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Plane figures made of thin wires of resistance $R = 50 mΩ / m$ are located in a uniform magnetic field perpendicular into the plane of the figures and which decreases at the rate dB/dt = 0.1 mT/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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$2 \times 10^{- 4} A$ (Anticlockwise), $10^{- 4} A$ (Anticlockwise)

$10^{- 4} A$ (Clockwise), $2 \times 10^{- 4} A$ (Clockwise)

$10^{- 4} A$ (Anticlockwise), $2 \times 10^{- 4} A$ (Clockwise)

$2 \times 10^{- 4} A$ (Clockwise), $10^{- 4} A$ (Anticlockwise)

answer is A.

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Detailed Solution

Current in the inner coil, $i = \frac{e}{R} = \frac{A_{1}}{R_{1}} \frac{dB}{dt}$

Length of the inner coil $= 2 π a$

So its resistance, $R_{1} = 50 \times 10^{- 3} \times 2 π (a)$

$∴ i_{1} = \frac{{πa}^{2}}{50 \times 10^{- 3} \times 2 π (a)} \times 0 .1 \times 10^{- 3} = 10^{- 4} A$

According to Lenz's law, direction of i₁ is clockwise.

Induced current in outer coil, $i_{2} = \frac{e_{2}}{R_{2}} = \frac{A_{2}}{R_{2}} \frac{d B}{d t}$

$\Rightarrow i_{2} = \frac{{πb}^{2}}{50 \times 10^{- 3} \times (2 πb)} \times 0 .1 \times 10^{- 3}$

$= 2 \times 10^{- 3} = 2 \times 10^{- 4} A (CW)$

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