y 2 = r 2. Also, that tan θ = y / x. 13. Two vectors can be added together geometrically. We take any one vector, move it without changing its direction so that both vectors start from the same point, and then make a parallelogram. The diagonal of the parallelogram is the resultant. G G G C = A+ B 14. Two vectors can also be added algebraically. In this case, we simply add the components of the two vectors along each axis separately. So, for example, Two vectors can be put together as (1.5,2.4) + (1, −1) = (2.5,1.4). © Copyright Virtual University of Pakistan 7 PHYSICS –PHY101 VU Summary of Lecture 3 – KINEMATICS II 1. The concept of the derivative of a function is exceedingly important. The derivative shows how fast a function changes when its argument is changed. (Remember that for f ( x ) we say that f is a function that depends upon the argument x. You should think of f as a machine that gives you the value f when you input x.) 2. Functions do not always have to be written as f ( x). x(t ) is also a function. It tells us where a body is at different times t. 3. The derivative of x (t ) at time t is defined as: dx Δx ≡ lim dt Δt → 0 Δt x(t + Δt ) − x(t ) = lim. Δt → 0 Δt Δx Δt 4. Let's see how to calculate the derivative of a simple function like x(t ) = t 2. We must first calculate the difference in x at two slightly different values, t and t + Δt , while remembering that we choose Δt to be extremely small: Δx = ( t + Δt ) − t 2 2 = t 2 + ( Δt ) + 2t Δt − t 2 2 Δx Δx dx = Δt + 2t ⇒ lim = =2 Δt Δt → 0 Δt dt 5. In exactly the same way you can show that if x(t ) = t n then: dx Δx ≡ lim = nt n −1 dt Δt →0 Δt This is an extremely useful result. © Copyright Virtual University of Pakistan 8 PHYSICS –PHY101 VU 6. Let us apply the above to the function x(t ) which represents the distance moved by a body with constant acceleration (see lecture 2): 1 x(t ) = x0 + v0t + at 2 2 dx 1 = 0 + v0 + a (2t ) = v0 + at dt 2 dv This clearly shows that = 0 + a = a (acceleration is constant) dt 7. A stone dropped from rest increases its speed in the downward direction according dv to = g ≈ 9.8 m/sec. This is true provided we are fairly close to the earth, otherwise dt the value of g decreases as we go further away from the earth. Also, note that if we measured distances from the ground up, then the acceleration would be negative. dv d ⎛ dx ⎞ d 2 x d 2x 8. A useful notation: write = ⎜ ⎟ = 2. We call 2 the second derivative dt dt ⎝ dt ⎠ dt dt of x with respect to t , or the rate of rate of change of x with respect to t. 9. It is easy to extend these ideas to a body moving in both the x and y directions. The position and velocity in 2 dimensions are: G r = x(t )iˆ + y (t ) ˆj G G dr dx ˆ dy ˆ v= = i+ j dt dt dt = v xiˆ + v y ˆj Here the unit vectors iˆ and ˆj are fixed, meaning that they do not depend upon time. G G 10. The scalar product of two vectors A and B is defined as: G G A ⋅ B = AB cosθ You can think of: G G A ⋅ B = ( A)( B cosθ ) G G G = (length of A) × (projection of B on A) OR, G G A ⋅ B = ( B)( A cosθ ) G G G = (length of B) × (projection of A on B). Remember that for unit vectors iˆ ⋅ iˆ = ˆj ⋅ ˆj = 1 and iˆ ⋅ ˆj = 0. © Copyright Virtual University of Pakistan 9 PHYSICS –PHY101 VU Summary of Lecture 4 – FORCE AND NEWTON’S LAWS 1. Ancient view: objects tend to stop if they are in motion; force is required to keep something moving. This was a natural thing to believe in because we see objects stop moving after some time; frictionless motion is possible to see only in rather special circumstances. 2. Modern view: objects tend to remain in their initial state; force is required to change motion. Resistance to changes in motion is called inertia. More inertia means it is harder to make a body accelerate or decelerate. 3. Newton’s First Law: An object will remain at rest or move with constant velocity unless acted upon by a net external force. (A non-accelerating reference frame is called an inertial frame; Newton’s First Law holds only in inertial frames.) 4. More force leads to more acceleration: ⇒ a∝F 5. The greater the mass of a body, the harder it is to change its state of motion. More mass means more inertia. In other words, more mass leads to less acceleration: 1 ⇒ a∝ m Combine both the above observations to conclude that: F a∝ m F 6. Newton's Second Law: a = (or, if you prefer, write as F = ma ). m 7. F = ma is one relation between three independent quantities (m, a, F ). For it to be useful, we must have separate ways of measuring mass, acceleration, and force. Acceleration is measured from observing the rate of change of velocity; mass is a measure of the amount of matter in a body (e.g. two identical cars have twice the mass of a single one). Forces (due to gravity, a stretched spring, repulsion of two like charges, etc) will be discussed later. 8. Force has dimensions of [mass] × [acceleration] = M LT -2. In the MKS system the unit of force is the Newton. It has the symbol N where: 1 Newton = 1 kilogram.metre/second 2. 9. Forces can be internal or external. For example the mutual attraction of atoms within a block of wood are called internal forces. Something pushing the wood © Copyright Virtual University of Pakistan 10 PHYSICS –PHY101 VU is an external force. In the application of F = ma, remember that F stands for the total external force upon the body. 10. Forces are vectors, and so they must be added vectorially: G G G G F = F1 + F2 + F3 + ⋅ ⋅ ⋅ ⋅ This means that the components in the xˆ direction must be added separately, those in the yˆ direction separately, etc. 11. Gravity acts directly on the mass of a body - this is a very important experimental observation due to Newton and does not follow from F = ma. So a body of mass m1 experiences a force F1 = m1 g while a body of mass m2 experiences a force F2 = m2 g , where g is the acceleration with which any body (big or small) falls under the influence of gravity. (Galileo had established this important fact when he dropped different masses from the famous leaning tower of Pisa!) 12. The weight of a body W is the force which gravity exerts upon it, W = mg. Mass and weight are two completely different quantities. So, for example, if you used a spring balance to weigh a kilo of grapes on earth, the same grapes would weigh only 1/7 kilo on the moon. 13. Newton's Third Law: for every action there is an equal and opposite reaction. More precisely, FAB = − FBA , where FAB is the force exerted by body B upon A whereas FBA is the force exerted by body A upon B. Ask yourself what would happen if this was not true. In that case, a system of two bodies, even if it is completely isolated from the surroundings, would have a net force acting upon it because the net force acting upon both bodies would be FAB + FBA ≠ 0. 