Question

In the given figure, a movable wire PQ starts moving on the fixed rail at t = 0 with constant speed of $\frac{3}{2}m/s$. Then work done by external force on wire PQ in 2 s is

Answer 1

${ϵ}_{pq}=BVL=2\times \frac{3}{2}\times 2=6V$
$i_1=\frac{ϵ}{R}=\frac{6}{2}=3A$
$ϵ=L\frac{d{i}_Z}{dt}\Rightarrow 6=Z\frac{d{i}_Z}{dt}\Rightarrow i_2=3t$
$F=F_B=ilB=3(1+t)\times 2\times 2=12(1+t)$
$W=\int fdx=12\int (1+t)dx\Rightarrow v=\frac{dx}{dt}\Rightarrow dx=\frac{3}{2}dt$
$W=12{\int }_0^2(1+t)\frac{3}{2}dt\Rightarrow W=72J$