SYBSC Sem IV Physics Magnetism Notes PDF

SYBSC Sem IV Physics Major `Unit II Chapter 1 Magnetism and magnetic field 1.1 Magnetism: Definition: Magnetism is the force exerted by magnets when they attract or repel each...

SYBSC Sem IV Physics Major `Unit II Chapter 1 Magnetism and magnetic field 1.1 Magnetism: Definition: Magnetism is the force exerted by magnets when they attract or repel each other. Magnetism is caused by the motion of electric charges. 1 SYBSC Sem IV Physics Major Suppose you cut off a compass needle and separate the two poles-red one and grey one (Shown in figure 25.5). If you drill a hole at the centre of red part and keep it on the pivot at the centre of the compass, do you get the same deflection if current is allowed to flow through the wire? Justify your answer. 1.2 Magnetic field: The basic difference between electric field and magnetic field is that an electric field is created by stationary or moving charge; but magnetic field can be created only by moving charge. 2 SYBSC Sem IV Physics Major Magnetic force on moving charges: 3 SYBSC Sem IV Physics Major 4 SYBSC Sem IV Physics Major 5 SYBSC Sem IV Physics Major Problem 1: Solution: Given: q = 1.6 X 10-19 C, v = 3 X 105 m/s, B = 2.0 T, ᶲ = 30° Problem 2 Solution: Given: m=0.195 g = 0.195 X 10-3 kg, Charge q = - 2.5 X 10-8 C, v = 4 X 104 m/s, B =? F = m.g = 0.195 X 10-3 X 9.8 = 1.911 X 10-3 N The particle will experience minimum force if the magnetic field and the velocity are in the same direction. The magnitude of the magnetic induction will be given by F = |q| v|| B F B= = 1.911 T q v|| Problem 3 Solution: Given: Charge on electron q = - 1.6 X 10-19 C, F= 4.60 X 10-15 N, φ=60°, B = 3.50X 10-3 T 6 SYBSC Sem IV Physics Major The force on electron due to magnetic field B is given by 𝐹 = |𝑞|𝑣𝐵 𝑠𝑖𝑛 𝜑 𝐹 𝑣= |𝑞|𝐵 sin 𝜑 4.60 𝑋 10−15 𝑣 = = 0.949 X 10 7 m/s 1.60 𝑋 10−19 𝑋 3.50𝑋10−3 𝑋𝑠𝑖𝑛 60 1.3 Magnetic field lines and magnetic flux: We can represent magnetic field by magnetic field lines which start from north pole and ends on south pole of any magnet. We draw the lines so that the line through any point is tangent to the magnetic field vector B at that point. If the field lines are close to each other, the magnitude of magnetic field is large and if the field lines are far apart, the magnitude of magnetic field is small. 7 SYBSC Sem IV Physics Major Magnetic flux and Gauss’s law for magnetism: But B⊥ = B cos φ ⃗. ⃗⃗⃗⃗⃗ Therefore, dɸB=B cos φ dA =𝐵 𝑑𝐴(27.5) The total magnetic flux through the surface is the sum of contributions from the individual area components. 8 SYBSC Sem IV Physics Major Gauss’s law for magnetism: Problem 4: Solution: 9 SYBSC Sem IV Physics Major Problem 5: Solution: Given: r = 6.50 cm= 6.50 X 10-2m, B = 0.230 T Area of circle = π r2 = 3.14 X (6.50 X 10-2)2 =132.67 X 10-4 m2 (a) B is in Z direction that means parallel to A. Hence φ = 0° = 0.230 X132.67 X 10-4X cos 0 = 30.51 X 10-4 Wb (b) At an angle 53.1° from the z- direction = 0.230 X132.67 X 10-4X cos 53.1 = 18.32 X 10-4 Wb (c) In the Y direction. Hence φ = 90° = 0.230 X132.67 X 10-4X cos 90 = 0 Wb Problem 6: Solution: Given: Area A = 3.20 X 2.80 cm2 = 9.184 X 10-4 m2, φ = 30°, ɸB= 4.20 X 10-4 Wb ɸ𝑩 𝟒. 