Question

A rod closing the circuit shown in the figure moves along a $\text{V-}$ shaped wire of a constant speed of $5\text{m/s}$ under the action of force $P$. The circuit is in a uniform magnetic field perpendicular to the plane. Calculate the force $P$ acting on the rod, if the rate of heat generation in the circuit is $15\text{J/s}$.

• A. $1\text{N}$
• B. $2\text{N}$
• C. $3\text{N}$
• D. $5\text{N}$

Answer 1

The correct option is C $3\text{N}$

Let the magnetic field is $B$ and the length of the rod is $l$.

$\therefore$ emf induced across the ends of the rod,

$E=Bvl$

$\therefore$ Current induced in the circuit,

$i=\frac{E}{R}=\frac{Bvl}{R}$

Where $R$ is the resistance of the circuit.

Magnetic force on the conductor,

$F^{\prime }=ilB=\left(\frac{Bvl}{R}\right)lB=\frac{B^2{l}^{2}v}{R}$

This force is equal to the applied force $P$ as there is no acceleration in the rod.

So, $F^{\prime }=P=\frac{B^2{l}^{2}v}{R}...........\left(1\right)$

Now, power induced in the circuit,

$\text{Power}=\frac{E^2}{R}=\left(\frac{B^2{l}^2{v}^2}{R}\right).......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we have:

$\frac{\text{Power}}{P}=v$

$\because \text{Power}=15\text{J/s},v=5\text{m/s}$

$\therefore P=\frac{15}{5}=3\text{N}$

Hence, $\left(\text{B}\right)$ is the correct answer.