QJEEM24Phd7c7: JEE Main 2024 Physics - Class 12 JEEM24Phd7c7 Solution: A rectangular loop of length and width is placed at to a…

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:

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

Where $\varepsilon$ 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:

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

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

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

$\varepsilon = -(-1 \mathrm{~Wb/s})$

$\varepsilon = +1 \mathrm{~V}$

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

Option C

$+1 \mathrm{~V}$