MHTCET 20 Apr 2026 Shift 1 Paper

Concept:A long straight wire produces a magnetic field $B = \frac{\mu_0 I_2}{2\pi r}$, and the force on a current-carrying side of the loop is $F = I_1 L B$. The net force is the difference between the forces on the two sides parallel to the wire.Explanation:The magnetic field due to wire $XY$ at a distance $r$ is $B = \frac{\mu_0 I_2}{2\pi r}$, directed into the plane of the loop.Side $AB$ is at distance $r = \frac{L}{3}$ from the wire.Current in $AB$ is in the same direction as $I_2$, so the force on $AB$ is attractive and acts towards the wire.$F_{AB} = I_1 L \cdot \frac{\mu_0 I_2}{2\pi \left(\frac{L}{3}\right)} = \frac{3\mu_0 I_1 I_2}{2\pi}$Side $CD$ is at distance $r = \frac{L}{3} + L = \frac{4L}{3}$ from the wire.Current in $CD$ is opposite to $I_2$, so the force on $CD$ is repulsive and acts away from the wire.$F_{CD} = I_1 L \cdot \frac{\mu_0 I_2}{2\pi \left(\frac{4L}{3}\right)} = \frac{3\mu_0 I_1 I_2}{8\pi}$Forces on sides $BC$ and $DA$ are equal and opposite, hence they cancel each other.Since $F_{AB}$ and $F_{CD}$ act in opposite directions, the net force is$F_{net} = F_{AB} - F_{CD} = \frac{3\mu_0 I_1 I_2}{2\pi} - \frac{3\mu_0 I_1 I_2}{8\pi}$$F_{net} = \frac{12\mu_0 I_1 I_2 - 3\mu_0 I_1 I_2}{8\pi} = \frac{9\mu_0 I_1 I_2}{8\pi}$Answer:$\frac{9\mu_0 I_1 I_2}{8\pi}$