Question

A square loop of length L and resistance R is pulled out of a magnetic field B which is perpendicular to the loop slowly and uniformly in t (sec). One edge of the loop was initially placed at the boundary of the magnetic field. Find the work done in pulling the loop out :

• A. $\frac{B^2{L}^4}{Rt}$
• B. $\frac{B^2{L}^3}{Rt}$
• C. $\frac{2B^2{L}^4}{Rt}$
• D. $\frac{B^2{L}^4}{2Rt}$

Answer 1

The correct option is A $\frac{B^2{L}^4}{Rt}$

We know that work done is given by
$W=q\Delta V..\left(1\right)$
Flux ${\varphi }_1=B{L}^2$
${\varphi }_2=0$
Change in flux $\Delta \varphi =B{L}^2$
Magnitude of induced emf $\in =\frac{\Delta \varphi }{\Delta t}$
$\in =\frac{B{L}^2}{t}....\left(2\right)$
$\therefore V=iR$ (From ohm's law)
$V=\frac{q}{t}R$
$q=\frac{Vt}{R}.....\left(3\right)$
Here $V=\in .....\left(4\right)$
From equations (1), (2) (3) & (4)
$W=\left(\frac{B{L}^2}{t}\right)\frac{t}{R}\times \frac{B{L}^2}{t}$
$W=\frac{B^2{L}^4}{Rt}$