Question

A coil has an area of $0.05m^2$ & it has$800$ turns. It is placed perpendicularly in a magnetic field of strength, $4\times {10}^{-5}\raisebox{1ex}{$wb$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$. It is rotated through ${90}^0$ in $0.1s$. The average $emf$ induced in the coil is

• A. $0.056V$
• B. $0.046V$
• C. $0.026V$
• D. $0.016V$

Answer 1

The correct option is D

$0.016V$

Step1: Given data.

Area $A=0.05m^2$

Number of turns $N=800$

Magnetic field strength $B=4\times {10}^{-5}\raisebox{1ex}{$wb$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$

Time taken to rotate through an angle of ${90}^0$ $t=0.1s$

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

Formula used:

$e=\frac{NBA}{t}$

Where $e$ is the $emf$ induced in the coil.

Now,

$\Rightarrow e=\frac{800\times 4\times {10}^{-5}\times 0.05}{0.1}$

$\Rightarrow e=0.016V$

The average $\mathit{e}\mathit{m}\mathit{f}$ induced in the coil is $\mathbf{0}\mathbf{.}\mathbf{016}\mathit{V}$.

Hence, option D is the correct answer.