Question

A coil of mean area $500 \, cm^{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
$e=-\frac{\Delta \phi}{\Delta t} n$ ... (i)
where $\Delta \phi$ is rate of change of magnetic flux, given by
$\Delta \phi=B \cdot A=B A \cos \theta$ ... (ii)
where $B$ is magnetic field intensity and $A$ is area.
Since coil is turned through $180^{\circ}$ change in flux is
$ \Delta \phi=B A-(-B A)=2 B A $
$\therefore e=-\frac{2 B A n}{\Delta t} $
Given, $ B =0.4 \times 10^{-4} T $
$ A =500 \,cm ^{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= 0.04\, V $