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Q.

Plane figures made of thin wires of resistance R = 50 milli ohm/metre are located in a uniform magnetic field perpendicular into the plane of the figures and which decrease at the rate dB/dt = 0.1 m T/s. Then currents in the inner and outer boundary are. (The inner radius a = 10 cm and outer radius b = 20 cm)

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a

10– 4 A (Clockwise), 2×10– 4 A (Clockwise)

b

10– 4 A (Anticlockwise), 2×10– 4 A (Clockwise)

c

2×10– 4 A (clockwise), 10– 4 A (Anticlockwise)

d

2×10– 4 A (Anticlockwise), 10– 4 A (Anticlockwise)

answer is A.

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Detailed Solution

Current in the inner coil i=eR=A1R1dBdtlength of the inner coil =2πaso it’s resistance R1=50×10−3×2π (a)∴i1=πa250×10−3×2π (a)×0.1×10−3=10−4AAccording to lenz’s law direction of i1 is clockwise. Induced current in outer coil i2=e2R2=A2R2dBdt⇒i2=πb250×10−3×(2πb)×0.1×10−3=2×10−4A (CW)

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Plane figures made of thin wires of resistance R = 50 milli ohm/metre are located in a uniform magnetic field perpendicular into the plane of the figures and which decrease at the rate dB/dt = 0.1 m T/s. Then currents in the inner and outer boundary are. (The inner radius a = 10 cm and outer radius b = 20 cm)