A solenoidal coil with 20 turns of wire is wound tightly around another coil wit

Question

A solenoidal coil with 20 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 24.0 cm long and has a diameter of 2.40 cm . At a certain time, the current in the inner solenoid is 0.130 A and is increasing at a rate of 1800 A/s .

Part A

For this time, calculate the average magnetic flux through each turn of the inner solenoid.

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Part B

For this time, calculate the mutual inductance of the two solenoids;

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Part C

For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

A solenoidal coil with 20 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 24.0 cm long and has a diameter of 2.40 cm . At a certain time, the current in the inner solenoid is 0.130 A and is increasing at a rate of 1800 A/s .

Part A

For this time, calculate the average magnetic flux through each turn of the inner solenoid.

B = Wb

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Part B

For this time, calculate the mutual inductance of the two solenoids;

M = H

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Part C

For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

E2 = V

Explanation / Answer

here,

no of turns, n1 = 20
n2 = 340

length of inner solenoid, Li = 24 cm = 0.24 m
Diameter of inner solenoid, Di = 2.40 cm = 0.0240 m
Current in inner solenoid, Ii = 0.130 A
rate of increasing, di/dt = 1800 A/s

magnatic field due to inner solenoid :

B = u0*N2*Ii/Li
B = (4 * pi * 10^-7 * 340 * 0.130) / 0.24
B = 2.314*10^-4 T

PART A:
the average magnetic flux will be :
Fi = B*area of inner
Fi = 2.3110^-4 * 3.14 * (0.0240/2)^2
Fi = 1.585 * 10^-5 Wb

PART B:
Mutual inductane,
M = uo * n1 *n2 * pi *r^2 *L
M = 4*pi*10^-7 * 20 * 340 * 3.14 * (0.0240/2)^2 * 0.24
M = 9.27 * 10^-7 H

PART C:
emf = M * di/dt
emf = 9.27 * 10^-7 * 1800
emf = 1.67 * 10^-3 V

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