Question

A square loop of side a lying a perpendicular magnetic field to its plane is changed to a circle. If change occusrs in t seconds in magnetic field B tesla, the induced emf is

• A. $\frac{4}{\pi }\frac{B{a}^2}{t}$
• B. $\frac{B{a}^2}{t}$
• C. $\frac{B{a}^2}{t}[\frac{4}{\pi }-1]$
• D. Zero

Answer 1

The correct option is C $\frac{B{a}^2}{t}[\frac{4}{\pi }-1]$

Given,

Square of sides dimension $a$ change to circle of radius $r$.

Change in Area, $\Delta A=a^2-\pi r^2$.....(I)

for the value of $r$

Area of square loop = Area of circle

$4a=2\pi r$

$r=\frac{4a}{2\pi }$

$r=\frac{2a}{\pi }$

Now, put the value of $r$ in equation (I)

$\Delta A=a^2-\pi \left(\frac{4a^2}{{\pi }^2}\right)$

$\Delta A=a^2(1-\frac{4}{\pi })$

Induced e m f

$\epsilon =\frac{-d\varphi }{dt}$

$\epsilon =\frac{-B\Delta A}{\Delta t}$

$\epsilon =\frac{-B\times \left(a^2(1-\frac{4}{\pi })\right)}{t}$

$\epsilon =\frac{-B{a}^2}{t}(1-\frac{4}{\pi })$

$\epsilon =\frac{B{a}^2}{t}(\frac{4}{\pi }-1)$

The induced emf is $\frac{B{a}^2}{t}(\frac{4}{\pi }-1)$