𝟐𝟎 𝐗 𝟏𝟎−𝟒 𝑩= = −𝟒 = 𝒐. 𝟓𝟐𝟖 𝑾𝒃/𝒎𝟐 𝑨 𝒄𝒐𝒔 𝝋 𝟗. 𝟏𝟖𝟒 𝐗 𝟏𝟎 𝑿𝒄𝒐𝒔 𝟑𝟎 _____________________________________________________________________ 10 SYBSC Sem IV Physics Major 1.4 Motion of charged particle in a magnetic field: 11 SYBSC Sem IV Physics Major 12 SYBSC Sem IV Physics Major 13 SYBSC Sem IV Physics Major Problem 1: Problem 2: Given: q= 6.40X10-19 C, R = 4.68 mm= 4.68 X10-3m, B=1.65 T (a) R= p/q B => Linear momentum p= R q B = (4.68 X 10-3)( 6.40X10-19)(1.65) = 49.42 X 10-22Kg m/s (b) L= R p => Angular Momentum L = 49.42 X 10-22 X4.68 X 10-3 =231.2 X 10-25 Kg m2/s 14 SYBSC Sem IV Physics Major 1. Velocity selector 15 SYBSC Sem IV Physics Major 16 SYBSC Sem IV Physics Major Problem 1: Solution: Consider a straight segment of conductor carrying current with length l and cross-sectional area A. The wire is in uniform magnetic field B perpendicular to the plane of diagram. Assume that the charge is positive. The current density J = nqVd. Therefore, 17 SYBSC Sem IV Physics Major F = JA l B But JA=I =Total current F=Ilb Problem 1: Given: I= 50 A, B = 1.20 T, ɸ = 45°, l = 1 m, F =? (a) F = I l B sin ɸ = 50 X 1 X 1.20 X sin 45 = 42.4 N 18 SYBSC Sem IV Physics Major 19 SYBSC Sem IV Physics Major zero. 20 SYBSC Sem IV Physics Major 21 SYBSC Sem IV Physics Major 22 SYBSC Sem IV Physics Major 𝝁𝟎 𝑰 𝑩= 𝟐𝝅𝒓 Problem 1: A long straight conductor is carrying current of 1.5 A. At what distance from the axis of the conductor does the resultant magnetic field have magnitude B = 0.65 X 10-4 T? Solution: Given: I = 1.5 A, B= 0.65 X 10-4 T, μ0= 4 πX 10-7 Tm/A 𝝁 𝑰 𝑩 = 𝟐𝝅𝟎 𝒓 ( 𝟒 𝛑𝐗 𝟏𝟎−𝟕 )(𝟏.𝟓) 𝒓 = 𝟐𝝅 𝑿 𝟎.𝟔𝟓 𝑿 𝟏𝟎−𝟒 = 𝟒. 𝟔 𝑿 𝟏𝟎−𝟑 𝒎 = 𝟒. 𝟔 𝒎𝒎 Problem 2: Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long straight wire. (a)If you were unfortunate enough to be 5.0 m away from such a lightning bolt, how large a magnetic field would you experience? (b) How does this field compare to one you would experience by being 5.0 cm from a long, straight household current of 10 A? Solution: (a) Given: I = 20 kA= 20 X 103 A, μ0= 4 πX 10-7 Tm/A, x = 5 m 𝝁 𝑰 (𝟒 𝝅𝑿 𝟏𝟎−𝟕 ) (𝟐𝟎 𝑿𝟏𝟎𝟑 ) 𝑩 = 𝟐𝝅𝟎 𝒓 = = 𝟖 𝑿𝟏𝟎−𝟒 𝑻 𝟐𝝅 𝑿 𝟓 23 SYBSC Sem IV Physics Major (b) Given: I = 10 A, μ0= 4 πX 10-7 Tm/A, x = 5 cm= 5 X 10-2 m 𝝁 𝑰 (𝟒 𝝅𝑿 𝟏𝟎−𝟕 ) (𝟏𝟎) 𝑩 = 𝟐𝝅𝟎 𝒓 = = 𝟎. 𝟒 𝑿𝟏𝟎−𝟒 𝑻 𝟐𝝅 𝑿 𝟓𝑿𝟏𝟎−𝟐 The magnetic field due to lightning current is much more than the magnetic field produced by the household currents. Problem 3: The body contains many small currents caused by the motion of ions in the organs and cells. Measurements of the magnetic field around the chest due to currents in the heart give values about 10 μG. Although the actual currents are rather complicated, we can gain a rough understanding of their magnitude if we model them as a long, straight wire. If the surface of the chest is 5 cm from this current, how large is the current in the heart? Ans: Given: B = 10 μG= 1 X 10-9 T μ0= 4 πX 10-7 Tm/A, r = 5 cm= 5 X 10-2m 𝝁 𝑰 (𝟐 𝝅𝑿𝟏 𝑿 𝟏𝟎−𝟗 ) (𝟓𝑿𝟏𝟎−𝟐 ) 𝑩 = 𝟐𝝅𝟎 𝒓 => 𝑰 = = 𝟐. 