Question

1) A circular loop of radius 3.53 cm contains 61 turns of tightly wound wire. If the current in the windings is 0.634 A and a constant magnetic field of 0.462 T makes an angle of 70.3 ◦ with a vector perpendicular with the loop, what torque acts on the loop?

2) The clockwise circulating current in a solenoid is increasing at a rate of 13 A/s. The crosssectional area of the solenoid is 3.14159 cm2 , and there are 124 turns on its 23.6 cm length. What is the magnitude of the self-induced emf E produced by the increasing current?

Answer 1

1.

Torque is given by:

T = N*i*A*B*sin theta

Using given values:

N = 61 turns

i = 0.634 Amp

B = 0.462 T

A = pi*r^2 = pi*0.0353^2 = 3.91*10^-3 m^2

sin 70.3 deg = 0.941

So,

T = 61*0.634*0.462*3.91*10^-3*0.941

T = 0.0657 N-m

2.

Magnetic field inside solenoid is given by:

B = u0*N*i/L

Now EMF is given by:

E = -N*d(phi)/dt

phi = B.A

E = -N*A*dB/dt

E = -N*A*d(u0*N*i/L)/dt

E = (-N^2*A*u0/L)*(di/dt)

Magnitude of EMF will be

|E| = (N^2*A*u0/L)*(di/dt)

Using given values:

|E| = (124^2*3.14159*10^-4*4*pi*10^-7/0.236)*(13)

|E| = 3.34*10^-4 V