JEE Main 2024 — Electromagnetic Induction Question with Solution

According to Faraday's Law of Electromagnetic Induction, the induced emf in a closed circuit is equal to the negative rate of change of magnetic flux through the circuit. Mathematically, it is expressed as:

$$\epsilon =-\frac{d\Phi }{dt}$$

Where $\epsilon$ is the induced emf, and $\Phi$ is the magnetic flux.

To find the magnetic flux $\Phi$ through the rectangular loop, we use the formula:

$$\Phi =B\cdot A\cdot \cos (\theta )$$

Where:

• $B$ is the magnetic field strength,
• $A$ is the area of the loop, and
• $\theta$ is the angle between the magnetic field lines and the normal (perpendicular) to the plane of the loop.

The area $A$ of the rectangular loop is:

$$A=\text{length}\times \text{width}=2.5\mathrm{m}\times 2\mathrm{m}=5{\mathrm{m}}^2$$

Given the angle $\theta =60^{\circ }$, we can calculate the initial magnetic flux ${\Phi }_{initial}$:

$${\Phi }_{initial}=B\cdot A\cdot \cos (60^{\circ })$$

$${\Phi }_{initial}=4\mathrm{T}\cdot 5{\mathrm{m}}^2\cdot \cos (60^{\circ })$$

$${\Phi }_{initial}=4\cdot 5\cdot \frac{1}{2}=10\mathrm{Wb}$$

(Since $\cos (60^{\circ })=\frac{1}{2}$)

When the loop is removed from the magnetic field, the final magnetic flux ${\Phi }_{final}$ is zero, because the loop is no longer within the magnetic field. Thus, the change in magnetic flux $\Delta \Phi$ is:

$$\Delta \Phi ={\Phi }_{final}-{\Phi }_{initial}=0-10\mathrm{Wb}=-10\mathrm{Wb}$$

The loop is removed from the field in $t=10$ seconds, so the rate of change of magnetic flux is:

$$\frac{d\Phi }{dt}=\frac{\Delta \Phi }{\Delta t}=\frac{-10\mathrm{Wb}}{10\mathrm{s}}=-1Wb/s$$

Now we can find the average induced emf $\epsilon$:

$$\epsilon =-\frac{d\Phi }{dt}$$

$$\epsilon =-(-1Wb/s)$$

$$\epsilon =+1\mathrm{V}$$

Therefore, the average induced emf in the loop during this time is $+1\mathrm{V}$. The correct answer is:

Option C

$$+1\mathrm{V}$$