14. If action and reaction are always equal, then why does a body accelerate at all? Students are often confused by this. The answer: in considering the acceleration of a body you must consider only the (net) force acting upon that body. So, for example, the earth pulls a stone towards it and causes it to accelerate because there is a net force acting upon the stone. On the other hand, by the Third Law, the stone also pulls the earth towards it and this causes the earth to accelerate towards the stone. However, because the mass of the earth is so large, we are only able to see the acceleration of the stone and not that of the earth. © Copyright Virtual University of Pakistan 11 PHYSICS –PHY101 VU Summary of Lecture 5 – APPLICATIONS OF NEWTON’S LAWS – I 1. An obvious conclusion from F = ma is that if F = 0 then a = 0 ! How simple, yet how powerful ! This says that for any body that is not accelerating the sum of all the forces acting upon it must vanish. 2. Examples of systems in equilibrium: a stone resting on the ground; a pencil balanced on your finger; a ladder placed against the wall, an aircraft flying at a constant speed and constant height. 3. Examples of systems out of equilibrium: a stone thrown upwards that is at its highest point; a plane diving downwards; a car at rest whose driver has just stepped on the car's accelerator. 4. If you know the acceleration of a body, it is easy to find the force that causes it to accelerate. Example: An aircraft of mass m has position vector, G r = (at + bt 3 )iˆ + (ct 2 + dt 4 ) ˆj What force is acting upon it? G d 2x d2y SOLUTION: F = m 2 iˆ + m 2 ˆj dt dt = 6 b mt i + m(2 c + 12d t 2 ) ˆj ˆ 5. The other way around is not so simple: suppose that you know F and you want to find x. For this you must solve the equation, d 2x F = dt 2 m This may or may not be easy, depending upon F (which may depend upon both x as well as t if the force is not constant). 6. Ropes are useful because you can pull from a distance to change the direction of a force. The tension, often denoted by T , is the force you would feel if you cut the rope and grabbed the ends. For a massless rope (which may be a very good approximation in many situations) the tension is the same at every point along the rope. Why? Because if you take any small slice of the rope it weighs nothing (or very little). So if the force on one side of the slice was any different from the force force on the other side, it would be accelerating hugely. All this was for the "ideal rope" which has no mass and never breaks. But this idealization if often good enough. © Copyright Virtual University of Pakistan 12 PHYSICS –PHY101 VU 7. We are all familiar with frictional force. When two bodies rub against each other, the frictional force acts upon each body separately opposite to its direction of motion (i.e it acts to slow down the motion). The harder you press two bodies against each G G G other, the greater the friction. Mathematically, F = μ N , where N is the force with which you press the two bodies against each other (normal force). The quantity μ is called the coefficient of friction (obviously!). It is large for rough surfaces, and small for smooth G G ones. Remember that F = μ N is an empirical relation and holds only approximately. This is obviously true: if you put a large enough mass on a table, the table will start to bend and will eventually break. 8. Friction is caused by roughness at a microscopic level - if you look at any surface with a powerful microscope you will see unevenness and jaggedness. If these big bumps are levelled somehow, friction will still not disappear because there will still be little bumps due to atoms. More precisely, atoms from the two bodies will interact each other because of the electrostatic interaction between their charges. Even if an atom is neutral, it can still exchange electrons and there will be a force because of surrounding atoms. 9. Consider the two blocks below on a frictionless surface: F m1 T T m2 We want to find the tension and acceleration: The total force on the first mass is F − T and so F − T = m1a. The force on the second mass is simply T and so T = m2 a. Solving the m2 F F above, we get: T = and a =. m1 + m2 m1 + m2 10. There is a general principle by which you solve equilibrium problems. For equilibrium, the sum of forces in every direction must vanish. So Fx = Fy = Fz = 0. You may always choose the x, y, z directions according to your convenience. So, for example, as in the lecture problem dealing with a body sliding down an inclined plane, you can choose the directions to be along and perpendicular to the surface of the plane. © Copyright Virtual University of Pakistan 13 PHYSICS –PHY101 VU Summary of Lecture 6 – APPLICATIONS OF NEWTON’S LAWS – II 1. As a body moves through a body it displaces the fluid. it has to exert a force on the fluid to push it out of the way. By Newton's third law, the fluid pushes back on the body with an equal and opposite force. The direction of the fluid resistance force on a body is always opposite to the direction of the body's velocity relative to the fluid. 2. The magnitude of the fluid resistance force usually increases with the speed of the body through the fluid. Typically, f = kv (an empirical law!). Imagine that you drop a ball bearing into a deep container filled with oil. After a while the ball bearing will approach its maximum (terminal) speed when the forces of gravity and friction balance each other: mg = kv from which vfinal = mg / k. 3. The above was a simple example of equilibrium under two forces. In general, while solving problems you should a)draw a diagram, b)define an origin for a system of coordinates, c)identify all forces (tension, normal, friction, weight, etc) and their x and y components, d)Apply Newton's law separately along the x and y axes. e) find the accelerations, then velocities, then displacements. This sounds very cook-book, and in fact it will occur to you naturally how to do this while solving actual problems. 