Question

A rectangular loop of size 25cm and 10cm carrying a current of 15 A is placed with its longer side parallel to a long straight conductor 2cm apart carrying a current of 25 A .What is the net force on the loop?​

Answer 1

Answer:

L= 25 cm= 0.25m

B=10 cm= 0.10 m

R1= 2 cm=0.02 m

R2=12 cm=0.12 m

Explanation:

Repulsive force= F1= μ0I(wire)I(loop)L/2πr

F1 = 4π×10−7×25×15×25/2π×2

F1 = 9.37×10−4 N away from the wire

F2 = 4π×10−7×25×15×25/2π×12

F2 = 1.56×10−4N towards the wire

Net force F = F1−F2 = 9.37×10−4 N – 1.56×10−4N

=7.81 ×10−4 N away from the wire.