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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2 x 10^{– 4} A (Anticlockwise), 10^{– 4}A (Anticlockwise)

10^{– 4} A (Clockwise), 2 x 10^{– 4} A (Clockwise)

10^{– 4} A (Anticlockwise), 2 x 10^{– 4} A (Clockwise)

2 x 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 $=2\pi a$
so its resistance ${R_1} = 50 \times {10^{ - 3}} \times 2\pi \;(a)$
$\therefore {i_1} = \frac{{\pi {a^2}}}{{50 \times {{10}^{ - 3}} \times 2\pi \;(a)}} \times 0.1 \times {10^{ - 3}} = {10^{ - 4}}A$
According to lenz’s law direction of i1 is clockwise.
Induced current in outer coil ${i_2} = \frac{{{e_2}}}{{{R_2}}} = \frac{{{A_2}}}{{{R_2}}}\frac{{dB}}{{dt}}$
$\Rightarrow {i_2} = \frac{{\pi {b^2}}}{{50 \times {{10}^{ - 3}} \times (2\pi b)}} \times 0.1 \times {10^{ - 3}} = 2 \times {10^{ - 4}}A\;(CW)$