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 $dB/dt = 0.1\ m\ T/s$. The current in the inner and outer boundary are inner radius $a = 10\ cm$ and outer radius $b = 20\ cm$)
A) $10^{- 4}A$ (Clockwise), $2 \times 10^{- 4}A$ (Clockwise)
B) $10^{- 4}A$ (Anticlockwise), $2 \times 10^{- 4}A$ (Clockwise)
C) $2 \times 10^{- 4}A$ (Clockwise), $10^{- 4}A$ (Anticlockwise)
D) $2 \times 10^{- 4}A$ (Anticlockwise), $10^{- 4}A$ (Anticlockwise)
Explanation
(a) 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 it’s 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 $i_{1}$ is clockwise
Induced current in outer coil $i_{2} = \frac{e_{2}}{R_{2}} = \frac{A_{2}}{R_{2}}\frac{dB}{dt}$
$\Rightarrow i_{2} = \frac{\pi b^{2}}{50 \times 10^{- 3} \times (2\pi b)} \times 0.1 \times 10^{- 3} = 2 \times 10^{- 4}A(CW)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.