4. Your weight in a lift: suppose you are in a lift that is at rest or moving at constant velocity. In either case a=0 and the normal force N and the force due to gravity are exactly equal, N − Mg = 0 ⇒ N = Mg. But if the lift is accelerating downwards then Mg − N = Ma or N = M ( g − a). So now the normal force (i.e. the force with which the floor of the lift is pushing on you) is decreased. Note that if the lift is accelerating downwards with acceleration a (which it will if the cable breaks!) then N=0 and you will experience weightlessness just like astronauts in space do. Finally, if the lift is accelerating upwards then a is negative and you will feel heavier. 5. Imagine that you are in a railway wagon and want to know how much you are accelerating. You are not able to look out of the windows. A mass is hung from the roof. Find the acceleration of the car from the angle made by the mass. a T © Copyright Virtual University of Pakistan 14 PHYSICS –PHY101 VU T We first balance the forces vertically: T cosθ = mg And then horizontally: T sin θ = ma θ a From these two equations we find that: tan θ = g Note that the mass m doesn't matter - it cancels out! mg 6. Friction is a funny kind of force. It doe not make up its mind which way to act until some other force compels it to decide. Imagine a block lying on the floor. If you push it forward, friction will act backward. And if you push it to the left, friction will act to the right. In other words, the direction of the frictional force is always in the opposite direction to the applied force. 7. Let us solve the following problem: a rope of total length L and mass per unit length m is put on a table with a length l hanging from one edge. What should be l such that the rope just begins to slip? N (L − l ) l μN ml g m( L − l ) g To solve this, look at the balance of forces in the diagram below: in the vertical direction, the normal force balances the weight of that part of the rope that lies on the table: N = m( L − l ) g. In the horizontal direction, the rope exerts a force mlg to the right, which is counteracted by the friction that acts to the left. Therefore μ N = mlg. Substituting N μL from the first equation we find that l =. Note that if μ isvery small then even a small μ +1 piece of string that hangs over the edge will cause the entire string to slip down. 9. In this problem, we would like to calculate the minimum force F such that the small block does not slip downwards. Clearly, F m M since the 2 bodies move together, F = (m + M )a. This gives F a=. We want the friction μ N to be at least as large as the downwards force, mg. (m + M ) ⎛ F ⎞ So, we put N = ma = m ⎜ ⎟ from which the minimum horizontal force needed to ⎝ (m + M ) ⎠ (m + M ) g prevent slippage is F =. μ © Copyright Virtual University of Pakistan 15 PHYSICS –PHY101 VU Summary of Lecture 7 – WORK AND ENERGY 1. Definition of work: force applied in direction of displacement × displacement. This means that if the force F acts at an angle θ with respect to the direction of motion, then G G W = F ⋅ d = Fd cos θ G F θ G d 2. a) Work is a scalar - it has magnitude but no direction. b) Work has dimensions: M × ( L T −2 ) × L = M L2T -2 c) Work has units: 1 Newton × 1 Metre ≡ 1 Joule (J) 3. Suppose you lift a mass of 20 kg through a distance of 2 metres. Then the work you do is 20 kg × 9.8 Newtons × 2 metres = 39.2 Joules. On the other hand, the force of gravity is directed opposite to the force you exert and the work done by gravity is - 39.2 Joules. 4. What if the force varies with distance (say, a spring pulls harder as it becomes longer). In that case, we should break up the distance over which the force acts into small pieces so that the force is approximately constant over each bit. As we make the pieces smaller and smaller, we will approach the exact result: 0 Δx x→ Now add up all the little pieces of work: N W = ΔW1 + ΔW2 + ⋅ ⋅ ⋅ + ΔWN = F1Δx + F2 Δx + ⋅ ⋅ ⋅ + FN Δx ≡ ∑ Fn Δx n =1 ∞ To get the exact result let Δx → 0 and the number of intervals N → ∞ : W = lim ∑ Fn Δx Δx → 0 n =1 xf ∞ Definition: W = lim ∑ Fn Δx ≡ ∫ F ( x)dx is called the integral of F with respect to x from Δx → 0 n =1 xi © Copyright Virtual University of Pakistan 16 PHYSICS –PHY101 VU xi to x f. This quantity is the work done by a force, constant or non-constant. So if the force is known as a function of position, we can always find the work done by calculating the definite integral. 5. Just to check what our result looks like for a constant force, let us calculate W if F = b, F F =b 1 2 3 4 0 Δx = a / 4 a x a 1 1 1 1 a(b) + a(b) + a(b) + a(b) = ab ∴ ∫ Fdx = ab 4 4 4 4 0 6. Now for a less trivial case: suppose that F=kx, i.e. the force increases linearly with x. F ( a, k a ) 4 3 2 1 0 a x→ Δx = a / 4 a 1 a2 a2 Area of shaded region = (a )(ka ) = k 2 2 ∴ ∫0 Fdx = k 2 7. Energy is the capacity of a physical system to do work: ƒ it comes in many forms – mechanical, electrical, chemical, nuclear, etc ƒ it can be stored ƒ it can be converted into different forms ƒ it can never be created or destroyed 8. Accepting the fact that energy is conserved, let us derive an expression for the kinetic energy of a body. Suppose a constant force accelerates a mass m from speed 0 to speed v over a distance d. What is the work done by the force? Obviously the answer is: mv 2 1 W = Fd. But F = ma and v = 2ad. This gives W = (ma)d = 2 d = mv 2. So, we 2d 2 conclude that the work done by the force has gone into creating kinetic energy. and that the amount of kinetic energy possessed by a body moving with speed v is 1 2 mv 2. 