JEE Physics 2023 Question with Solution

Given: rod length $\ell = 1 \, \text{m}$, magnetic field $B = 0.15 \, \text{T}$, resistance $R = 5 \, \Omega$, speed $v = 4 \, \text{m/s}$.

Find: the force required to move the rod at constant speed.

From motional emf,

$$e = B\ell v$$

The current in the circuit is

$$I = \frac{e}{R} = \frac{B\ell v}{R}$$

The magnetic force on the rod is

$$F = B I \ell$$

Substituting for $I$,

$$F = B\ell \left(\frac{B\ell v}{R}\right) = \frac{B^2 \ell^2 v}{R}$$

Now substitute the given values,

$$F = \frac{(0.15)^2 \cdot (1)^2 \cdot 4}{5}$$ $$F = \frac{0.0225 \cdot 4}{5} = \frac{0.09}{5} = 0.018 \, \text{N}$$

Therefore, the force needed is $18 \times 10^{-3} \, \text{N}$.

The provided working shows $B = 15 \, \text{T}$ and concludes $18 \, \text{N}$, but the question states the magnetic flux density is $0.15 \, \text{T}$. Using the question data and the same formula,

$$F = \frac{B^2 \ell^2 v}{R}$$

we get

$$F = \frac{(0.15)^2 \cdot 1^2 \cdot 4}{5} = 0.018 \, \text{N}$$

So the correct numerical answer is $18 \times 10^{-3}$, which matches the answer field. The $18 \, \text{N}$ result comes from an incorrect substitution of the magnetic field value in the provided the solution.