Moving Charges and Magnetism PDF

MOVING CHARGES AND MAGNETISM In the year 1820, Oersted noticed deflection in a magnetic needle placed near a current carrying conductor. He investigated this phenomenon further. He noticed that the deflection in the needle got increased when the current was made la...

MOVING CHARGES AND MAGNETISM In the year 1820, Oersted noticed deflection in a magnetic needle placed near a current carrying conductor. He investigated this phenomenon further. He noticed that the deflection in the needle got increased when the current was made large or the needle was placed closer to the conductor. On reversing the direction of the current, deflection in the needle got reversed. Oersted concluded that moving charges or electric currents produced a magnetic field in the surrounding space. Magnetic field is the space around a current carrying conductor (or a magnet) in which its magnetic effects can be felt. Magnetic field around a current carrying conductor disappears as soon as current in the conductor is turned off. It means a moving charge is the source of magnetic field. Force on a moving charge in a Magnetic field ⃗⃗ such that the A positive charge q is moving with a velocity 𝑣⃗ in a magnetic field of strength 𝐵 angle between its velocity and magnetic field is 𝜃. As a consequence of the interaction between the magnetic field produced by the moving charge and the applied magnetic field, the charge experiences a force which depends on Magnitude of the applied magnetic field Magnitude of the charge The component of velocity perpendicular to the applied magnetic field. Combining the above factors, we get 𝐹 ∝ 𝑞𝐵𝑣 sin 𝜃 𝐹 = 𝐾𝑞𝐵𝑣 sin 𝜃 The value of constant of proportionality (k) = 1. 𝐹 = 𝑞𝐵𝑣 sin 𝜃 ⃗⃗⃗⃗ In vector form, the force is given by 𝐹 = 𝑞(𝑣⃗ × 𝐵 ⃗⃗ ) The magnetic force on moving charge q includes a vector product of velocity and magnetic field. Its direction can be found by Right handed screw rule or Right hand rule. Accordingly, the direction of magnetic force is perpendicular to the plane containing 𝑣 ⃗⃗. ⃗⃗⃗⃗and 𝐵 Special cases If a charge is at rest in a magnetic field, then its velocity 𝑣 ⃗⃗⃗⃗ = 0. Therefore, ⃗⃗⃗⃗ 𝐹 = 0. Thus a stationary charge will never experience a force in magnetic field. If the charge moves parallel or anti-parallel to the direction of magnetic field, thenθ=00or 1800. In either cases sin 𝜃 = 0. So ⃗⃗⃗⃗ 𝐹 = 0. Therefore, a charge moving parallel or anti-parallel to a magnetic field experiences no force. If the charge moves perpendicular to the applied magnetic field, then,θ=900the charge will experience maximum force given by 𝐹𝑚𝑎𝑥 = 𝐵𝑞𝑣. The direction of this force is given by Fleming’s left hand rule. Definition of Magnetic field If 𝑞 = 1, 𝑣 = 1 𝑎𝑛𝑑 sin 𝜃 = 1 then 𝐹 = 𝐵.Thus, Magnetic field strength at a given point is defined as the force experienced by a unit charge moving with unit velocity at right angles to the magnetic field. S.I unit of Magnetic field is tesla (T) Lorentz force Suppose a charge 𝑞 is moving with a velocity 𝑣 ⃗⃗⃗⃗) and ⃗⃗⃗⃗ in the presence of both electric(𝐸 magnetic(𝐵 ⃗⃗⃗⃗)field. The force experienced by the moving charge in both electric and magnetic fields is called Lorentz force. It is given by ⃗⃗⃗⃗ 𝐹 = 𝑞𝐸 ⃗⃗⃗⃗ + 𝑞(𝑣 ⃗⃗ ), where𝑞𝐸 ⃗⃗⃗⃗ × 𝐵 ⃗⃗⃗⃗ is the force on the charge due to electric field and 𝑞(𝑣 ⃗⃗ ) is the force on the moving charge due to magnetic field. ⃗⃗⃗⃗ × 𝐵 MOTION OF A CHARGE IN A MAGNETIC FIELD: A charged particle q having a mass m enters a uniform magnetic field 𝐵 ⃗⃗ with velocity 𝑣⃗ making an angle 𝜃 with the direction of magnetic field as shown in the diagram. Velocity 𝑣 ⃗⃗⃗⃗ of the charge is resolved into its rectangular components. 