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Magnetic Effects & Magnetism

JEE Main level, single correct, medium

200 of 276

Question 200 of 276

A current carrying rectangular loop $PQRS$ is made of uniform wire. The length $PR=QS=5cm\text{and}PQ=RS=100cm$ . If ammeter current reading changes from $I$ to $2I$ , the ratio of magnetic forces per unit length on the wire $PQ$ due to wire $RS$ in the two cases respectively $\left({f}_{PQ}^{I}:{f}_{PQ}^{2I}\right)$ is:

Show answer and solution

Answer: (B) $1:4$

Force acting per unit length between two parallel wires is given by,

f=\dfrac{{\mu }_{0}{I}_{1}{I}_{2}}{2\pi d}

.
Clearly,

f\propto {I}_{1}{I}_{2}

Therefore,

{f}_{PQ}^{I}:{f}_{PQ}^{2I}=\dfrac{I\times I}{\left(2I\right)\times \left(2I\right)}=1:4

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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