Physics · Electromagnetic Induction

JEE Advanced 2025 — Paper 1 — Question 3

A conducting square loop of side $L$ , mass $M$ and resistance $R$ is moving in the $X Y$ plane with its edges parallel to the $X$ and $Y$ axes. The region $y \geq 0$ has a uniform magnetic

field, $\vec{B}=B_{0} k$ . The magnetic field is zero everywhere else. At time $t=0$ , the loop starts to enter the magnetic field with an initial velocity $v_{0} \hat{\jmath} \mathrm{~m} / \mathrm{s}$ , as shown in the figure.

Considering the quantity $K=\frac{B_{0}^{2} L^{2}}{R M}$ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct

Option A:
If $v_{0}=1.5 K L$ , the loop will stop before it enters completely inside the region of magnetic field
Option B:
When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.
Correct
Option C:
If $v_{0}=\frac{K L}{10}$ , the loop comes to rest at $t=\left(\frac{1}{K}\right) \ln \left(\frac{5}{2}\right)$ .
Option D:
If $v_{0}=3 K L$ , the complete loop enters inside the region of magnetic field at time $t=\left(\frac{1}{K}\right) \ln \left(\frac{3}{2}\right)$ .
Correct

Answer: B, D

Step-by-step solution

$\Rightarrow \frac{-\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{B}_{0} \times \ell \times \mathrm{y}\right)=\mathrm{BV} \ell$ $\overrightarrow{\mathrm{F}}=\mathrm{B}(\hat{\mathrm{i}})(\ell)(-\hat{\mathrm{j}})$

$\mathrm{ma}=-\mathrm{B}_{0}\left[\frac{\mathrm{~B}_{0} \mathrm{~V} \ell}{\mathrm{R}}\right](\ell)$

$\mathrm{a}=-\frac{\mathrm{B}_{0}^{2} \ell^{2} \mathrm{~V}}{\mathrm{mR}}$

Also $\mathrm{K}=\frac{\mathrm{B}_{0}^{2} \ell^{2} \mathrm{~V}}{\mathrm{RM}}$

So $[\mathrm{a}=-\mathrm{kv}]$

$\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{kv}$

$\int_{v_{0}}^{v} \frac{d v}{d t}=\int_{0}^{t}-k d t$

$\ell \mathrm{n} \frac{\mathrm{V}}{\mathrm{V}_{0}}=-\mathrm{kt}$

$\left[\mathrm{v}=\mathrm{v}_{0} \mathrm{e}^{-\mathrm{kt}}\right]$

$\frac{d x}{d t}=v_{0} e^{-k t} \quad(x \leq \ell)$

$\int_{0}^{x} d x=\int_{0}^{t} v_{0} e^{-k t} d t$

$=\frac{\mathrm{V}_{0}}{\mathrm{k}}\left(1-\mathrm{e}^{-\mathrm{kt}}\right)$

When $\mathrm{x}=\ell$

$\ell=\frac{\mathrm{v}_{0}}{\mathrm{k}}\left(1-\mathrm{e}^{-\mathrm{kt}}\right)$

Option (D) $\left(\mathrm{v}_{0}=3 \mathrm{k} \ell\right)$

$\ell=\frac{3 \mathrm{k} \ell}{\mathrm{k}}\left(1-\mathrm{e}^{-\mathrm{kt}}\right)$

$\frac{1}{3}=1-\mathrm{e}^{-\mathrm{kt}}$

$f \frac{2}{3}=2 e^{-k t}$

$-\mathrm{kt}=8 \mathrm{n}\left(\frac{2}{3}\right)$

$\mathrm{t}=\frac{1}{\mathrm{k}} \ln \left(\frac{2}{3}\right)$

Complete loop will enter at $\mathrm{t}=\frac{1}{\mathrm{k}} \ln \left(\frac{2}{3}\right)$

Option (B) $\frac{\mathrm{d} \phi}{\mathrm{dt}}=0, \underline{\underline{\mathrm{e}}}=0, \mathrm{i}=0, \mathrm{~F}=0$

Ans. B,D)

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 1
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Motional EMF