Question

A magnetic field $\overrightarrow{B}=B_o\frac{y}{a}\hat{k}$ is out of the x-y plane, were $B_o$ and $a$ are positive constants. A square loop PQRS of side $a$, mass $m$ and resistance $R$ in x-y plane starts falling under gravity. The terminal velocity of the loop then must be :

(Given acceleration due to gravity $=g$)

• A. $\frac{B_o^2{a}^2}{mgR}$
• B. $\frac{mg{a}^2}{B_o^{2}R}$
• C. $\frac{mgR}{B_o^2{a}^2}$
• D. $\frac{B_o{a}^2}{mgR}$

Answer 1

Answer: (C)

$\frac{mgR}{B_o^2{a}^2}$

Terminal velocity where force are equal $\frac{dy}{dt}=-V$

$F_G=mg$

$\varphi =\frac{B_o}{a}{\int }_y^{a+y}y\left(ady\right)=\frac{B_o}{2}(2ay+a^2)=\frac{B_{o}a}{2}(2y+a)$

$r=-\frac{d\varphi }{dt}=\frac{-B_{o}a}{2}\left(2\frac{dy}{dt}\right)=a{B}_{o}V$

$i=\frac{a{B}_{o}V}{R}$

Forces on wire$=ia\left(\frac{B_o}{a}\right(y+a)-\frac{B_{o}y}{a})$

$=i{B}_{o}a=a^2\frac{B_o^{2}V}{R}$

$\therefore F_G=F_M$

$mg=\frac{a^2{B}_o^{2}V}{R}$

$V=\frac{mgR}{a^2{B}_o^2}$