Question
A wire $88 \ cm$ long is bent into a circular loop and kept with its plane perpendicular to a magnetic field of induction $2.5 Wb / m ^2$. Within $0.5$ second, the coil is changed to a square and the magnetic induction is increased by $0.5 Wb / m ^2$. Calculate the emf induced in the wire.

Answer

Data: $I=88 \ cm , B_i=2.5 Wb / m ^2,$
$ B _f=3 Wb / m ^2, \Delta t =0.5 s$
For the circular loop, $1=2 \pi r$
$\therefore r=\frac{l}{2 \pi}=\frac{88}{2 \times(22 / 7)}=14 \ cm =0.14 m$
Area of the circular loop, $A_i=\pi r^2$
$=\frac{22}{7}(0.14)^2=0.0616 m ^2$
Initial magnetic flux $\Phi_i=A_i B_i$
$=0.0616 \times 2.5=0.154 Wb$
For the square loop, length of each side
$=\frac{88}{4} \ cm =22 \ cm =0.22 m 4$
Area of the square loop, $A_f=(0.22)^2$
$=0.0484 m ^2$
$\therefore \text { Final magnetic flux } \Phi_f=A_f B_f$
$=0.0484 \times 3=0.1452 Wb$
$\text { Induced emf, e }=-\frac{\Phi_r-\Phi_1}{\Delta t}$
$=\frac{\Phi_1-\Phi_r}{\Delta t}$
$\therefore e =\frac{0.154-0.1452}{0.5}$
$=8.8 \times 10^{-3} \times 2$
$=1.76 \times 10^{-2} V$

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