𝑣∥ = 𝑣 cos 𝜃 and𝑣⊥ = 𝑣 sin 𝜃. Here the component of velocity 𝑣 sin 𝜃 is perpendicular to⃗⃗⃗⃗ 𝐵. Hence the magnetic force on the charge is 𝐹 ⃗⃗⃗⃗ = 𝑞(𝑣⃗⃗⃗⃗⃗ ⃗⃗ ⊥ × 𝐵 ). The direction of this magnetic force is perpendicular to the plane containing 𝐵 ⃗⃗ and ⃗⃗⃗⃗⃗ 𝑣⊥ and is given by right hand rule. Since this force is perpendicular to 𝑣 sin 𝜃it does not perform any work and hence the magnitude of ⃗⃗⃗⃗⃗ 𝑣⊥ (𝑣 sin 𝜃) does not change with time. However, this magnetic force 𝐹 ⃗⃗⃗⃗ changes the direction of motion of the charge, Consequently, the charge moves in a circular path as shown in the diagram. The centripetal force required for the charge to move in circular path is provided by the magnetic force acting on it. 2 𝑚𝑣⊥ Mathematically, = 𝐵𝑞𝑣⊥ 𝑟 𝐵𝑞𝑟 𝑚𝑣⊥ 𝑣⊥ = and 𝑟 = 𝑚 𝑞𝐵 The angular velocity of the charge in magnetic field is given 𝑣 𝐵𝑞𝑟 𝐵𝑞 by, 𝜔 = 𝑟 = 𝑟𝑚 = 𝑚 Clearly 𝜔 is independent of velocity of the charge or its kinetic energy. Further,frequency of revolution of the charge 𝜔 𝐵𝑞 in magnetic field is 𝑓 = 2𝜋 = 2𝜋𝑚 2𝜋𝑚 Time period of revolution of the charge in magnetic field is 𝑇 = 𝐵𝑞 For component velocity𝑣 cos 𝜃, charge will not experience any force in the magnetic field because the angle between 𝑣⃗∥ and 𝐵⃗⃗ is zero (0). Thus the charge will cover a linear distance with a constant speed 𝑣 cos 𝜃. Due to the combined effect of the two component velocities the charged particle in magnetic field covers linear path as well as circular path. As a result, path of the charge is helical, whose axis is parallel to the magnetic field. The linear distance covered by the charged particle in the magnetic field in one revolution is called pitch of the helix. Mathematically, pitch of the helix is given by, 2𝜋𝑚 𝑑 = 𝑣∥ 𝑇 = 𝑣 cos 𝜃 𝐵𝑞 NOTE: Magnetic field does no work on a charged particle.i.e. kinetic energy of a moving charge remains constant in a magnetic field. Unlike electric field magnetic field can neither speed up nor slow down a charged particle. A moving charge experiences maximum force in a magnetic field when it enters the field at 𝑚𝑣 right angles to it. The charge will move in a circular path whose radius is given by𝑟 = 𝐵𝑞. The larger the momentum(𝑚𝑣) of the charge, larger is the radius and bigger the circle described. VELOCITY FILTER OR VELOCITY SELECTOR: A charged particle q is moving with velocity 𝑣 ⃗⃗⃗⃗ in presence of both electric and magnetic fields experiences a force given by ⃗⃗⃗⃗⃗ 𝐹 = ⃗⃗ 𝑞(𝐸 + 𝑣 ⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗ × 𝐵 )= 𝐹𝐸 + 𝐹𝐵 ⃗ We shall consider a simple case where electric and magnetic fields are perpendicular to each other and also to the velocity of the particle as shown in the figure. We have, 𝐹⃗𝐸 = 𝑞𝐸⃗⃗ = 𝑞𝐸 𝑗̂ , 𝐹⃗𝐵 = 𝑞(𝑣 ⃗⃗⃗⃗ × 𝐵 ⃗⃗ ) = 𝑞(𝑣 𝑖 × 𝐵𝑘) = −𝑞𝐵𝑗̂ Therefore, 𝐹 ⃗⃗⃗⃗ = 𝑞(𝐸 − 𝑣𝐵)𝑗̂ Thus, electric and magnetic fields are in opposite directions. If the values of electric and magnetic fields are adjusted such that the magnitudes of the two forces are equal, then the net force acting on the charge will be zero and the charge will move undeflected in