A conducting loop (as shown) has total resistance $R$. A uniform magnetic field $B=\gamma t$ is applied perpendicular to plane of the loop where $\gamma$ is a constant and $t$ is time. The induced current flowing through loop is

• A

  $\frac{\left(b^2+a^2\right)\gamma t}{R}$
• Correct answer
  B

  $\frac{\left(b^2-a^2\right)\gamma }{R}$
• C

  $\frac{\left(b^2-a^2\right)\gamma t}{R}$
• D

  $\frac{\left(b^2+a^2\right)\gamma }{R}$

Solution

Given,

Uniform magnetic field $\left(B\right)=\gamma t$



Total flux, $\overrightarrow{B_1}. \overrightarrow{A_1}+\overrightarrow{B_2}.\overrightarrow{A_2}$

$=B{b}^2\cos 0^o+B{a}^2\cos {180}^o$

$=B{b}^2-B{a}^2$

$\varphi =B(b^2-a^2)$

$\varphi =\gamma t(b^2-a^2)$

We know that,

Induced current (i) $=\frac{\left|e\right|}{R}=\frac{\left|\frac{d\varphi }{dt}\right|}{R}$

From Equation (i),

$\left(i\right)=\frac{\frac{d}{dt}\left[\gamma t\left(b^2-a^2\right)\right]}{R}$

$\left(i\right)=\frac{\left(b^2-a^2\right)\gamma }{R}$

Hence, the correct answer is option: 'B'.