Question

A coil having an area $2m^2$ is placed in a magnetic field which changes from $1\raisebox{1ex}{$wb$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$ to $4\raisebox{1ex}{$wb$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$ in an interval of $2sec$. The $e.m.f$ induced in the coil will be

• A. $4V$
• B. $3V$
• C. $1.5V$
• D. $2V$

Answer 1

The correct option is B

$3V$

Step1: Given data.

Area of the coil, $A=2m^2$

Change in magnetic field strength at time interval $2sec$ is $1\raisebox{1ex}{$wb$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$ to $4\raisebox{1ex}{$wb$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$.

Step2: Finding $\mathit{e}\mathbf{.}\mathit{m}\mathbf{.}\mathit{f}$ in the coil.

We know that

Magnetic flux, $\varphi$$=A\times B$ $....\left(i\right)$

And $e=\frac{d\varphi }{dt}$ $....\left(ii\right)$

Where, $e$ is the $e.m.f$ in the coil.

Now,

Change in magnetic field with respect to time

$\Rightarrow \frac{dB}{dt}=\frac{4-1}{2}$

$\Rightarrow \frac{dB}{dt}=\frac{3}{2}\raisebox{1ex}{$wb$}\!\left/ \!\raisebox{-1ex}{$m^{2}s$}\right.$ $....\left(iii\right)$

Again,

Differentiate equation $....\left(i\right)$ with respect to $t$ both side we get.

$\Rightarrow$ $\frac{d\varphi }{dt}=A\frac{dB}{dt}$

$\Rightarrow$$e=2m^2\times \frac{3}{2}\raisebox{1ex}{$wb$}\!\left/ \!\raisebox{-1ex}{$m^{2}s$}\right.$ $\left[Fromequation\left(ii\right)and\left(iii\right)\right]$

$\Rightarrow$$e=3V$

Hence, option B is correct.