JEE Physics 2025 Question with Solution

Given: A conductor frame ABCDE has each arm $10 \, \text{cm}$. It moves to the right with speed $v = 10 \, \text{cm s}^{-1} = 0.1 \, \text{m s}^{-1}$ in a uniform magnetic field $B = \dfrac{1}{\sqrt{2}} \, \text{T}$ perpendicular to the plane.

Find: The induced emf between A and E.

For a rigid conductor moving with uniform velocity $\vec v$ in a uniform magnetic field $\vec B$ perpendicular to the plane, the motional electric field is

$$\vec E_m = \vec v \times \vec B$$

Hence the emf between two points depends only on their displacement:

$$\varepsilon_{AE} = \int_A^E (\vec v \times \vec B) \cdot d\vec \ell = (\vec v \times \vec B) \cdot \vec{AE} = Bv \, \Delta y$$

where $\Delta y$ is the vertical separation between A and E.

Step 1: Find the vertical separation from the geometry.

Arms AB and DE are horizontal, so they give no vertical contribution. The two slanted arms are at $45^\circ$, so each contributes a vertical drop of

$$10 \cos 45^\circ = \frac{10}{\sqrt{2}} \, \text{cm}$$

Therefore,

$$\Delta y = \frac{10}{\sqrt{2}} + \frac{10}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, \text{cm} = 0.1\sqrt{2} \, \text{m}$$

Step 2: Use $\varepsilon = Bv\Delta y$.

$$\varepsilon = \left(\frac{1}{\sqrt{2}}\right)(0.1)(0.1\sqrt{2}) = 0.01 \, \text{V}$$

Thus,

$$0.01 \, \text{V} = 10 \, \text{mV}$$

Therefore, the induced emf between A and E is $10 \, \text{mV}$.

Given: The conductor moves horizontally in a magnetic field perpendicular to the plane.

Find: The emf between A and E.

Since $\vec v$ is horizontal and $\vec B$ is perpendicular to the plane, $\vec v \times \vec B$ is vertical. Therefore only the vertical separation between A and E matters.

Directly,

$$\varepsilon = Bv \times \left[2(10 \sin 45^\circ \text{ cm})\right]$$

Converting to SI,

$$\varepsilon = \frac{1}{\sqrt{2}} \times 0.1 \times 0.1\sqrt{2} = 0.01 \, \text{V}$$

Therefore, the correct numerical answer is 10.