Question

A small piece of metal wire is dragged across the gap between the poles of a magnet is $0.4\, s$. If the change in magnetic flux in the wire is $ 8\times {{10}^{-4}} \, Wb$, then emf induced in the wire is:

• A. $ 8\times {{10}^{-3}}\,V $
• B. $ 6\times {{10}^{-3}}\,V $
• C. $ 4\times {{10}^{-3}}\,V $
• D. $ 2\times {{10}^{-3}}\,V $

Answer 1

Answer: (D)

From Faraday's law of electromagnetic induction, when the magnetic flux through a circuit is changing an induced emf $(e)$ is set up in the circuit whose magnitude is equal to the negative rate of change of magnetic flux $(\phi)$
$e=-\frac{d \phi}{d t}$
Given, $d \phi=8 \times 10^{-4}\, Wb ,$
$ d t=0.4\, s$
$\Rightarrow e =-\frac{8 \times 10^{-4}}{0.4} $
$=-2 \times 10^{-3}\, V$
Note : The direction of induced emf is such as to oppose the change that produced it.