CBSE 2025, Set 2

Question

A rectangular loop carries a current of $1	ext{ A}$. A straight long wire carrying $2	ext{ A}$ current is kept near the loop in the same plane as shown in the figure.

![Rectangular loop and straight wire](https://www.edvaya.com/images/cbse/physics/2025/set2/q27-rectangular-loop-straight-wire.webp)

Find:
(i) the torque acting on the loop, and
(ii) the magnitude and direction of the net force on the loop.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Torque acting on the loop:**
The torque is given by $\vec{	au} = \vec{m} 	imes \vec{B} = mB \sin	heta$.
The magnetic field $\vec{B}$ due to the long straight wire is perpendicular to the plane of the paper (inwards) at the location of the loop.
The magnetic moment $\vec{m}$ of the loop (with clockwise current) is also perpendicular to the plane of the paper (inwards).
Since $\vec{m}$ and $\vec{B}$ are in the same direction, the angle $	heta = 0^\circ$.
Therefore, $	au = mB \sin(0^\circ) = \boxed{0}$.

**(ii) Net force on the loop:**
The magnetic forces on the top and bottom horizontal segments of the loop are equal and opposite, and their lines of action align, so they cancel out completely.
The net force is the difference between the attractive force on the near vertical wire and the repulsive force on the far vertical wire.

Force formula between two parallel wires of length $l$: $F = \frac{\mu_0 I_1 I_2 l}{2\pi r}$

Force on near wire ($F_1$, attractive towards the straight wire):
$r_1 = 1	ext{ cm} = 10^{-2}	ext{ m}$
$F_1 = \frac{\mu_0 I_1 I_2 l}{2\pi r_1}$

Force on far wire ($F_2$, repulsive away from the straight wire):
$r_2 = 1	ext{ cm} + 1	ext{ cm} = 2	ext{ cm} = 2 	imes 10^{-2}	ext{ m}$
$F_2 = \frac{\mu_0 I_1 I_2 l}{2\pi r_2}$

Net Force $F_{net} = F_1 - F_2$ (towards the straight wire)
$$F_{net} = \frac{\mu_0 I_1 I_2 l}{2\pi} \left( \frac{1}{r_1} - \frac{1}{r_2} ight)$$
$$F_{net} = \frac{4\pi 	imes 10^{-7} 	imes 2 	imes 1 	imes (5 	imes 10^{-2})}{2\pi} \left( \frac{1}{10^{-2}} - \frac{1}{2 	imes 10^{-2}} ight)$$
$$F_{net} = 2 	imes 10^{-7} 	imes 10 	imes 10^{-2} 	imes 10^2 \left( 1 - \frac{1}{2} ight)$$
$$F_{net} = 20 	imes 10^{-7} 	imes \frac{1}{2}$$
$$F_{net} = 10 	imes 10^{-7}	ext{ N} = \boxed{1 	imes 10^{-6}	ext{ N}}$$

**Direction:** Since the attractive force $F_1$ is stronger than the repulsive force $F_2$, the net force on the loop is **towards the long straight wire**.

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Physics > Magnetism Induction And Alternating Current > Moving Charges And Magnetism

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