Question

A circular loop of radius 18 cm carries a current of 10 A. A flat coil of radius 0.62 cm, having 39 turns and a current of 1.6 A, is concentric with the loop. The plane of the loop is perpendicular to the plane of the coil. Assume the loop's magnetic field is uniform across the coil. What is the magnitude of (a) the magnetic field produced by the loop at its center and (b) the torque on the coil due to the loop?

Answer 1

Solution)

Given,

Radius, r=0.18 m

Current, I=10 A

Part a)

We know, Magnetic Field, B=uo*I/2R

Substitute values,

B=(4pi*10^-7)*(10)/(2*0.18)= 3.48*10^-5 T (Ans)

====

Part b)

Torque, T= BI*pi*r^2*N

Substitute values,

T=(3.48*10^-5)*(1.6)*pi*(0.0062)^2*39

T= 2.62*10^-7 Nm (Ans)

==========