Question

Solve all parts of Question 1 clearly:

1. Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 800 turns, and solenoid 2 has 500 turns. When the current in solenoid 1 is 6.52 A, the average flux through each turn of solenoid 2 is 0.0320 Wb.

1. What is the mutual inductance of the pair of solenoids?
2. When the current in solenoid 2 is 6.52 A, what is the average flux through each turn of solenoid 1?
   Question 1: a) Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 800 turns, and solenoid 2 has 500 turns. When the current in solenoid 1 is 6.52 A, the average flux through each turn of solenoid 2 is 0.0320 Wb. i) What is the mutual inductance of the pair of solenoids? (8 points) ii) When the current in solenoid 2 is 6.52 A, what is the average flux through each turn of solenoid 1? (8 points) b) A 100 cm long solenoid of diameter 0.400 cm is wound around uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils? (9 points)
   

2. A 100 cm long solenoid of diameter 0.400 cm is wound around uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

Answer 1

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# N=800 , N₂ = 500, I, = 6.52A R = 0.032 wb a) Mutual Inductance M = N₂ Be = 500 * 0.032 6.82 M= 2.454 H b) M=N,OB, I, w M= Iz oo * PB, PB 7 2.8154 = 800 6.52 | DB = 0·02 Wb

the mutual inductance M = μ0 N1*N2*A/L

the No. of turns in first solenoid N1 = 800

the No. of turns in second solenid N2 = 50.0

the Length of solenoid L = 100 cm = 1 m
the Area of cross section A = π* r2
= 3.14* ( 0.400 * 10(-2) / 2)2
= 1.256* 10(-5) m2
M = 4π* 10(-7) * 800 * 50.0 * 1.256 * 10(-5 )/1
= 6.31* 10^(-7) H