9. The work done by a force is just the force multiplied by the distance – it does not depend upon time. But suppose that the same amount of work is done in half the time. © Copyright Virtual University of Pakistan 17 PHYSICS –PHY101 VU We then say that the power is twice as much. We define: Work done Power = Time taken Work F Δx If the force does not depend on time: = = F v. Therefore, Power = F v. Time Δt 10.Let's work out an example. A constant force accelerates a bus (mass m) from speed v1 to speed v 2 over a distance d. What work is done by the engine? Recall that for constant acceleration, v22 − v12 = 2a ( x2 − x1 ) where: v2 = final velocity, v 22 − v12 x 2 = final position, v1 = initial velocity, x1 = initial position. Hence, a =. Now 2d v 22 − v12 1 1 calculate the work done: W = Fd = mad = m d = m v 2 2 − m v12. So the 2d 2 2 1 work done has resulted in an increase in the quantity m v 2 , which is kinetic energy. 2 © Copyright Virtual University of Pakistan 18 PHYSICS –PHY101 VU Summary of Lecture 8 – CONSERVATION OF ENERGY 1. Potential energy is, as the word suggests, the energy “locked up” up somewhere and which can do work. Potential energy kam karnay ki salahiat hai! Potential energy can be converted into kinetic energy, 12 mv 2. As I showed you earlier, this follows directly from Newton’s Laws. 2. If you lift a stone of mass m from the ground up a distance x, you have to do work against gravity. The (constant) force is mg , and so W = mgx. By conservation of energy, the work done by you was transformed into gravitational potential energy whose values is exactly equal to mgx. Where is the energy stored? Answer: it is stored neither in the mass or in the earth - it is stored in the gravitational field of the combined system of stone+earth. 3. Suppose you pull on a spring and stretch it by an amount x away from its normal (equilibrium) position. How much energy is stored in the spring? Obviously, the spring gets harder and harder to pull as it becomes longer. When it is extended by length x and you pull it a further distance dx, the small amount of work done is dW = Fdx = kxdx. Adding up all the small bits of work gives the total work: x x 1 2 W = ∫ Fdx = ∫ k xdx = kx 0 0 2 This is the work you did. Maybe you got tired working so hard. What was the result of your working so hard? Answer: this work was transformed into energy stored in the spring. The spring contains energy exactly equal to 1 2 k x2. 4. Kinetic energy obviously depends on the frame you choose to measure it in. If you are running with a ball, it has zero kinetic energy with respect to you. But someone who is standing will see that it has kinetic energy! Now consider the following situation: a box of mass 12kg is pushed with a constant force so that so that its speed goes from zero to 1.5m/sec (as measured by the person at rest on the cart) and it covers a distance of 2.4m. Assume there is no friction. v =1.5m/ s 2.4m 15m / s mass of box = 12 kg © Copyright Virtual University of Pakistan 19 PHYSICS –PHY101 VU Let's first calculate the change in kinetic energy: 1 ΔK = K f − K i = (12kg )(1.5m / s ) 2 − 0 = 13.5 J 2 And then the (constant) acceleration: v 2f − vi2 (1.5m / s ) 2 − 0 a= = = 0.469m / s 2 2( x f − xi ) 2(2 ⋅ 4m) This acceleration results from a constant net force given by: F = ma = (12kg )(0.469 m / s 2 ) = 5.63N From this, the work done on the crate is: W = F Δx = (5.63 N )(2.4m) = 13.5 J (same as ΔK = 13.5 J !) 5. Now suppose there is somebody standing on the ground, and that the trolley moves at 15 m/sec relative to the ground: 15 m / s 16.5m / s 50.4m 15m / s 48.0m Let us repeat the same calculation: 1 1 ΔK ′ = K ′f − K i′ = mvf 2 − mvi 2 2 2 1 1 = (12kg )(16.5m / s ) 2 − (12kg )(15.0m / s ) 2 = 284 J 2 2 This example clearly shows that work and energy have different values in different frames. 6. The total mechanical energy is: Emech = KE + PE. If there is no friction then Emech is conserved. This means that the sum does not change with time. For example: a ball is thrown upwards at speed v0. How high will it go before it stops? The loss of potential 1 2 v2 energy is equal to the gain of potential energy. Hence, mv0 = mg h ⇒ h = 0. 2 2g Now look at the smooth, frictionless motion of a car over the hills below: A v0 B C h h h/2 D © Copyright Virtual University of Pakistan 20 PHYSICS –PHY101 VU Even though the motion is complicated, we can use the fact that the total energy is a constant to get the speeds at the points B,C,D: 1 1 At point A: mv A 2 + mgh = mv B 2 + mgh ⇒ v B = vC 2 2 1 1 h At point C: mv A 2 + mgh = mvC 2 + mg ⇒ v C = v A 2 + gh 2 2 2 1 1 At point D: mv A2 + mgh = mv D 2 ⇒ v D = v A 2 + 2 gh 2 2 7. Remember that potential energy has meaning only for a force that is conservative. A conservative force is that for which the work done in going from point A to point B is independent of the path chosen. Friction is an example of a non-conservative force dV and a potential energy cannot be defined. For a conservative force, F = −. So, for dx 1 2 a spring, V = kx and so F = −kx. 2 dV 8. Derivation of F = − : If the particle moves distance Δx in a potential V , then dx ΔV change in PE is ΔV where, ΔV = − F Δx. From this, F = −. Now let Δx → 0. Δx ΔV dV Hence, F = − lim =−. Δx → 0 Δx dx © Copyright Virtual University of Pakistan 21 PHYSICS –PHY101 VU Summary of Lecture 9 – MOMENTUM 1. Momentum is the "quantity of motion" possessed by a body. More precisely, it is defined as: Mass of the body × Velocity of the body. The dimensions of momentum are MLT -1 and the units of momentum are kg-m/s. G 2. Momentum is a vector quantity and has both magnitude and direction, p = mv. We can easily see that Newton's Second Law can be reexpressed in terms of momentum. When G G G dv G I wrote it down originally, it was in the form ma = F. But since = a, this can also be dt G dp G written as = F (new form). In words, the rate of change of momentum of a body equals dt the total force acting upon it. Of course, the old and new are exactly the same, G G G d p d (m v) dv G G = =m = ma = F. dt dt dt G 3. When there are many particles, then the total momentum P is, G G G G P = p1 + p 2 + ⋅ ⋅ ⋅ p N G G G d G d p1 d p 2 dp P= + + ⋅⋅⋅ N dt dt dt dt G G G G = F1 + F2 + ⋅ ⋅ ⋅ FN = F This shows that when there are several particles, the rate at which the total momentum changes is equal to the total force. It makes sense! 4. A very important conclusion of the above is that if the sum of the total external forces G d G vanishes, then the total momentum is conserved, ∑ Fext = 0 ⇒ P = 0. This is quite dt independent of what sort of forces act between the bodies - electric, gravitational, etc. - or how complicated these are. We shall see why this is so important from the following examples. 