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Physics · Electromagnetic Induction

JEE Main 2026 — 21 January, Morning Shift — Question 39

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T . If the resistance of the total circuit is 2Ω2 \Omega then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s1.5 \mathrm{~m} / \mathrm{s} is ____\_\_\_\_ N.

Question figure
  1. Option A:

    7.5×10−27.5 \times 10^{-2}

  2. Option B:

    5.7×10−35.7 \times 10^{-3}

  3. Option C:

    5.7×10−25.7 \times 10^{-2}

  4. Option D:

    7.5×10−37.5 \times 10^{-3}

    Correct

Answer: D

Step-by-step solution

To maintain constant speed

Fext=FB⇒ Fext=ilB=(vBlR)l B=B2l2vR=(0.1)2×(1)2×1.52=7.5×10−3 N\begin{aligned} & \mathrm{F}_{\mathrm{ext}}=\mathrm{F}_{\mathrm{B}} & \Rightarrow \mathrm{~F}_{\mathrm{ext}}=\mathrm{ilB} & =\left(\frac{\mathrm{vBl}}{\mathrm{R}}\right) l \mathrm{~B} & =\frac{\mathrm{B}^{2} l^{2} \mathrm{v}}{\mathrm{R}} & =\frac{(0.1)^{2} \times(1)^{2} \times 1.5}{2} & =7.5 \times 10^{-3} \mathrm{~N} \end{aligned}
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Motional EMF

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