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Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

198 of 276

Question 198 of 276

A straight wire AB of mass $40g$ and length $50cm$ is suspended by a pair of flexible leads in uniform magnetic field of magnitude $0.40T$ 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 $\left.g=10m{s}^{-2}\right)$ .

Show answer and solution

Answer: 2

The formula to calculate the magnetic force on the conductor is given by

{F}_{B}=ILB...\left(1\right)

The formula to calculate the gravitational force on the conductor is given by

{F}_{G}=mg...\left(2\right)

Equate both the equations and simplify to obtain the required current through the conductor.

\begin{matrix}ILB & = & mg \\ & \Rightarrow & I=\dfrac{mg}{LB}...\left(3\right)\end{matrix}

Substitute the values of the known parameters into equation (3) to calculate the required current through the conductor.

\begin{matrix}I & = & \dfrac{\dfrac{40}{1000}kg\times 10m{s}^{-2}}{0.5m\times 0.40T} \\ & = & 2A\end{matrix}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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