Question

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 690 turns and solenoid 2 has 450 turns. When the current in solenoid 1 is 6.60 A, the average flux through each turn of solenoid 2 is 3.50×10-2 Wb.
A: What is the mutual inductance of the pair of solenoids?

B: When the current in solenoid 2 is 2.60 A, what is the average flux through each turn of solenoid 1?

Answer 1

The concept that we need here to give a proper solution is mutual inductance.

The mutual inductance  is given by the expression

$M=\frac{N\Phi}{I}$

Where,

I = current

N = Number of turns

$\Phi =$Flux through the solenoid.

Part A) Then we have in our values that,

$I=6.6A$

$\Phi= 3.50*10^{-2}Wb$

$N=450$

Replacing in the equation,

$M = \frac{450*350*10^{-2}}{6.60}$

$M = 2.39H$

Part B) Here is required the Flux, then using the same expression we have that

$\Phi = \frac{IM'}{N}$

We conserve the same value for the Inductance but now we have a current of 2.6, then

$\Phi = \frac{2.6*2.39}{690}$

$\Phi = 9*10^{-3}Wb$

Therefore the flux in Solenoid 1 is $9*10^{-3}Wb$