JEE Advanced 2025 — Paper 1 — PHY-II solution

JEE Advanced2025Paper 1PHY-II

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, $B = B_{0} \hat{k}$ . The magnetic field is zero everywhere else. At time $t = 0$ , the loop starts to enter the magnetic field with an initial velocity $v_{0} \hat{j}$ m/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:

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

Solution

While the loop is partially inside the field, only its leading edge cuts field lines. The induced EMF is $ε = B_{0} Lv$ , producing current $i = B_{0} Lv / R$ and a retarding force $F = - B_{0}^{2} L^{2} v / R = - K M v$ .

Equation of motion: $\frac{d v}{d t} = - K v \Rightarrow v (t) = v_{0} e^{- K t}$ .

Integrating: $x (t) = \frac{v _{0}}{K} (1 - e^{- K t})$ , valid until $x = L$ .

Option (A): The maximum distance the loop can ever travel inside is $x_{\infty} = v_{0} / K$ . For $v_{0} = 1.5 K L$ , $x_{\infty} = 1.5 L > L$ , so the loop fully enters. Statement (A) is incorrect.

Option (B): Once the loop is entirely inside the field, the flux is constant, no EMF is induced, no current flows, and the net force vanishes. Statement (B) is correct.

Option (C): The motion is exponential decay, so $v$ asymptotically approaches zero but never reaches it in finite time. Statement (C) is incorrect.

Option (D): Setting $x = L$ with $v_{0} = 3 K L$ :

$L = \frac{3 K L}{K} (1 - e^{- K t}) \Rightarrow 1 - e^{- K t} = \frac{1}{3} \Rightarrow e^{- K t} = \frac{2}{3}$

$t = \frac{1}{K} ln (\frac{3}{2})$ . Statement (D) is correct.

$∴$ Correct options are (B) and (D).

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