the fields. This happens when, 𝐸 𝑞𝐸 = 𝑞𝐵𝑉 or 𝑣 = 𝐵 This is the condition used to select charged particles of a particular velocity out of a beam of charges moving with different speeds. Velocity filter is an arrangement of crossed electric and magnetic fields in a region which helps to select charged particles of a particular velocity from a beam of charged particles having different velocities. CYCLOTRON: Cyclotron is a machine to accelerate charged particles to high energies. The cyclotron uses both electric and magnetic fields in combination to increase the energy of charged particles. As the two fields are perpendicular to each other they are called crossed fields. Cyclotron uses the fact that frequency of revolution of a charged particle in a magnetic field is independent of its velocity. The arrangement consists of two D- shaped hollow evacuated metal chambers D1 and D2 called the dees. The two dees are connected to an oscillator (A.C source). The whole assembly is enclosed in an evacuated chamber. Positively charged particles are released at the centre P. They move in a semi-circular path in one of the dees and arrive in the gap between the dees in a time interval equal to half the time period of A.C. The frequency of the applied voltage is adjusted so that the polarity of the dees is reversed in the same time that it takes the charges to complete one half of the revolution. Each time when the charges arrive in between the two dees the electric field accelerates them where as the magnetic field ensures that the charges move in circular paths. 𝑚𝑣 Radius of the circular path described by a charged particle 𝑞 of mass 𝑚 is 𝑟 = 𝐵𝑞. Clearly, as 𝑣 increases,𝑟 increases. Time taken by the charge to describe a semi-circular path is 𝜋𝑟 𝜋𝑚 𝜋 𝑡 = 𝑣 = 𝐵𝑞 = 𝑞⁄ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. 𝐵( 𝑚) This time is independent of velocity of the charge and radius of the circular path. Consequently the charge takes the same time to arrive in between the two dees irrespective of its velocity and radius of its path. It gets repeatedly accelerated by the electric field until it acquires energy to have a radius approximately that of dees. Then the magnetic field deflects the charge and it leaves the system via an exit slit.  2𝜋𝑚 If 𝑇 is the time period of A.C, then 𝑇 = 2𝑡 = 𝐵𝑞 𝐵𝑞 Cyclotron frequency is given by 𝑓 = 2𝜋𝑚 If R is the maximum radius of the circular path followed and if 𝑣𝑜 is the maximum velocity gained 1 by the charge, then maximum kinetic energy attained by the charge is 𝐾𝐸𝑚𝑎𝑥 = 2 𝑚𝑣𝑜2 𝑚𝑣 2 𝐵𝑞𝑅 𝐵2 𝑞 2 𝑅2 Since 𝑅 𝑜 = 𝐵𝑞𝑣𝑜 , so 𝑣𝑜 = 𝑚 substituting this in the above equation we get, 𝐾𝐸𝑚𝑎𝑥 = 2𝑚 Uses: The charged particles accelerated by cyclotron are used to bombard certain nuclei for studying the resulting nuclear reactions. It is used in hospitals to produce radioactive substances which can be used in diagnosis and treatment. Limitations It cannot be used to accelerate uncharged particles. Cyclotron cannot speed up a charge beyond a certain limit, reason being increase in mass of the 𝑚𝑜 charge as its speed approaches that of light.