Question

A rectangular loop with a sliding connector of length $l=1.0\text{m}$ is situated in a uniform magnetic field $B=2\text{T}$ 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 $v=2\text{m/s}$ is

• A. $6\text{N}$
• B. $4\text{N}$
• C. $2\text{N}$
• D. $1\text{N}$

Answer 1

The correct option is C $2\text{N}$

Motional emf $\epsilon =Bvl$
$\epsilon =\left(2\right)\left(2\right)\left(1\right)=4\text{V}$
This acts as a cell of emf $E=4\text{V}$ and internal resistance $r=2\Omega$. The simple circuit can be drawn as follows:

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