Question

# 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

A
104A(Clockwise),2×104A(Clockwise)
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B
104A(Anticlockwise),2×104A(Clockwise)
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C
2×104A(clockwise),104A(Anticlockwise)
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D
2×104A(Anticlockwise),104A(Anticlockwise)
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Solution

## The correct option is A 10–4A(Clockwise),2×10–4A(Clockwise) Current in the inner coil i=eR=A1R1dBdt length of the inner coil =2πα so it's resistance R1=50×10−3×2π(α) ∴ i1=πα250×10−3×2π(α)×0.1×10−3=10−4A According 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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