JEE Advanced 2025 · Paper 1 · Physics Q5

← JEE Advanced 2025 · Paper 1

PhysicsMultiple correct+4 / -2Question 21 of 48

A conducting square loop of side $L$ , mass $M$ and resistance $R$ is moving in the $XY$ plane with its edges parallel to the $X$ and $Y$ axes. The region $y \ge 0$ has a uniform magnetic field, $\vec B = B_0\hat k$ . 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 $K = \dfrac{B_0^2L^2}{RM}$ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct:

3D axes: Z up, Y to the right, X towards the lower left. The region y ≥ 0 of the XY plane is shaded, with an upward arrow B along Z. A square loop of side L lies in the XY plane at y < 0 with its leading edge on the X-axis (y = 0), moving with velocity v₀ along +Y into the field region.

v_0 = 1.5KL

, the loop will stop before it enters completely inside the region of magnetic field.

When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.

v_0 = \dfrac{KL}{10}

, the loop comes to rest at

t = \left(\dfrac1K\right)\ln\left(\dfrac52\right)

v_0 = 3KL

, the complete loop enters inside the region of magnetic field at time

t = \left(\dfrac1K\right)\ln\left(\dfrac32\right)

Answer and worked solution

Official answer: B, D

Equation of motion while entering. The emf is $B_0Lv$ , so the current is $\frac{B_0Lv}R$ and the retarding force is $\frac{B_0^2L^2v}R$ :
$M\frac{dv}{dt}=-\frac{B_0^2L^2}{R}v \implies v=v_0e^{-Kt},$
and, using $v\frac{dv}{dx}=-Kv$ , $v=v_0-Kx$ .

Options.

(A) False. With $v_0=1.5KL$ , after moving $L$ the speed is $v_0-KL=0.5KL>0$ , so the loop enters completely.
(B) True. Fully inside a uniform field there is no flux change, no current and no force.
(C) False. With $v_0=\frac{KL}{10}<KL$ , $v=v_0e^{-Kt}$ only tends to zero; it never stops in finite time.
(D) True. With $v_0=3KL$ , $x=\frac{v_0}{K}(1-e^{-Kt})=L$ gives $e^{-Kt}=\frac23$ , so $t=\frac1K\ln\frac32$ .

Topics:Motional EMF Newton's laws

← Previous question Next question →

JEE Advanced 2025 · Paper 1, Physics question 5