JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2026
- Hard
- MCQ
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 \(2 \Omega\) then the force needed to move the rod towards right with constant speed (v) of \(1.5 m / s\) is _________ N.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \( 7.5 \times 10^{-3} \)
Step-by-step Solution
Detailed explanation
To maintain constant speed
\(F _{ ext }= F _{ B }\)
\(=\left(\frac{ vB l}{ R }\right) l B\)

\(=\frac{ B ^2 l^2 v }{ R }\)
\(=\frac{(0.1)^2 \times(1)^2 \times 1.5}{2}\)
\(=7.5 \times 10^{-3} N\)
\(F _{ ext }= F _{ B }\)
\(=\left(\frac{ vB l}{ R }\right) l B\)

\(=\frac{ B ^2 l^2 v }{ R }\)
\(=\frac{(0.1)^2 \times(1)^2 \times 1.5}{2}\)
\(=7.5 \times 10^{-3} N\)
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