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

Provided Information: - Straight wire current: \( I_w = 2\, \text{A} \) - Loop current: \( I_l = 1\, \text{A} \) - Loop width: \( 1\, \text{cm} = 0.01\, \text{m} \) - Loop height: \( 5\, \text{cm} = 0.05\, \text{m} \) - Distance from wire to loop's left edge: \( 1\, \text{cm} = 0.01\, \text{m} \) - Distance from wire to loop's right edge: \( 2\, \text{cm} = 0.02\, \text{m} \)\vspace{0.3cm}(i) Torque on the loop:
The net torque on the loop is zero.Justification: - Magnetic forces on the vertical sides of the loop are equal in magnitude and opposite in direction. They act at equal distances from the loop's center, resulting in no net torque. - Forces on the horizontal sides are also opposite but collinear, thus not forming a couple that would produce torque.\[\Rightarrow \tau = 0\]\vspace{0.3cm}(ii) Net force on the loop:
The net magnetic force on the loop due to the straight wire's current is calculated as follows:Magnetic field from a long wire at distance \( r \):\[B = \frac{\mu_0 I}{2\pi r}\]Force on a current segment in a magnetic field:\[F = I_l \cdot L \cdot B\]Force on the left side of the loop (distance = 0.01 m):\[F_1 = I_l \cdot h \cdot \frac{\mu_0 I_w}{2\pi \cdot 0.01}\]Force on the right side of the loop (distance = 0.02 m):\[F_2 = I_l \cdot h \cdot \frac{\mu_0 I_w}{2\pi \cdot 0.02}\]Force directions:- Left vertical side: attractive force (towards the wire).- Right vertical side: repulsive force (away from the wire).Net force calculation:\[F_{\text{net}} = F_1 - F_2 = I_l h \frac{\mu_0 I_w}{2\pi} \left( \frac{1}{0.01} - \frac{1}{0.02} \right)= I_l h \frac{\mu_0 I_w}{2\pi} \cdot \frac{1}{0.02}\]Substituting the given values:\[I_l = 1\, \text{A},\quad h = 0.05\, \text{m},\quad I_w = 2\, \text{A},\quad \mu_0 = 4\pi \times 10^{-7}\, \text{T}\cdot\text{m/A}\]\[F_{\text{net}} = 1 \cdot 0.05 \cdot \frac{4\pi \times 10^{-7} \cdot 2}{2\pi} \cdot \left( \frac{1}{0.01} - \frac{1}{0.02} \right)= 0.05 \cdot (4 \times 10^{-7}) \cdot (100 - 50)= 0.05 \cdot 4 \times 10^{-7} \cdot 50= 1 \times 10^{-6}\, \text{N}\]Direction of net force: The net force is directed towards the wire. This is because the attractive force on the closer vertical side is stronger than the repulsive force on the farther vertical side.\vspace{0.3cm}Final Results:
(i) Torque on the loop: \( 0 \)
(ii) Net force on the loop: \( 1 \times 10^{-6}\, \text{N} \) towards the wire