Question

A coil of mean area $500c{m}^2$ and having $1000$ turns is held perpendicular to a uniform field of $0.4$ gauss. The coil is turned through $180^{\circ }$ in $\frac{1}{10}s$ the average induced emf will be

• A. 0.08 V
• B. 0.16 V
• C. 1.6 V
• D. 0.04 V

Answer 1

Answer: (D)

From Faraday's law of electromagnetic induction, the induced emf $(e)$ is given by
$e=-\frac{\Delta \varphi }{\Delta t}n$ ... (i)
where $\Delta \varphi$ is rate of change of magnetic flux, given by
$\Delta \varphi =B\cdot A=BA\cos \theta$ ... (ii)
where $B$ is magnetic field intensity and $A$ is area.
Since coil is turned through $180^{\circ }$ change in flux is
$\Delta \varphi =BA-(-BA)=2BA$
$\therefore e=-\frac{2BAn}{\Delta t}$
Given, $B=0.4\times 10^{-4}T$
$A=500c{m}^2$
$=500\times {\left(10^{-2}\right)}^2{m}^2$
$\Delta t=\frac{1}{10}s,n=1000$
$e=\frac{2\times 0.4\times 500\times 10^{-4}\times 1000\times 10^{-4}}{1/10}$
$\Rightarrow e=0.04V$