Question

If the radius of a coil is changing at the rate $10^{-2}$ units in a normal magnetic field $10^{-3}$ units, the induced emf is $1 \,\mu V$. What is the final radius of the coil?

• A. 1.6cm
• B. 16cm
• C. 12cm
• D. 1.2cm

Answer 1

Answer: (A)

Induced emf is given by
$|e|=\frac{d \phi}{d t}=B \frac{d A}{d t} $
$(\because \phi=B A)$
$=\pi B \frac{d}{d t}\left(r^{2}\right)=\pi B .2 r \frac{d r}{d t} $
Given, $\frac{d r}{d t}=10^{-2}$ units,
$B=10^{-3} $ units, $ e=1 \mu V$
$\therefore 1 \times 10^{-6}=3.14 \times 10^{-3} \times 2 \times r \times 10^{-2} $
or $r=\frac{10^{-6}}{3.14 \times 10^{-5} \times 2}$
$=0.016\, m=1.6\, cm$