(𝑚 = 2 ). It is for the same reason cyclotron is √1−𝑣2 𝑐 not suitable for accelerating electrons as mass of the electron is very less. It will very quickly go out of the step with the oscillating electric field. BIOT-SAVART LAW Biot-savart’s law deals with the magnetic field at a point due a current element (a small portion of conductor carrying current). Consider a conductor carrying a current 𝐼. Let P be a point at a distance 𝑟 from the mid-point of current element. Let 𝜃 be the angle between 𝐼𝑑𝑙 ⃗⃗⃗⃗ and𝑟⃗. The magnetic field ⃗⃗⃗⃗⃗⃗ 𝑑𝐵 at point P is to be determined due the current element. According to Biot-Savart’s law, the magnitude of magnetic field (𝑑𝐵) is directly proportional to the current (𝐼), the element length (𝑑𝑙), sine of the angle between the line joining point P and mid- point of current element (sin 𝜃) and inversely proportional to square of the distance (𝑟). 𝐼𝑑𝑙 sin 𝜃 Mathematically, 𝑑𝐵 ∝ 𝑟 2 𝐼𝑑𝑙 sin 𝜃 𝑑𝐵 = 𝑘 𝑟 2 , the value of 𝑘 depends on the units chosen for measurement and the medium between point P and the current element. When there is free space between point P and the 𝜇 current element and in SI units, 𝑘 = 4𝜋𝑜 where, 𝜇𝑜 is absolute permeability of free space. Its value is 4π x 10-7 TA-1m 𝜇𝑜 𝐼𝑑𝑙 sin 𝜃 Therefore, 𝑑𝐵 = 4𝜋 𝑟2 𝜇𝑜 𝐼𝑑𝑙 sin 𝜃 In vector form,⃗⃗⃗⃗⃗⃗ 𝑑𝐵 = 𝑟̂ 4𝜋 𝑟2 ⃗⃗⃗⃗ × 𝑟⃗) 𝜇𝑜 (𝐼𝑑𝑙 ⃗⃗⃗⃗⃗⃗ 𝑑𝐵 = 4𝜋 𝑟3 ⃗⃗⃗⃗⃗⃗ is determined by right hand thumb rule or right hand screw rule. The direction of 𝑑𝐵 NOTE: If 𝜃 = 00 or 1800 ,then the magnetic field dB= 0 𝜇 𝐼𝑑𝑙 If 𝜃 = 900,then the magnetic field 𝑑𝐵 = 4𝜋𝑜 𝑟 2 MAGNETIC FIELD AT A POINT ON THE AXIS OF A CIRCULAR COIL CARRYING CURRENT: Consider a circular coil of insulated copper wire having N turns each of radius ‘a’ carrying current I in the anticlockwise direction. The plane of the coil is vertical and is perpendicular to the paper. Let P be a point on the axis of the coil at a distance ‘x’ from the centre O of the loop. Let C be the current element of length dl. The distance of the point P from the current element is r. 𝑎 r= √𝑎2 + 𝑥 2 and sinα = 𝑟 According to Biot- Savart’s law, the magnitude of magnetic field at P due to current element C is 𝜇0 𝐼𝑑𝑙𝑠𝑖𝑛𝜃 dB= 4𝜋 𝑟2 Where, θ is the angle between ⃗⃗⃗⃗ 𝑑𝑙 and 𝑟⃗ 𝜇0 𝐼𝑑𝑙 dB= along PM (sinθ= 900) 4𝜋 𝑎2 +𝑥 2 Similarly, the magnetic field due to the diametrically opposite current element of C at a point P is 𝜇0 𝐼𝑑𝑙 dB= along PN 4𝜋 𝑎2 +𝑥 2 The magnetic field dB is resolved into horizontal and vertical component dBsinα and dBcosα respectively, where α is the angle between the line joining the point to the current element and axis of the coil. The vertical components of magnetic field cancel each other and horizontal component gets added up. Hence, the magnetic field at P due to two elements = 2dBsin𝛼 𝜇0 𝐼𝑑𝑙 𝑎 𝑎 =2 (In ∆𝐶𝑂𝑃 sinα=√𝑎2 ) 4𝜋 (𝑎2 +𝑥 2 ) √𝑎2 +𝑥 2 +𝑥 2 𝜇0 𝐼𝑎 = 2 dl 4𝜋 (𝑎2 +𝑥 2 )3/2 The magnetic field B at P due to the entire coil is 𝜇0 𝐼𝑟 B= ∑ 𝑑𝐵 = ∑ 2dl 4𝜋 (𝑎2 +𝑥 2 )3/2 𝜇0 𝐼𝑟 B= ∑ 2𝑑𝑙 4𝜋 (𝑎2 +𝑥 2 )3/2 But ∑ 2𝑑𝑙 = 2πa for circular loop 𝜇0 𝐼𝑟 B= (N × 2πa) for N turns 4𝜋 (𝑎2 +𝑥 2 )3/2 𝝁𝟎 𝟐𝝅𝑵𝑰𝒓𝟐 B= along PX 𝟒𝝅 (𝒂𝟐 +𝒙𝟐 )𝟑/𝟐 Special cases: 1) If P lies far away from the centre of the coil i.e., x >> a then, a2 +x2 ≈x2 𝜇 2𝑁𝐼𝐴 Hence, B= 0 3 4𝜋 𝑥 𝜇 2𝜋𝑁𝐼 2) If point P lies at the centre of the coil i.e., x=0 then, B = 4𝜋0 𝑎 NOTE: The magnetic field is maximum at the centre of the coil. It decreases along the axis on either side. AMPERE’S CIRCUITAL LAW Statement: According to Ampere’s circuital law, the line integral of the magnetic field along a closed path or a circuit enclosing an arbitrary area is equals to μ0 times the current flowing normally through the area.  → The line integral of magnetic field is  ( B. dl ) = μoInet Applications: THE MAGNETIC FIELD DUE TO A STRAIGHT CONDUCTOR CARRYING CURRENT Consider a long straight conductor carrying a current I as shown in the figure. Let ‘P’ be any point in the field of the conductor on a closed path as shown in the figure at a distance ‘a’. The line element and magnetic field at that point both are directed along the same line. Therefore the angle between them is 00. From the ampere circuital law  → I B in o  ( B. dl ) = μ0I  =0 dl ∮ 𝐵𝑑𝑙 𝑐𝑜𝑠𝜃 = μ0I B ∮ 𝑑𝑙 = μ0I (θ=00) B 2πa = μ0I μ I μ 2I B = 0 or B = 0 2π a 4π a This is the field due to a long straight conductor carrying current. NOTE: The magnetic field is directly proportional to the current and inversely proportional to the distance from the current source. Direction of magnetic field due to a long wire is given by right hand rule. SOLENOID: Solenoid is a coil of large number of turns of insulated wire wound very closely on an insulating cylinder. Every turn in a solenoid produces its own magnetic field along its axis. Therefore, the net field is a vector sum of the fields due to all the turns, and is directed along the axis as shown in the figure. On the surface of the solenoid, the fields due to two adjacent turns are in the opposite direction and equal. Hence they cancel out resulting zero field. Outside the solenoid the field is negligibly weak. Therefore, the entire field of the solenoid is uniform within the solenoid and parallel to its axis. Along any line normal to the axis, the component of magnetic field is zero. MAGNETIC FIELD DUE TO A SOLENOID: Consider an ideal solenoid whose length is d B 0 very large compared to its radius. Let L be the length of the solenoid and ‘P’ be any c point inside it. Let the current flowing B=0 B=0 through the solenoid wire be ‘i’. To apply the Ampere’s law draws an amperian loop ‘abcda’, such that ab passes through the i i point ‘P’ and ab= l According to Ampere’s Law, the line a dl B b integral of the magnetic field along ‘abcda’ is equals to μ0 times the total current through abcda.  →  ( B. dl ) = μ0 I Where I–total current through the path ‘abcda’ The line integral of magnetic field 𝐵 ⃗⃗ over the closed loop abcda is  → 𝑏 ⃗⃗⃗⃗ + ∫𝑐 𝐵 ⃗⃗⃗⃗ + ∫𝑑 𝐵 ⃗⃗⃗⃗ + ∫𝑎 𝐵 ( B. dl ) = ∫ 𝐵 𝑎 ⃗⃗. 𝑑𝑙 𝑏 ⃗⃗. 𝑑𝑙 𝑐 ⃗⃗. 𝑑𝑙 𝑑 ⃗⃗⃗⃗ ⃗⃗. 𝑑𝑙 Since the field normal to the ideal solenoid along bc and da are zero and along cd it is negligibly weak, the total field is along ab only. b B  dl = μ I a 0 Bl = μ0 I But the net current through the loop abcd is I = nlI Hence, B =μ0n I N Since = n is number of turns per unit length. L 𝝁𝟎 𝑵𝑰 B= This is the magnetic field due to a solenoid. 𝑳 Note: i) Magnetic field of solenoid is inside it along its axis. ii) Solenoid acts as a magnetic dipole. iii) Magnetic field is directly proportional to the number of turns per unit length and current. TOROID: Toroid is a long solenoid taken in a circular form. It is a coil made by winding large number of turns without any spacing between adjacent i turns, with an insulated wire on a circular tube made up of an insulator. i MAGNETIC FIELD DUE TO A TOROID: Let ‘P’ be a point inside a toroid at a distance ‘r’ from centre of the toroid. Using Amperes circuital law we can find out magnetic field at this point by drawing an amperian loop that passes through ‘P’. i r This closed path is