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. ${10}^{–4}A\left(Clockwise\right),2\times {10}^{–4}A\left(Clockwise\right)$
  
• B. ${10}^{–4}A\left(Anticlockwise\right),2\times {10}^{–4}A\left(Clockwise\right)$
  
• C. $2\times {10}^{–4}A\left(clockwise\right),{10}^{–4}A\left(Anticlockwise\right)$
  
• D. $2\times {10}^{–4}A\left(Anticlockwise\right),{10}^{–4}A\left(Anticlockwise\right)$

Answer 1

The correct option is A ${10}^{–4}A\left(Clockwise\right),2\times {10}^{–4}A\left(Clockwise\right)$


Current in the inner coil $i=\frac{e}{R}=\frac{A_1}{R_1}\frac{dB}{dt}$
length of the inner coil $=2\pi \alpha$
so it's resistance $R_1=50\times {10}^{-3}\times 2\pi \left(\alpha \right)$
$\therefore i_1=\frac{\pi {\alpha }^2}{50\times {10}^{-3}\times 2\pi \left(\alpha \right)}\times 0.1\times {10}^{-3}={10}^{-4}A$
According to lenz's law direction of $i_1$ is clockwise.
Induced current in outer coil $i_2=\frac{e_2}{R_2}=\frac{A_2}{R_2}\frac{dB}{dt}$
$\Rightarrow i_2=\frac{\pi b^2}{50\times {10}^{-3}\times \left(2\pi b\right)}\times 0.1\times {10}^{-3}=2\times {10}^{-4}A\left(CW\right)$