5. Two balls, which can only move along a straight line, collide with each other. The initial momentum is Pi = m1u1 + m2u2 and the final momentum is Pf = m1v1 + m2 v 2. Obviously one ball exerts a force on the other when they collide, so its momentum changes. But, from the fact that there is no external force acting on the balls, Pi = Pf , or m1u1 + m2u2 = m1v1 + m2 v 2. 6. A bomb at rest explodes into two fragments. Before the explosion the total momentum is © Copyright Virtual University of Pakistan 22 PHYSICS –PHY101 VU zero. So obviously it is zero after the explosion as well, P f = 0. During the time that the explosion happens, the forces acting upon the pieces are very complicated and changing rapidly with time. But when all is said and done, there are two pieces flying away with a total zero final momentum P f = m1v1 + m2 v 2. Hence m1v1 = − m2 v2. In other words, the fragments fly apart with equal momentum but in opposite directions. The centre-of-mass stays at rest. So, knowing the velocity of one fragment permits knowing the velocity of the other fragment. 7. If air resistance can be ignored, then we can do some interesting calculations with what we have learned. So, suppose a shell is fired from a cannon with a speed 10 m/s at an angle 600 with the horizontal. At the highest point in its path it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. Let us find the velocity of the other piece immediately after the explosion. M Solution: After the explosion: P1x = −5 (why?). But P1x + P2 x = Px = M × 10cos 60 2 M M ⇒ P2 x = 5 M + 5. Now use: P2 x = v 2 x ⇒ v2 x = 15 m / s. 2 2 8. When you hit your thumb with a hammer it hurts, doesn't it? Why? Because a large amount of momentum has been destroyed in a short amount of time. If you wrap your thumb with foam, it will hurt less. To understand this better, remember that force is the dp rate of change of momentum: F = ⇒ dp = Fdt. Now define the impulse I as: dt force × time over which the force acts. If the force changes with time between the limits , then one should define I as, t2 t2 pf I = ∫ Fdt. Since ∫ Fdt = ∫ dp, therefore I = p f − pi. In words, the change of momentum t1 t1 pi equals the impulse, which is equal to the area under the curve of force versus time. Even if you wrap your thumb in foam, the impulse is the same. But the force is definitely not! © Copyright Virtual University of Pakistan 23 PHYSICS –PHY101 VU 9. Sometimes we only know the force numerically (i.e. there is no expression like F=something). F But we still know what the integral means: it is the area under the curve of force versus time. The curve here is that of a hammer striking a table. Before the hammer strikes, the force is zero, reaches a peak, and goes back to zero. Δt © Copyright Virtual University of Pakistan 24 PHYSICS –PHY101 VU Summary of Lecture 10 – COLLISIONS 1. Collisions are extremely important to understand because they happen all the time - electrons collide with atoms, a bat with a ball, cars with trucks, star galaxies with other galaxies,...In every case, the sum of the initial momenta equals the sum of the final momenta. This follows directly from Newton's Second Law, as we have already seen. 2. Take the simplest collision: two bodies of mass m1 and m2 moving with velocities u1 and u2. After the collision they are moving with velocities v1 and v 2. From momentum conservation, m1u1 + m2u2 = m1v1 + m2 v 2 ⇒ m1 (u1 − v1 ) = m2 (v 2 − u2 ) This is as far as we can go. There are two unknowns but only one equation. However, if the collision is elastic then, 1 2 m1u12 + 12 m2u22 = 1 2 m1v12 + 12 m2 v 22 ⇒ 12 m1 (u12 − v12 ) = 12 m2 (v 2 2 − u2 2 ). Combine the two equations above, u1 + v1 = v 2 + u2 ⇒ u1 − u2 = v 2 − v1. In words, this says that in an elastic collision the relative speed of the incoming particles equals the relative speed of the outgoing particles. 3. One can solve for v1 and v 2 (please do it!) easily and find that: m1 − m 2 2m 2 v1 = ( )u1 + ( )u2 m1 + m 2 m1 + m 2 2m1 m 2 − m1 v2 = ( )u1 + ( )u2 m1 + m 2 m1 + m 2 Notice that if m1 = m2 , then v1 = u2 and v 2 = u1. So this says that after the collision, the bodies will just reverse their velocities and move on as before. 4. What if one of the bodies is much heavier than the other body, and the heavier body is at rest? In this case, m2 >> m1 and u2 = 0. We can immediately see that v1 = −u1 and v 2 = 0. This makes a lot of sense: the heavy body continues to stay at rest and the light body just bounces back with the same speed. In the lecture, you saw a demonstration of this! 5. And what if the lighter body (rickshaw) is at rest and is hit by the heavier body (truck)? In this case, m2 > d. from the diagram you can see that r2 − r1 ≈ d cosθ and that r1r2 ≈ r 2. Hence, q d cosθ 1 p cosθ V≈ =. So we have calculated the potential at any θ with such 4πε 0 r 2 4πε 0 r 2 π little difficulty. Note that V = 0 at θ =. 2 +q r1 P θ d r r2 −q r2 − r1 = d cos θ 8. Now let us calculate the potential which comes from charges that are uniformly spread over a ring. This is the same problem as in the previous P lecture, but simpler. Give the small amount of potential coming from the small amount of charge θ 1 dq z dq = λ ds some name, dV =. Then obviously 4πε 0 r 2 y 1 dq 1 q R V = ∫ dV = ∫ =. 