nothing but a circle of radius ‘r’ p inside the toroid. From Ampere’s circuital law. r  →  dl ) = μ0 I ( B. B∮ 𝑑𝑙= μ0 I (∮ 𝑑𝑙 = 2πr = circumference) B2πr = μoI = μon 2πr I B = μo n I 𝜇 𝑁𝐼 OR B= 2𝜋0 𝑟 From the above expression, it is clear that the toroid produces magnetic field similar to a solenoid. MAGNETIC FORCE ON CURRENT CARRYING CONDUCTOR IN A MAGNETIC FIELD: Consider a straight conductor of length ‘L’, area of cross section A carrying a current ‘I’ in a uniform magnetic field of strength, ‘B’. Let ‘n’ be the number of electrons per unit volume of the conductor, ‘q’ be the charge of an electron and vd be its drift velocity. The number of electrons in conductor of length l is N= nAL. The charges flowing inside the conductor experience a magnetic forces. The force on each charge is given by Fcharge = B Vd q sin θ If each charge experiences this force then the magnetic force on conductor of length L is given by F = N B vd q sin θ = nAlqvdB sinθ = nAlq (𝑉 ⃗⃗⃗⃗⃗ ⃗⃗ 𝑑 ×𝐵 ) The steady current through the conductor I= nAqvd 𝐼 Current density 𝑗⃗ = 𝐴 = nqvd Hence, F= BILsinθ where, θ is the angle between B and I ⃗ 𝐹 = AL ( 𝑗⃗×𝐵 ) ⃗⃗ 𝐹⃗ = I( 𝐿 ⃗⃗× 𝐵⃗⃗ ) Where, 𝐿 ⃗⃗ = L 𝐿̂ , 𝐿̂ is a unit vector in the direction of I Case 1): when the conductor is placed parallel to the field, θ = 00, then 𝐹⃗ = 0 which is minimum. Case 2): when the conductor is placed perpendicular to the field ,θ = 900 , then 𝐹⃗ =BIL and is maximum. FORCE BETWEEN TWO PARALLEL CONDUCTORS CARRYING CURRENT: Consider two parallel straight conductors each of length L separated by a distance‘d’ carrying currents I1& I2 as shown in fig. As these 2 conductors carry currents, they produce I1 I2 their own magnetic fields around them. If ‘d’ is small, both B1 the conductors are present in the magnetic fields of each other. As a result each conductor experiences a magnetic force. We know the force on conductor carrying current in a F12 F21 magnetic field is given by I2 I1 𝐹⃗ = I ( 𝐿 ⃗⃗× 𝐵 ⃗⃗ ) B2 The force on the 2nd conductor with current I2 exerted by 1st conductor with current I1, is given by F21 = B1 I2 L sin θ F21 = BI2 L (θ = 900) Magnetic field B due to a straight conductor from Amperes circuital law is µ0 2𝐼1 B1 = 4𝜋 𝑑 µ0 2𝐼1  F21= { } I2 L 4𝜋 𝑑 µ0 2𝐼1 𝐼2 F21 = L …………………… (a) 4𝜋 𝑑 According to Fleming’s left hand rule, this force is directed towards the conductor 1 Similarly, force on first conductor exerted by second conductor is µ0 2𝐼1 𝐼2 F12 = L …………………… (b) 4𝜋 𝑑 This force is directed towards the conductor 2 From equations (a) and (b) F12 = - F21 The force per unit length is given by 𝐹 µ0 2𝐼1 𝐼2 = …………………….(c) 𝐿 4𝜋 𝑑 Definition of Ampere: In the above expression of force per unit length, ifd= 1m and I1 = I2 = 1A, then F 4π×10-7 2 × 1 × 1 = L 4π 1 F = 2x10-7 N m-1 L When two long straight parallel conductors kept at unit distance in free space carry equal currents in the same direction and experience a force of 2x10-7 N m-1, then the current is said to be one ampere. TORQUE ON A RECTANGULAR CURRENT LOOP IN A UNIFORM MAGNETIC FIELD: Consider a rectangular loop ABCD of length ‘b’, and breadth ‘a’, placed in a uniform magnetic field 𝐵⃗⃗. Let I be the current through the loop in clockwise direction. Case 1) magnetic field is parallel to the plane of the loop The magnetic field exerts no force on the side BC and DA The side CD and AB are perpendicular to the