JEE Physics 2024 Question with Solution

Given: Side of square loop = $15 \, \text{cm}$, speed = $2 \, \text{cm/s}$, magnetic-field width = $50 \, \text{cm}$, and the front edge enters at $t = 0$.

Find: Induced emf at $t = 10 \, \text{s}$.

Using Faraday's law, induced emf exists only when magnetic flux through the loop changes.

The distance moved by the loop in $10 \, \text{s}$ is

Since the loop side is $15 \, \text{cm}$, after moving $20 \, \text{cm}$ the entire loop is completely inside the $50 \, \text{cm}$ wide magnetic field region. Therefore, the area of the loop inside the field is constant, so magnetic flux does not change.

Hence,

Therefore, the induced emf is zero, so the correct option is C.

Given: The loop enters the field at $t = 0$ and moves uniformly.

Find: Whether flux is changing at $t = 10 \, \text{s}$.

Magnetic flux through the loop is

where $A$ is the portion of loop area inside the field.

While the loop is entering or leaving the magnetic field, $A$ changes with time, so emf is induced. But when the whole loop is inside the field, $A$ remains constant.

At $t = 10 \, \text{s}$, the front edge has moved $20 \, \text{cm}$, which is more than the loop side $15 \, \text{cm}$. So the rear edge has also entered the field, and the full loop lies inside the field region of width $50 \, \text{cm}$.

Thus,

and the induced emf is $0 \, \text{V}$. Therefore, the correct option is C.