4πε 0 r 2 4πε 0 R2 + z 2 λds x © Copyright Virtual University of Pakistan 69 PHYSICS –PHY101 VU Summary of Lecture 25 – CAPACITORS AND CURRENTS 1. Two conductors isolated from one another and from their surroundings, form a capacitor. These conductors may be of any shape and size, and at any distance from each other. If a potential difference is created between the conductors (say, by connecting the terminals of a battery to them), then there is an electric field in the space between them. The electric field comes from the charges that have been pushed to the plates by the battery. The amount of charge pushed on to the conductors is proportional to the potential difference between the battery terminals (which is the same as between the capacitor plates). Hence, Q ∝ V. To convert this into an equality, we write Q = CV. This provides the definition of Q capacitance, C =. V 2. Using the above definition, let us calculate the capacitance of two parallel plates separated by a distance d as in the figure below. + + + ++q + + + + + + + + + + + d Gaussian surface − − − − − − − − − −q G G q Recall Gauss's Law: Φ ≡ ∫ E ⋅ dA = enclosed. Draw any Gaussian surface. Since the electric ε0 Q field is zero above the top plate, the flux through the area A of the plate is Φ = EA = , ε0 Q where Q is the total charge on the plate. Thus, E = is the electric field in the gap ε0 A E Q ε A between the plates. The potential difference is V = , and so C = = 0. You can see d V d that the capacitance will be large if the plates are close to each other, and if the plates have a large area. We have simplified the calculation here by assuming that the electric field is strictly directed downwards. This is only true if the plates are infinitely long. But we can usually neglect the side effects. Note that any arrangement with two plates forms a capacitor: plane, cylindrical, spherical, etc. The capacitance depends upon the geometry, the size of plates and the gap between them. © Copyright Virtual University of Pakistan 70 PHYSICS –PHY101 VU 3. One can take two (or more) capacitors in various ways and thus change the amount of charge they can contain. Consider first two capacitors connected in parallel with each other. The same voltage exists across both. For each capacitor, q1 = C1V , q2 = C2V where V is the potential between terminals a and b. The total charge is: Q = q1 + q2 = C1V + C2V = (C1 + C2 )V Q Now, let us define an "effective" or "equivalent" capacitance as Ceq =. Then we can V immediately see that for 2 capacitors Ceq = C1 + C2 , and Ceq = ∑ Cn ( for n capacitors ). C1 Ceq V C2 4. We can repeat the analysis above when the capacitors are put in series. Here the difference is that now we must start with V = V1 + V2 , where V1 and V2 are the voltages across the two. Clearly the same charge had to cross both the capacitors. Hence, Q Q 1 1 V = V1 + V2 = + = Q( + ). C1 C2 C1 C2 Q 1 1 1 From our definition, Ceq = , it follows that = +. The total capacitance is now V Ceq C1 C2 1 1 less than if they were in parallel. In general, =∑ ( for n capacitors ). Ceq Cn V1 V2 Ceq C1 C2 V V 5. When a battery is connected to a capacitor, positive and negative charges appear on the opposite plates. Some energy has been transferred from the battery to the capacitor, and now been stored in it. When the capacitor is discharged, the energy is recovered. Now let us calculate the energy required to charge a capacitor from zero to V volts. Begin: the amount of energy required to transfer a small charge dq to the plates is dU = vdq, where v is the voltage at a time when the charge is q = Cv. As time goes on, the total charge increases until it reaches the final charge Q (at which point the voltage becomes V ). So, © Copyright Virtual University of Pakistan 71 PHYSICS –PHY101 VU Q q q Q2 1 dU = vdq = dq ⇒ U = ∫ dU = ∫ dq = = CV 2. C 0 C 2C 2 But where in the capacitor is the energy stored? Answer, it is present in the electric field in the volume between the two plates. We can calculate the energy density: 1 CV 2 2 energy stored in capacitor U 2 ε ⎛V ⎞ 1 u= = = = 0 ⎜ ⎟ = ε 0 E 2. volume of capacitor Ad Ad 2 ⎝d⎠ 2 ε A In the above we have used C = 0 , derived earlier. The important result here is that d u ∝ E. Turning it around, wherever there is an electric field, there is energy available. 2 1 Q 6. Dielectrics. Consider a free charge + Q. Around it is an electric field, E =. 4πε 0ε r r 2 Now suppose this charge is placed among water molecules. These molecules will polarise, i.e. the centre of positive charge and centre of negative charge will be slightly displaced. The negative part of the water molecule will be attracted toward the positive charge + Q. 1 1 Q So, in effect, the electric field is weakened by and becomes,. Here I have εr 4πε 0ε r r 2 introduced a new quantity ε r called "dielectric constant". This is a number that is usually bigger than one and measures the strength of the polarization induced in the material. For air, ε r = 1.0003 while ε r ≈ 80 for pure water. The effect of a dielectric is to increase the capacitance of a capacitor: if the air between the plates of a capacitor is replaced by a ε A ε A dielectric, C = 0 → ε r 0. d d © Copyright Virtual University of Pakistan 72 PHYSICS –PHY101 VU Summary of Lecture 26 – ELECTRIC POTENTIAL ENERGY 1. Electric current is the flow of electrical charge. If a small amount of charge dq flows in dq time dq, then the current is i =. If the current is constant in time, then in time t , the dt current that flows is q = i × t. The unit of charge is ampere, which is define as: 1 coulomb 1 ampere = second A car's battery supplies upto 50 amperes when starting the car, but often we need to deal with smaller values: 1 milliampere = 1 ma = 10−3 A 1 microampere = 1 μ A = 10−6 A 1 nanoampere = 1 nA = 10−9 A 1 picoampere = 1 pA = 10−12 A 2. The direction of current flow is the direction in which positive charges move. However, in a typical wire, the positive charges are fixed to the atoms and it is really the negative charges (electrons) that move. In that case the direction of current flow is reversed. electron flow conventional current flow device + − 3. Current flows because something forces it around a circuit. That "something" is EMF, electromotive force. But remember that we are using bad terminology and that EMF is not a force - it is actually the difference in electric potentials between two parts of a circuit. So, in the figure below, V = Va − Vb is the EMF which causes current to flow in the resistor. How much current? Generally, the larger V is , the more current will flow and we expect I ∝ V. In general this relation will not be completely accurate but when it holds, we say V V Ohm's Law applies: I =. Here, R = is called the resistance. R I R I I V © Copyright Virtual University of Pakistan 73 PHYSICS –PHY101 VU 4. Be careful in understanding Ohm's Law. In general the current may depend upon the applied voltage in a complicated way. Another way of saying this is that the resistance may depend upon the current. Example: when current passes through a resistor, it gets hot and its resistance