Question

Explain mutual inductance in coil due solenoid

Answer 1

Mutual inductance describes the phenomenon where a changing current in one coil induces an EMF in a nearby coil. It depends on the geometry of the coils and the permeability of the medium.

Answer 2

Mutual inductance is the phenomenon where a change in current in one coil (the primary coil) induces an electromotive force (EMF) in a nearby coil (the secondary coil). This occurs because the magnetic field produced by the primary coil links with the secondary coil, and if the current in the primary coil changes, the magnetic flux through the secondary coil also changes, inducing an EMF according to Faraday's Law of electromagnetic induction.

Derivation for Two Long Coaxial Solenoids

Consider two long, coaxial solenoids, S1 (primary coil) and S2 (secondary coil), of the same length $l$.

Let:

• $N_1$ be the number of turns in solenoid S1.
• $N_2$ be the number of turns in solenoid S2.
• $r_1$ be the radius of solenoid S1.
• $r_2$ be the radius of solenoid S2. (Assume $r_1$ is the radius of the inner solenoid, or the radius that defines the effective area for flux linkage, typically the smaller radius if one is inside the other).

1. Magnetic Field due to the Primary Solenoid (S1):

If a current $I_1$ flows through the primary solenoid S1, the magnetic field $B_1$ produced inside it (assuming it's a long solenoid) is given by:

$B_1 = \mu_0 n_1 I_1 = \mu_0 \frac{N_1}{l} I_1$

where $\mu_0$ is the permeability of free space and $n_1 = N_1/l$ is the number of turns per unit length of S1.

2. Magnetic Flux through the Secondary Solenoid (S2):

The magnetic field $B_1$ produced by S1 passes through the cross-sectional area of S1 ($A_1 = \pi r_1^2$). Since the solenoids are coaxial and long, this field is uniform inside S1.

The flux linking each turn of the secondary solenoid S2 is $B_1 A_1$.

Since there are $N_2$ turns in S2, the total magnetic flux $\Phi_2$ through S2 is:

$\Phi_2 = N_2 (B_1 A_1)$

Substitute the expression for $B_1$ and $A_1$:

$\Phi_2 = N_2 \left( \mu_0 \frac{N_1}{l} I_1 \right) (\pi r_1^2)$

$\Phi_2 = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} I_1$

3. Definition of Mutual Inductance (M):

By definition, the magnetic flux linked with the secondary coil is directly proportional to the current in the primary coil:

$\Phi_2 = M I_1$

where $M$ is the mutual inductance of the two coils.

4. Expression for Mutual Inductance (M):

Comparing the two expressions for $\Phi_2$:

$M I_1 = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} I_1$

Therefore, the mutual inductance $M$ of the two coaxial solenoids is:

$M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l}$

This formula shows that mutual inductance depends on the geometric configuration of the coils (number of turns, length, radius) and the magnetic properties of the medium ($\mu_0$).

Induced EMF

According to Faraday's Law, the induced EMF ($E_2$) in the secondary coil is the negative rate of change of magnetic flux through it:

$E_2 = -\frac{d\Phi_2}{dt}$

Substituting $\Phi_2 = M I_1$:

$E_2 = -\frac{d(M I_1)}{dt}$

Since $M$ is a constant for a given geometry:

$E_2 = -M \frac{dI_1}{dt}$

This equation quantifies the induced EMF in the secondary coil due to the changing current in the primary coil.