Question

A rectangular loop of sides 15 and 10 cm carrying a current of 1a, is placed with its longer side parallel to long straight wire carrying a current of 2a placed at a distance of 2 cm. The net force experienced by the loop is

Answer 1

Hey buddy,

◆ Answer-
F = 4.167×10^-6

◆ Explaination-
# Given-
l = 25 cm = 0.25 m
b = 10 cm = 0.1 m
r1 = 2 cm = 0.02 m
r2 = 0.1 + 0.02 = 0.12 m
I1 = 1 A
I2 = 2 A

# Solution-
The force will act only on sides parallel to wire length.

Repulsive force is -
F1 = (μ0.I1.I2.l) / (2π.r1)
F1 = (4π×10^-7 × 1 × 2 × 0.25) / 2π(0.02)
F1 = 50×10^-7 N

Attractive force is -
F2 = (μ0.I1.I2.l) / (2π.r2)
F2 = (4π×10^-7 × 1 × 2 × 0.25) / 2π(0.12)
F2 = 8.33×10^-7 N

Net force acting on rectangular loop is -
F = F1 - F2
F = 50×10^-7 - 8.33×10^-7
F = 4.167×10^-6 N

Therefore, resultant force acting on the loop is 4.167×10^-6 N.

Hope it helps...