increases. Only when the graph of current versus voltage is a straight line does Ohm's Law hold. Else, we can only define the "incremental resistance". I ( Amps ) R= Δi ΔV does not obey Ohm's Law obeys Ohm's Law V ( volts ) 5. Charge is always conserved, and therefore current is conserved as well. This means that when a current splits into two currents the sum remains constant, i1 = i2 + i3. i2 i1 i3 6. When resistors are put in series with each other, the same current flows through both. So, V1 = iR1 and V2 = iR2. The total potential drop across the pair is V = V1 + V2 = i ( R1 + R2 ). ⇒ Req = R1 + R2. So resistors in series add up. R1 R2 V1 V2 7. Resistors can also be put in parallel. This means that the same V V voltage V is across both. So the currents are i1 = , i2 =. R1 R2 V V V Since i = i1 + i2 it follows that i = = + a R1 b Req R1 R2 R2 1 1 1 RR ⇒ = + or Req = 1 2. a Req b Req R1 R2 R1 + R2 This makes sense: with two possible paths the current will find less resistance than if only one was present. 8. When current flows in a circuit work is done. Suppose a small amount of charge dq is moved through a potential difference V. Then the work done is dW = Vdq = V idt. Hence © Copyright Virtual University of Pakistan 74 PHYSICS –PHY101 VU V dW Vidt = i 2 Rdt ( because i = ). The rate of doing work, i.e. power, is P = = i 2 R. R dt V2 This is an important formula. It can also be written as P = , or as P = iV. The unit of R joule coulomb joule power is: 1 volt-ampere = 1 ⋅ =1 = 1 watt. coulomb second second 9. Kirchoff's Law: The sum of the potential differences encountered in moving around a closed circuit is zero. This law is easy to prove: since the electric field is conservative, therefore no work is done in taking a charge all around a circuit and putting it back where it was. However, it is very useful in solving problems. As a trivial example, consider the circuit below. The statement that, starting from any point a we get back to the same potential after going around is: Va − iR + ε = Va. This says − iR + ε = 0. a b + i ε i R − i d c 10. We can apply Kirchoff's Law to a circuit that consists of a resistor and capacitor in order to see how current flows q through it. Since q = VC , we can see that + iR = 0. Now C 1 dq di differentiate with respect to time to find + R = 0, − + C dt dt di 1 − t C i or =− i. This equation has solution: i = i0e RC. The dt RC R product RC is called the time constant τ , and it gives the time by which the current has fallen to 1/ e ≈ 1/ 2.7 of the initial value. x 2 x3 x 4 A reminder about the exponential function, e x ≡ 1 + x + + + + ⋅⋅⋅⋅ 2! 3! 4! d x 2 x 3x 2 4 x3 d −x From this, e =1+ + + + ⋅ ⋅ ⋅ = e x Similarly, e = −e − x dx 2! 3! 4! dx 11. Circuits often have two or more loops. To find the voltages and currents in such situations, it is best to apply Kirchoff's Law. In the figure below, you see that there are 3 loops and you can see that: © Copyright Virtual University of Pakistan 75 PHYSICS –PHY101 VU ε1 ε2 i1 + i3 = i2 ε1 − i1R1 + i3 R3 = 0 i i2 −i3 R3 − i2 R2 + ε 2 = 0 R1 i1 R3 i3 R2 Make sure that you understand each of these, and then check that the solution is: ε1 ( R2 + R3 ) − ε 2 R3 ε1R3 − ε 2 ( R1 + R3 ) −ε1R2 − ε 2 R1 i1 = i2 = i3 = R1R2 + R2 R3 + R1R3 R1R2 + R2 R3 + R1R3 R1R2 + R2 R3 + R1R3 12. A charge inside a wire moves under the influence of the applied electric field and suffers many collisions that cause it to move on a highly irregular, jagged path as shown below. Nevertheless, it moves on the average to the right at the "drift velocity" (or speed). 13. Consider a wire through which charge is flowing. Suppose that the number of charges per unit volume is n. If we multiply n by the crossectional area of the wire A and the length L, then the charge in this section of the wire is q = ( nAL ) e. If the drift velocity of the charges is v d , then the time taken for the charge to move through the wire is L q nALe t=. hence the current is i = = = nAevd. From this we can calculate the drift vd t L / vd i velocity of the charges in terms of the measured current, vd =. The current density, nAe i which is the current per unit crossectional area is defined as j = = nevd. If j varies A G G inside a volume, then we can easily generalize and write, i = ∫ j ⋅ dA. L I A I V © Copyright Virtual University of Pakistan 76 PHYSICS –PHY101 VU Summary of Lecture 27 – THE MAGNETIC FIELD 1. The magnetic field exerts a force upon any charge that moves in the field. The greater the size of the charge, and the faster G G G F = qv × B it moves, the larger the force. The direction of the force is perpendicular to both the direction of motion and the magnetic G G G B field. If θ is the angle between v and B, then F = qvB sin θ is the + θ G G magnitude of the force. This vanishes when v and B are q G parallel (θ = 0), and is maximum when they are perpendicular. v 2. The unit of magnetic field that is used most commonly is the tesla. A charge of one coulomb moving at 1 metre per second perpendicularly to a field of one tesla experiences a force of 1 newton. Equivalently, newton newton 1 tesla = 1 =1 = 104 gauss (CGS unit) coulomb ⋅ meter/second ampere ⋅ meter In order to have an appreciation for how much a tesla is, here are some typical values of the magnetic field in these units: Earth's surface 10-4 T Bar magnet 10-2 T Powerful electromagnet 1T Superconducting magnet 5T 3. When both magnetic and electric fields are present at a point, the total force acting upon G G G G a charge is the vector sum of the electric and magnetic forces, F = qE + qv × B. This is known as the Lorentz Force. Note that the electric force and magnetic force are very different. The electric force is non-zero even if the charge is stationary, and it is in the G same direction as E. 4. The Lorentz Force can be used to select charged particles of whichever velocity we want. In the diagram below, particles enter from the left with velocity v. They experience a force due to the perpendicular magnetic field, as well as force downwards because of an electric field. Only particles with speed v = E / B are undeflected and keep going straight. + + ++ + + FB = q v B ⊗ ⊗ ⊗ ⊗ + ⊗ ⊗ ⊗ ⊗ +⊗ ⊗ ⊗ ⊗ + ⊗ ⊗ ⊗ ⊗ −−−−−− FE = qE E qE = qvB ⇒ v = velocity selector !! B © Copyright Virtual University of Pakistan 77 PHYSICS –PHY101 VU 5. A magnetic field can be strong enough to lift an elephant, but it can never increase or G decrease the energy of a particle. Proof: suppose the magnetic force F moves a particle G through a displacement d r. Then the small amount of work done is, G G G G G dW = F ⋅ d r = q (v × B) ⋅ d r G G G dr = q (v × B) ⋅ dt dt G G G = q (v × B) ⋅ v dt = 0. Basically the force and direction of force are orthogonal, and hence there can be no work done on the particle or an increase in its energy. 