JEE Physics 2023 Question with Solution

Given: wire length $l = 1 \, \text{m}$, velocity $v = 8 \, \text{m/s}$, magnetic field $B = 2 \, \text{T}$. The wire moves at right angles to the magnetic field.

Find: the induced emf between the ends of the wire.

For a straight conductor moving perpendicular to a magnetic field, the induced emf is

$$\varepsilon = Bvl$$

Substituting the given values,

$$\varepsilon = 2 \times 8 \times 1 = 16 \, \text{V}$$

Therefore, the induced emf is $16 \, \text{V}$. The correct option is D.

The solution also shows a conflicting label "B", but the worked value is $16 \, \text{V}$, which matches option D.

Given: $B = 2 \, \text{T}$, $v = 8 \, \text{m/s}$, $l = 1 \, \text{m}$.

Find: induced emf $\varepsilon$.

The induced electromotive force in a wire moving perpendicularly through a magnetic field is given by

$$\varepsilon = Bvl$$

Here the motion is perpendicular to the field, so the full formula applies directly. Hence,

$$\varepsilon = 2 \, \text{T} \times 8 \, \text{m/s} \times 1 \, \text{m} = 16 \, \text{V}$$

Therefore, the magnitude of induced emf is $16 \, \text{V}$.