Question

A rectangular loop with a sliding connector of length $l=1.0\, m$ is situated in a uniform magnetic field $B=2\, T$ 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 figure. The external force required to keep the connector moving with a constant velocity $v=2\, m / s$ is

• A. 6 N
• B. 4 N
• C. 2 N
• D. 1 N

Answer 1

Answer: (C)

Motional e.m.f., $e=B v l$
$e=(2)(2)(1)=4\, V$
This acts as a cell of e.m.f. $E=4\, V$ and internal resistance $r=2\, \Omega$.
The simple circuit can be drawn as follows:

$\therefore $ Current through the connector $i=\frac{4}{2+2}=1\, A$
Magnetic force on connector, $F_{m}=i l B$
$=(1)(1)(2)$
$=2\, N$ (towards left) Therefore, to keep the connector moving with a constant velocity a force of $2\, N$ will have to be applied towards right.