6. A magnetic field bends a charged particle into a circular orbit because the particle feels a force that is directed perpendicular to the magnetic field. As we saw above, the particle cannot change its speed, but it certainly does change direction! So it keeps bending and bending until it makes a full circle. The radius of orbit can be easily calculated: the magnetic v2 and centrifugal forces must balance each other for equilibrium. So, qvB = m and we r mv find that r =. A strong B forces the particle into a tighter orbit, as you can see. We qB v qB can also calculate the angular frequency, ω = =. This shows that a strong B makes r m the particle go around many times in unit time. There are a very large number of applications of these facts. ⊗ ⊗ ⊗ ⊗G − v G ⊗ ⊗ F⊗ G ⊗ v2 mv F qv B = m ⇒ r=. ⊗ ⊗ ⊗ r qB ⊗ G ⊗ ⊗ F⊗ ⊗ G G v v ⊗ ⊗ ⊗ ⊗ 7. The fact that a magnetic field bends charged particles is responsible for shielding the earth from harmful effects of the "solar wind". A large number of charged particles are released from the sun and reach the earth. These can destroy life. Fortunately the earth's magnetic field deflects these particles, which are then trapped in the "Van Allen" belt around the earth. © Copyright Virtual University of Pakistan 78 PHYSICS –PHY101 VU 8. The mass spectrometer is an extremely important equipment + v that works on the above principle. Ions are made from atoms by stripping away one electron. Then they pass through a mv + velocity selector so that they all have the same speed. In a r= qB beam of many different ions, the heavier ones bend less, and + lighter ones more, when they are passed through a B field. + 9. A wire carries current, and current is flowing charges. Since each charge experiences a force when placed in a magnetic field, you might expect the same for the current. Indeed, that is exactly the case, and we can easily calculate the force on a wire from the force on individual charges. Suppose N is the total number of charges and they are moving at the G G G G average (or drift) velocity v d. Then the total force is F = Nevd × B. Now suppose that the wire has length L, crossectional area A, and it has n charges per unit volume. Then clearly G G G N = nAL, and so F = nALevd × B. Remember that the current is the charge that flows G G through the wire per unit time, and so nAev d = I. We get the important result that the G G G force per unit length on the wire is F = I × B. 10. A current that goes around a loop (any shape) produces a G magnetic field. We define the magnetic moment as the μ G product of current and area, μ = IAzˆ. Here A is the area of the loop and I the current flowing around it. The direction I is perpendicular to the plane of the loop, as shown. 11. Magnetic fields are produced by currents. Every small bit of current produces a small amount of the B field. Ampere's Law, illustrated below, says that if one goes around a loop (of any shape or size) then the integral of the B field around the loop is equal to the enclosed current. In the loop below I = I1 + I 2. Here I 3 is excluded as it lies outside. G G i1 ∫ B ⋅ ds = μ0 I enclosed i3 Amperian loop G i2 G ds θ B © Copyright Virtual University of Pakistan 79 PHYSICS –PHY101 VU 12. Let us apply Ampere's Law to a circular loop of radius r outside an infinitely long wire G G carrying current I through it. The magnetic field goes around in circles, and so B and ds G G are both in the same direction. Hence, ∫ B ⋅ ds = B ∫ ds = B ( 2π r ) = μ0 I. We get the μ0 I important result that B =. 2π r 13. Assuming that the current flows uniformly over the crossection, we can use Ampere's Law to calculate the magnetic field at distance r , where r now lies inside the wire. ⎛ π r2 ⎞ μ Ir B ( 2π r ) = μ 0 I ⎜ 2 ⎟ ⇒B= 0 2 ⎝πR ⎠ 2π R Here is a sketch of the B field inside and outside the wire as a function of distance r. G G B B ds r inside outside R r B r=R © Copyright Virtual University of Pakistan 80 PHYSICS –PHY101 VU Summary of Lecture 28 – ELECTROMAGNETIC INDUCTION 1. Earlier we had defined the flux of any vector field. For a magnetic field, this means that flux of a uniform magnetic field (see figure) is Φ = B⊥ A = BA cosθ. If the field is not A constant over the area then we must add up all the little G G pieces of flux: Φ B = ∫ B ⋅ dA. The dimension of flux is G B magnetic field × area, and the unit is called weber, where θ 1 weber = 1 tesla ⋅ metre2. 2. A fundamental law of magnetism states that the net flux through a closed surface is always G G zero, Φ B = ∫ B ⋅ dA = 0. Note that this is very different from what you learned earlier in electrostatics where the flux is essentially the electric charge. There is no such thing as a magnetic charge! What we call the magnetic north (or south) pole of a magnet are actually due to the particular electronic currents, not magnetic charges. In the bar magnet below, no matter which closed surface you draw, the amount of flux leaving the surface is equal to that entering it. N S Example : A sphere of radius R is placed near a long, straight wire that carries a steady current I. The magnetic field generated by the current is B. Find the total magnetic flux passing through the sphere. © Copyright Virtual University of Pakistan 81 PHYSICS –PHY101 VU Answer: zero, of course! 3. Faraday's Law for Induced EMF: when the magnetic flux changes in a circuit, an electro- motive force is induced which is proportional to the rate of change of flux. Mathematically, ε = − d Φ B where ε is the induced emf. If the coil consists of N turns, then ε = − N d Φ B. dt dt How does the flux through a coil change? Consider a coil and magnet. We can: a) move the magnet, b) change the size and shape of the