Question

A small thin coil with N2 loops, each of area A2, is placed inside a long solenoid, near its center. The solenoid has N1 loops in its length l and has area A1. Determine the mutual inductance as a function of q, the angle between the plane of the small coil and the axis of the solenoid. Express your answer in terms of some or all of the variables A1, A2, N1, N2, l, q, and appropriate constants.

Answer 1

Answer:

the mutual inductance as a function of q is   $M = \frac{\mu_o N_1 N_2 A_2 sin(q)}{L}$

Explanation:

The Magnetic field  that is being produced by the inner coil is mathematically given as

               $B_i = \mu_o (\frac{N_2}{L} )L$

The magnetic flux is mathematically represented as

             $\o _B = BA_2 sin(q)$

             $\o_B = \mu_o \frac{N_2}{L} L *A_2 sin (q)$

Where q is given in the question as the angle between the plane of the small coil and the axis of the solenoid

       The mutual inductance is mathematically represented as

                 $M =\frac{N_1 \o_B}{L}$

Substituting the for $\o_B$ in the equation

                  $M = \frac{N_1 *\mu_0 * \frac{N_2}{L} L A_2 sin(q) }{L}$

                       $M = \frac{\mu_o N_1 N_2 A_2 sin(q)}{L}$

Where $\mu_o$ is permeability of free space

Answer 2

The mutual inductance as a function of q, the angle between the plane of the small coil and the axis of the solenoid is $\frac{\mu_0 N_1N_2 A_2 \times sin(q)}{L}$.

The given parameters;

• Number of thin coil turns, = N₂
• Area of the coil,= A₂
• Number of turns of solenoid, = N₁
• Length of the loop, = l
• Area of the loop, = A₁

The magnetic field produced in the inner coil is calculated as follows;

$B = \mu_0 (\frac{N_2}{L} )L$

The magnetic flux is calculated as follows;

$\phi = BA_2 \times sin(\theta)\\\\\phi = \mu_0 (\frac{N_2}{L} )L A_2 \times sin(q)$

The mutual inductance is calculated as follows;

$M = \frac{N_1 \phi }{L} \\\\M = \frac{N_1 \times \mu_0 (\frac{N_2}{L} )LA_2 \times sin(q)}{L} \\\\M = \frac{\mu_0 N_1N_2 A_2 \times sin(q)}{L}$

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