JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2023
- Medium
- Numerical
A straight wire \(AB\) of mass \(40\,g\) and length \(50\,cm\) is suspended by a pair of flexible leads in uniform magnetic field of magnitude \(0.40\,T\) as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is ...........\(A\). (Take \(g=10\,ms ^{-2}\) ).

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
For equilibrium
\(Mg = I \ell B\)
\(I =\frac{ mg }{\ell B }=\frac{40 \times 10^{-3} \times 10}{50 \times 10^{-2} \times 0.4}=2\,A\)
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