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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2026 · 21 January · Morning Shift · Physics · Electromagnetic Induction

A 1 m long metal rod AB completes the circuit as shown in figure....

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)of1.5 m/sof 1.5 \mathrm{~m} / \mathrm{s} is ____\_\_\_\_ N.
A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic

Options

  1. A5.7×10−25.7 \times 10^{-2}
  2. B7.5×10−27.5 \times 10^{-2}
  3. C7.5×10−37.5 \times 10^{-3}
  4. D5.7×10−35.7 \times 10^{-3}

Correct answer

C. 7.5×10−37.5 \times 10^{-3}

Step-by-step solution

Given: l=1l = 1 m, B=0.10B = 0.10 T, R=2R = 2 Ω, v=1.5v = 1.5 m/s
Motional EMF: ε=Blv=0.10×1×1.5=0.15\varepsilon = Blv = 0.10 \times 1 \times 1.5 = 0.15 V
Current: I=εR=0.152=0.075I = \frac{\varepsilon}{R} = \frac{0.15}{2} = 0.075 A
Magnetic force on rod: F=BIl=0.10×0.075×1=7.5×10−3F = BIl = 0.10 \times 0.075 \times 1 = 7.5 \times 10^{-3} N
For constant velocity, applied force equals magnetic force.
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