Question

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

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

Hint:

Always compare magnetic field strength at both sides of the loop. Closer side experiences a stronger field, so net force is toward the wire, but torque cancels due to symmetry.

Answer 1

Given: - Current in straight wire: \( I_w = 2\, \text{A} \) - Current in loop: \( I_l = 1\, \text{A} \) - Width of loop: \( 1\, \text{cm} = 0.01\, \text{m} \) - Height of loop: \( 5\, \text{cm} = 0.05\, \text{m} \) - Distance from wire to left side of loop: \( 1\, \text{cm} = 0.01\, \text{m} \) - Distance to right side of loop: \( 2\, \text{cm} = 0.02\, \text{m} \) \vspace{0.3cm} (i) Torque acting on the loop:
There is no net torque acting on the loop. Reason: - Each vertical side of the loop experiences a magnetic force due to the magnetic field of the straight wire. - The forces on the two vertical sides (left and right) are equal in magnitude and opposite in direction, but they act at equal distances from the center, and hence produce no net torque (they cancel each other out). - The horizontal sides (top and bottom) experience forces in opposite directions, but being collinear, they do not form a couple either. \[ \Rightarrow \tau = 0 \] \vspace{0.3cm} (ii) Net force on the loop:
Let’s calculate the net magnetic force on the loop due to the current in the straight wire. Magnetic field due to long wire at a distance \( r \): \[ B = \frac{\mu_0 I}{2\pi r} \] Force on a current-carrying wire in magnetic field: \[ F = I_l \cdot L \cdot B \] Calculate force on left side (distance = 1 cm = 0.01 m): \[ F_1 = I_l \cdot h \cdot \frac{\mu_0 I_w}{2\pi \cdot 0.01} \] Calculate force on right side (distance = 2 cm = 0.02 m): \[ F_2 = I_l \cdot h \cdot \frac{\mu_0 I_w}{2\pi \cdot 0.02} \] Direction: - On left vertical side: force is attractive (toward the wire). - On right vertical side: force is repulsive (away from the wire). Net force: \[ F_{\text{net}} = F_1 - F_2 = I_l h \frac{\mu_0 I_w}{2\pi} \left( \frac{1}{0.01} - \frac{1}{0.02} \right) = I_l h \frac{\mu_0 I_w}{2\pi} \cdot \frac{1}{0.02} \] Substitute values: \[ I_l = 1\, \text{A},\quad h = 0.05\, \text{m},\quad I_w = 2\, \text{A},\quad \mu_0 = 4\pi \times 10^{-7} \] \[ F_{\text{net}} = 1 \cdot 0.05 \cdot \frac{4\pi \times 10^{-7} \cdot 2}{2\pi} \cdot \left( \frac{1}{0.01} - \frac{1}{0.02} \right) = 0.05 \cdot (4 \times 10^{-7}) \cdot (100 - 50) = 0.05 \cdot 4 \times 10^{-7} \cdot 50 = 1 \times 10^{-6}\, \text{N} \] Direction: The net force is toward the wire, because the attractive force on the closer side is greater than the repulsive force on the farther side. \vspace{0.3cm} Final Answer:
(i) Torque on the loop: \( 0 \)
(ii) Net force: \( 1 \times 10^{-6}\, \text{N} \) toward the wire