Question

Figure shows a movable rigid and conducting connector $MN$ between a fixed $L$-shaped conducting wire. The whole system is in horizontal plane having vertical magnetic field $B$ and point $M$ is pulled horizontally with constant speed $v$ starting from $t=0$ by some external agent. Initially, at $t=0$. if $\theta ={90}^o$, induced current in the loop $OMN$ is found to be zero at time $t=\frac{l}{\sqrt{K}V}$. Find $K$ ( Given that $M$ and $N$ are always in contact with the $L$ -shaped wire)

Answer 1

$x^2+y^2=l^2$

Differentiating with respect to ${}^{\prime }{t}^{\prime }$

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

$\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

flux at time $t\varphi =\frac{1}{2}\times x\times y\times B$

emf induced is $e=\frac{d\varphi }{dt}=\frac{1}{2}B[x\frac{dy}{dt}+y\frac{dx}{dt}]$

$e=\frac{B}{2}[-x.\frac{x}{y}+y]v$

$e=\frac{B}{2}r[y-\frac{x^2}{y}]$

for zero induced current

$emf=0$

$\Rightarrow y=\frac{x^2}{y}\Rightarrow [y=x]$

$\Rightarrow \theta ={45}^{\circ }$

$x=vt\Rightarrow t=\frac{x}{v}=\frac{lcos45}{v}$

$t=\frac{l/\sqrt{2}}{v}=\frac{l}{\sqrt{2}v}$

Ans $K=2$