Question

Question 4.7: Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-169

Answer 1

force of the attraction per unit length on two parallel wires carrying current in same direction is given by $F/l=\frac{\mu_0}{4\pi}\frac{2I_1I_2}{r}$
here, I1 and I2 are the current passing through given two parallel wires.
r is the distance between two parallel wires.

here, r = 4cm = 0.04 m I1 = 8A , I2 = 5A
now, F/l = 10^-7 × 2 × 8 × 5/(4 × 10^-2) N/m
= 20 × 10^-5 N/m

hence, attractive force on 10cm section of wire A
F =(F/l ) × l = 20 × 10^-5 × 10 × 10^-2 = 2 × 10^-5 N

Answer 2

Answer:

Here your answer...

Current in wire A, I1=8 A

Current in wire B, I2=5 A

Force exerted on 10cm section of wire A,

B=μo2I1I2/4πr=2×10^−5T

Attractive force normal to A towards B, because current direction same.