field. The force on AB is F1= I b B and is directed into the loop. The force on CD is F2 = I b B and is directed out of the plane of the paper. The forces F1 and F2 are equal in magnitude, opposite in direction and are not collinear. Hence they constitute a couple. This torque on loop tends to rotate the loop in anti- clockwise direction. Τorque, τ= force × perpendicular distance =BIb×a = BIA Where, A = ab is the area of the rectangle. Case 2) when the uniform magnetic field makes angle θ w.r.t axis of the loop: Let θ be the angle between field and the normal to the coil. Forces acting on sides BC and DA are equal and opposite. They are collinear and don’t produce torque. But the forces on sides AB and CD are F1 and F2. They are too equal and opposite, with magnitude but not collinear and thus, constitute a couple. F1 = F2=I b B Magnitude of torque(τ) due to forces F1 & F2 is τ = force × arm of the couple =I B b× a sinθ τ = B I A sinθ If loop has N turns, then τ =N B I A sinθ MAGNETIC DIPOLE MOMENT OF A CURRENT LOOP: The magnetic moment is defined as the product of current I and the area vector of the loop. Magnetic moment, 𝑚 ⃗⃗⃗= I 𝐴⃗ The direction of 𝐴⃗ is along the normal to the plane of the loop and hence is the direction of 𝑚 ⃗⃗⃗. 2 Unit of magnetic dipole moment is Am Hence, 𝑤𝑒 ℎ𝑎𝑣𝑒 𝜏⃗ = 𝑚 ⃗⃗ ⃗⃗⃗ × 𝐵 CIRCULAR CURRENT LOOP AS A MAGNETIC DIPOLE: W.k.t the magnetic field on the axis of a circular loop of radius R carrying a current I has a magnitude µ 2𝜋𝐼𝑅2 B= 4𝜋0 (𝑎2 +𝑥2 )3/2 Where, x is the distance along the axis from the centre of loop. When x>> a, µ 2𝜋𝐼𝑎 2 µ 2𝐼𝐴 B = 4𝜋0𝑥3 = 4𝜋0 𝑥3 where, A = πa2 is the area of the loop ⃗⃗⃗⃗ µ0 2𝑚 ⃗⃗ = = Hence, 𝐵 ⃗⃗⃗= I 𝐴⃗ where, 𝑚 4𝜋 𝑥 3 THE MAGNETIC DIPOLE MOMENT OF A REVOLVING ELECTRON: According to Bohr atom model, electrons revolve around the positively charged nucleus. The electron of charge (-e) performs uniform circular motion around the nucleus of charge +Ze. Let r be the radius of the orbit of the electron and v be the orbital speed, then the current due to motion 𝑒 of electron is I = 𝑇 where, T is the time period of revolution. 𝑒𝑣 2𝜋𝑟 I= 2𝜋𝑟 since, T = 𝑣 There will be a magnetic moment associated with this current and is given by, + Ze e 𝑒𝑣𝑟 r Magnetic moment ,μl = IA = 2 GYROMAGNETIC RATIO: The ratio of magnetic moment of electron to the orbital angular momentum of the electron is called gyromagnetic ratio. 𝑒𝑣𝑟 We have, Magnetic moment,μl = 2 Multiply and divide above equation by mass of the electron, me 𝑒 𝑒 μl = 𝑚𝑒 vr = l 2𝑚𝑒 2𝑚𝑒 where, l= 𝑚𝑒 vr is the orbital angular momentum. µ 𝑒 Gyromagnetic ratio= 𝑙l = 2𝑚 for electron. It value is 8.8× 10 10 Ckg-1 𝑒 BOHR MAGNETON: 𝑛ℎ According to Bohr’s quantisation rule, l = 2𝜋 where, n= 1,2,3.... 𝑒 𝑛ℎ Hence, μl = 2𝑚 𝑒 2𝜋 𝑒ℎ For n=1, μl = 4𝜋𝑚𝑒 = 9.27 × 10-24 Am2 This value is called Bohr magneton of an electron. Bohr magneton is the magnetic moment of the electron in the first orbit of the Hydrogen atom MOVING COIL GALVANOMETER: Moving coil galvanometer or a pointer galvanometer is a device used to detect the presence of a current in an electrical circuit. Principle: It works on the principle that a current carrying conductor experiences a mechanical force in a direction perpendicular to the direction of both the current and the applied magnetic field or a current carrying loop placed in a magnetic field experiences