Question

A rectangular loop with a sliding connector of length $l=1.0m$ is situated in a uniform magnetic field $B=2T$ perpendicular to the plane of loop. Resistance of connector is $r=2\Omega$.

Two resistances of $6\Omega$ and $3\Omega$ are connected as shown in the figure. Find the external force (in newton) required to keep the connector moving with a constant velocity $v=2m/s$.

Answer 1

Answer: 2

$R_{\text{eff }}=\frac{3\times 6}{3+6}+2=4\Omega$
$\therefore I=\frac{\epsilon }{R_{cff}}=\frac{Blv}{R_{\text{eff }}}=\frac{2\times 1\times 2}{4}=1A$
$\therefore F=IlB=1\times 1\times 2=2N$