Question

A square-shaped wire loop of mass $m$, resistance $R$ and side a moving with speed $v_0$, parallel to the x-axis, enters a region of uniform magnetic field $B$, which is perpendicular to the plane of the loop. The speed of the loop changes with distance $x(x<a)$ in the field, as

• A. $v_0-\frac{B^2{a}^2}{Rm}x$
• B. $v_0-\frac{B^2{a}^2}{2Rm}x$
• C. $v_0+\frac{B^{2}a}{Rm}x$
• D. $v_0$

Answer 1

The correct option is A $v_0-\frac{B^2{a}^2}{Rm}x$

At distance $x$, let speed by $v$

Magnetic flux, $\varphi =BA=Bax$

Using faraday's laws, induced emf is

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

Using ohm's law,

$I=\frac{\epsilon }{R}=\frac{Bav}{R}$

$v=v_o+$

Force acting on the loop is only due to leading edge.

$F=-BIa$

$mv\frac{dv}{dx}=-\frac{B^2{a}^2}{R}v$

${\int }_{v_o}^{v}dv=-{\int }_0^x\frac{B^2{a}^2}{mR}dx$

$v=v_o-\frac{B^2{a}^2}{mR}x$