𝟓 𝑿𝟏𝟎−𝟒 𝑨 = 𝟎. 𝟐𝟓 𝒎𝑨 𝟒𝝅 𝑿 𝟏𝟎−𝟕 24 SYBSC Sem IV Physics Major Problem 1: Two straight, parallel, superconducting wires 4.5 mm apart carry equal currents of 15000 A in opposite directions. What force, per unit length, does each wire exert on the other? Solution: Given: r = 4.5 mm = 4.5 X 10-3 m, I = 15000 A, I’= 15000 A, μ0= 4 πX 10-7 Tm/A Problem 2: Two long, parallel wires are separated by a distance of 2.50 cm. The force per unit length that each wire exerts on the other is 4 X 10-5 N/m and the wires repel each other. The current in one wire is 0.6000 A. (a) What is the current in the second wire? (b) Are the two currents in the same direction or in the opposite directions? Solution: Given: r = 2.5 cm =2.5 X 10-2 m, F = 4 X 10-5 N/m, I= 0.6000 A, μ0= 4 πX 10-7 Tm/A 𝑭 𝝁𝟎 𝑰 𝑰′ (a) = 𝑳 𝟐𝝅𝒓 𝑭.𝟐𝝅𝒓 𝟒𝑿𝟏𝟎−𝟓 𝑿𝟐 𝝅𝑿𝟐.𝟓𝑿 𝟏𝟎−𝟐 𝑰′ = = = 𝟖. 𝟑 𝑨 𝑳𝝁𝟎 𝑰 𝟒𝝅𝑿𝟏𝟎−𝟕 𝑿𝟎.𝟔 (b) As the wires repel each other, the currents are in opposite directions. 25 SYBSC Sem IV Physics Major 26 SYBSC Sem IV Physics Major Problem 1: Solution: Given: N =100, a = 0.60 m, I = 5 A, μ0= 4 πX 10-7 Tm/A 27 SYBSC Sem IV Physics Major Problem 2: A closely wound circular coil with a diameter of 4.00 cm has 600 turns and carries a current of 0.500 A. What is the magnitude of the magnetic field (a) at the center of the coil and (b) at a point on the axis of the coil 8.00 cm from its center? Solution: Given: N =600, a= 2 cm=0.02 m, I = 0.500 A, μ0= 4 πX 10-7 Tm/A (a) At the centre of the coil x =0 (𝟒𝝅𝑿𝟏𝟎−𝟕 )(𝟔𝟎𝟎)(𝟎. 𝟓𝟎𝟎)(𝟎. 𝟎𝟐)𝟐 𝑩𝒙 = 𝟑 = 𝟗𝟒. 𝟐𝑿𝟏𝟎−𝟒 𝑻 𝟐 (𝟎𝟐 + 𝟎. 𝟎𝟐𝟐 ) 𝟐 (b) At a point 8 cm from its center x = 8 cm = 0.08 m (𝟒𝝅𝑿𝟏𝟎−𝟕 )(𝟔𝟎𝟎)(𝟎. 𝟓𝟎𝟎)(𝟎. 𝟎𝟐)𝟐 𝑩𝒙 = 𝟑 = 𝟏. 𝟑𝟒𝑿𝟏𝟎−𝟒 𝑻 𝟐 (𝟎. 𝟎𝟖𝟐 + 𝟎. 𝟎𝟐𝟐 ) 𝟐 28 SYBSC Sem IV Physics Major 29 SYBSC Sem IV Physics Major 30  Electromagnetic induction Problem 1: Problem 2: Solution:  𝒅𝑩 Given: N = 500, r = 4 cm = 4 X 10-2 m, 𝒅𝑻 = 𝟎. 𝟐𝟎𝟎 𝑻/𝒔, ε = ? Problem 3: 𝒅𝑩 Solution: Given: A = 0.0900 m2, B = 3.80 T, = −𝟎. 𝟏𝟗𝟎 𝑻/𝒔, R= 0.600 Ω 𝒅𝒕 𝒅∅ 𝒅(𝑩𝑨) 𝒅𝑩 (a) = =.𝑨 𝒅𝒕 𝒅𝒕 𝒅𝒕 =(-0.190) (0.0900) = -0.0171 V 𝒅∅ Therefore, ε = − 𝒅𝒕 = +𝟎. 𝟎𝟏𝟕𝟏 𝑽 𝜺 𝟎.𝟎𝟏𝟕𝟏 (b) 𝑰 = 𝑹 = = 𝟎. 𝟎𝟐𝟖𝟓 𝑽 𝟎.𝟔𝟎𝟎 When a conductor is placed in the changing magnetic field, the induced current in the conductor is termed as Eddy current. We can define it as: Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday’s law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field. The magnitude of the eddy currents produces a large current since the resistance of the metallic conductor becomes low. They may produce undesirable effects if the large eddy current is allowed into the core of a choke coil, transformer, etc. By making use of a laminated core, the eddy currents produced in the core of a transformer are reduced. In the present day, electric brakes are provided on the trains in addition to those vacuum brakes. Eddy currents look like eddies or whirlpools.