JEE Physics 2024 Question with Solution

Given: A rigid wire has a semicircular portion of radius $R$ and two straight sections. It carries current $i$ and is placed in a perpendicular magnetic field $\vec{B}$.

Find: The net magnetic force on the wire.

For a current-carrying wire in a uniform magnetic field,

$$\vec{F} = i\,\vec{L} \times \vec{B}$$

where $\vec{L}$ is the effective length vector from one end of the segment to the other.

For the semicircular part, the effective length equals the diameter, so its magnitude is

$$L = 2R$$

The two straight sections are parallel to the magnetic field, so the force on each straight segment is zero.

$$\vec{F}_{\text{straight}} = 0$$

Hence the total force is due to the semicircular part alone. Substituting the effective length $2R$,

$$\vec{F} = i\,(2R)\,B\,$$

Using the right-hand rule for the given current direction and magnetic field, the force is along $-\hat{j}$. Therefore,

$$\vec{F} = -2iBR\,\hat{j}$$

Therefore, the correct option is D.

Given: Only the portion of wire inside the field contributes to magnetic force.

Find: Net force on the wire.

A wire segment in uniform magnetic field can be treated using the end-to-end vector of that segment:

$$\vec{F} = i\,\vec{l} \times \vec{B}$$

For a semicircular arc, the end-to-end distance is the diameter,

$$l = 2R$$

The straight sections are parallel to $\vec{B}$, so they contribute no force. Thus directly,

$$\vec{F} = i\,(2R)\,B$$

With direction from the right-hand rule, the force is $-\hat{j}$. Hence,

$$\vec{F} = -2iBR\,\hat{j}$$

So the correct option is D.