Question

A squre loop of side l, mass m and resistance R falls vertically into a uniform magnetic field directed perpendicular to the plane of the coil. One of the sides of the coil is always outside the magnetic field. The height h through which the loop falls so that it attains terminal velocity on entering the region of magnetic field is given by

• A. $\frac{mgR}{2Bl}$
• B. $\frac{m^{2}g{R}^2}{2B^2{l}^2}$
• C. $\frac{mg{R}^2}{4B^3{l}^3}$
• D. $\frac{m^{2}g{R}^2}{2B^4{l}^4}$

Answer 1

The correct option is D

$\frac{m^{2}g{R}^2}{2B^4{l}^4}$

Velocity $v=\sqrt{2gh}$. Induced emf $e=Blv=Bl\sqrt{2gh}.$ Therefore, the induced currect in the loop is
$I=\frac{Bl\sqrt{2gh}}{R}\therefore ForceF=BIl=\frac{B^2{l}^2\sqrt{2gh}}{R}$
The loop will attain terminal velocity if this force equals mg, i.e. if
$\frac{B^2{l}^2\sqrt{2gh}}{R}=mgwhichgivesh=\frac{m^{2}g{R}^2}{2B^4{l}^4}$