Please show detail and explain 6. A current loop with sides 5.0 long carries 0.4

Question

Please show detail and explain

6. A current loop with sides 5.0 long carries 0.45A current as shown below. The loop is in a 12T uniform external magnetic field. The axis of loop, perpendicular to the plane of the loop, is 30.0° away from the field direction. 30, 13 Bar ext (a) Find the magnitude of the torque on the current loop. (b) What is the total magnetic flux through this square 15-turn current loop? (c) What is the final orientation of the axis of loop relative to the direction of the magnetic field if the loop can rotate freely? (d) If the loop spends 0.25s to reach the final orientation, what is the induced emf during this rotation?

Explanation / Answer

N=15

l=b=5cm

I=0.45 A

B=1.2T

theta=30 degrees

a)

Torque,T=N*I*A*B*sin(theta)

=15*0.45*(5*5)*10^-4*1.2*sin(30)

=10.12*10^-3 N.m

b)

magnetic flux=N*B*A*cos(theta)

=15*1.2*(5*5*10^-4)*cos(30)

=38.97*10^-3 Wb

c)

the direction of axis of the loop becomes parallel to the dierction of the magnetic field

d)

emf=flux/time

=38.97*10^-3/0.25

=155.9*10^-3 V

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