Question

The strength of the magnetic field within a solenoid is B = 2.1 × 10-2 T...

The strength of the magnetic field within a solenoid is B = 2.1 × 10-2 T (outside the solenoid B = 0). A smaller, single loop is placed in the solenoid parallel to the plane of each loop in the solenoid. The resistance of the solenoid is 5.9 W, the resistance of the loop is 0.27 W, the diameter of the solenoid is 0.09 m, and the diameter of the loop is 0.05 m. An emf of 12 V is placed across the ends of the solenoid and the length of the solenoid is 0.6 m.(a) What is the total number of turns of the solenoid?

N = turns     

(b) What is the magnitude of the magnetic flux through the loop inside the solenoid (remember, it only has one turn and the normal to the loop is parallel to the axis of the solenoid.)

flux = T-m2     

(c) If the emf across the solenoid is varied so that the current in the solenoid now decreases to zero at a constant rate over a time interval of 6 seconds, what is the current through the loop during this time? (If the direction of the induced current in the loop is in the same direction as the current in the solenoid, give your answer as a positive value; if the direction of the induced current in the loop is in the opposite direction compared to the current in the solenoid, give your answer as a negative value.)

I = A

Homework Answers

Answer #1

given

B = 2.1*10^-2 T
R = 5.9 ohms
R' = 0.27 ohms
d = 0.09 m
d' = 0.05 m

V = 12 V

L = 0.6 m

a) current through the solenoid, I = V/R

= 12/5.9

= 2.03 A

we know, B = mue*N*I/L

==> N = B*L/(mue*I)

= 2.1*10^-2*0.6/(4*pi*10^-7*2.03)

= 4939 turns

b) magnetic flux through the loop inside = B*A'

= B*pi*d'^2/4

= 2.1*10^-2*pi*0.05^2/4

= 4.12*10^-5 T.m^2

c) induced emf in the loop = change in megnctic flux/time

= (4.12*10^-5 - 0)/6

= 6.87*10^-6 V

induced current in the loop = induced emf/R'

= 6.87*10^-6/0.27

= 2.54*10^-5 A or 25.4 micro A

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