Question

A coil of area $100{\text{cm}}^2$ has $500$ turns. Magnetc field of $0.1{\text{weber/m}}^2$ is perpendicular to the plane of coil. The magnetic field is reduced to zero in $0.1\text{s}$. Then induced e.m.f. in the coil is

• A. $1\text{V}$
• B. $5\text{V}$
• C. $50\text{V}$
• D. $0\text{V}$

Answer 1

The correct option is B $5\text{V}$

Magnetic flux linked with coil is ( taking area $\left(S\right)$ in same direction as $\left(B\right)$ hence $\cos \theta =\cos 0^{\circ }=1$),
$\varphi =BS\cos \theta =BS$
Rate of change of flux is:
$\frac{\Delta \varphi }{\Delta t}=S\left(\frac{\Delta B}{\Delta t}\right)$
$S=10\times {10}^{-4}{\text{m}}^2$
$\Rightarrow \frac{\Delta \varphi }{\Delta t}=100\times (10{)}^{-4}\times \left(\frac{0-0.1}{0.1-0}\right)$
$\therefore \frac{\Delta \varphi }{\Delta t}=-{10}^{-2}\text{weber/s}...\left(i\right)$
From Faraday's law:
$\epsilon =-N\left(\frac{\Delta \varphi }{\Delta t}\right)...\left(ii\right)$
$\epsilon =-500\times (-{10}^{-2})$
$\therefore \epsilon =5\text{V}$