Question

The figure shows rectangular loop MNOP of wire of size $5\,\text{cm} \times 10\,\text{cm}$ carrying a steady current of $1.0\,\text{A}$. A straight long wire carrying $2.0\,\text{A}$ current is kept near the loop. If the loop and the wire are coplanar, find:

(i) the torque acting on the loop, and

(ii) the magnitude and direction of the net force on the loop.

Hint:

The field of the wire is perpendicular to the loop plane, so the torque is zero. Find the forces on the two long sides using the force between parallel currents and subtract them.

Answer 1

Step 1: Read the figure and the data:
The long straight wire carries 2.0 A upward. The loop MNOP carries 1.0 A, and the arrow on side PM shows the current going upward in PM.
So the current runs P to M, then M to N, then N to O, then O to P.
Side PM is 1 cm from the wire and side NO is 1 + 5 = 6 cm from the wire. Each long side is 10 cm. The loop and the wire lie in one plane.

Step 2: Torque on the loop:
The magnetic field of the straight wire is perpendicular to the plane of the page at every point of the loop.
The magnetic moment \(\vec{m}\) of a plane loop is also perpendicular to the plane of the loop. So \(\vec{m}\) is parallel (or antiparallel) to \(\vec{B}\).
The torque is \(\vec{\tau} = \vec{m} \times \vec{B}\), and the angle between them is \(0^{\circ}\) or \(180^{\circ}\).
\[ \tau = mB\sin\theta = 0 \]
So the torque on the loop is zero.

Step 3: Forces on sides MN and OP:
Sides MN and OP carry equal currents in opposite directions. They are at the same distances from the wire at their two ends.
So the forces on them are equal in size and opposite in direction. They cancel each other.

Step 4: Force on side PM:
Current in PM is parallel to the current in the wire, so the force is attractive (towards the wire).
\[ F_{PM} = \frac{\mu_0 I_1 I_2 l}{2\pi d_1} = \frac{(2\times 10^{-7})(2.0)(1.0)(0.10)}{0.01} \]
\[ F_{PM} = 4.0 \times 10^{-6} \text{ N (towards the wire)} \]

Step 5: Force on side NO:
Current in NO is opposite to the current in the wire, so the force is repulsive (away from the wire).
\[ F_{NO} = \frac{(2\times 10^{-7})(2.0)(1.0)(0.10)}{0.06} = 6.7 \times 10^{-7} \text{ N (away from the wire)} \]

Step 6: Net force:
The two forces act along one line in opposite directions. The nearer side has the larger force.
\[ F_{net} = 4.0\times 10^{-6} - 0.67\times 10^{-6} = 3.3 \times 10^{-6} \text{ N} \]
The net force is directed towards the wire, so the loop is attracted to the wire.

Final Answer:
(i) The torque on the loop is zero. (ii) The net force is about \(3.3\times 10^{-6}\) N, directed towards the straight wire. \[ \boxed{\tau = 0, \quad F_{net} \approx 3.3\times 10^{-6}\ \text{N towards the wire}} \]