torque. Construction: Moving coil galvanometer consists of rectangular coil of copper wire wound on an insulating rectangular frame mounted between two bearings and two (phosphore bronze) helical springs as shown in the figure. This arrangement is fixed along the axis of two cylindrical pole pieces of strong magnets. The two ends of the coil are taken through the helical springs through which the current flows to the galvanometer. A soft ion cylinder is mounted along the common axis of the coil, which increases the magnetic field strength in that region. Theory When a current flows through the coil, a torque acts on it. Let N be the number of turns, A be the area of cross section of the coil and B be the radial magnetic field then, torque acting on it is given by τ = NI AB (Since the field is radial, θ = 900 and sinθ = 1). The magnetic torque (NIAB) tends to rotate the coil. A spring Sp provides a counter torque equal to (k  ) that balances the magnetic torque (NIAB) resulting in a steady angular deflection . In equilibrium (k  ) = (NI AB). Where, k is the torsional constant of the spring (the restoring torque per unit twist). The deflection  is indicated on the scale by a pointer attached to the spring. Angle of deflection is given by  = ( NAB ) I k SENSITIVITY OF MOVING COIL GALVANOMETER: A galvanometer is said to be sensitive if a small current in the galvanometer produces a large deflection. 1) Current sensitivity: It is defined as the deflection produced in the galvanometer when an unit current flows through it. 𝜙 𝑁𝐵𝐴 Hence, current sensitivity, 𝐼 = 𝐾 Its unit is rad A-1 or div A-1 2) Voltage sensitivity: It is defined as the deflection produced in the galvanometer when a unit potential difference is applied across the two terminals of the galvanometer. 𝜙 𝑁𝐵𝐴 𝐼 𝑁𝐵𝐴 1 Voltage sensitivity, 𝑉 = 𝐾 𝑉 = 𝐾 𝑅 Where, R is the resistance of the galvanometer. Current sensitivity Hence, voltage sensitivity = 𝑅 Unit of voltage sensitivity is radV-1 or divV-1 CONVERSION OF GALVANOMETER 1) CONVERSION OF GALVANOMETER INTO AMMETER: Ammeter is a current measuring device. Any current measuring device is connected in series in the circuit. When the current to be measured flows through, it should not reduce the actual value of current in the circuit. Hence its resistance should be very low. The resistance of ideal ammeter is zero. If galvanometer has to be converted in to ammeter, a suitable low resistance called shunt resistance should be connected parallel to the coil of the galvanometer. This resistance depends on main maximum current to be measured. VG = VS Ig G = S (I – Ig) Ig G S= I-Ig Where, G - resistance of Galvanometer Ig - Current which produce full scale deflection in the galvanometer. I - main current 2) CONVERSION OF GALVANOMETER INTO VOLTMETER: Voltmeter is a device that measures voltage across the ends of a circuit element. Any voltage measuring device should not draw any appreciable current into it, to measure the actual voltage across the circuit element. Hence an ideal galvanometer should have infinite resistance. Practically it should have a large resistance. To convert a galvanometer into voltmeter, the galvanometer resistance should be raised suitably. And hence a large resistance is connected in series with the galvanometer. This can be calculated as follows. V = Vg + VR V = Ig G + Ig R V = Ig (G + R) V  R= -G Ig