Learn mutual inductance formula with definition, practice to find the magnitude of average induced EMF in coil and given the magnetic field of double inner and outer solenoid.

- Consider two closely wound coils of wire, as shown in figure.

- The current \(I_1\) in coil 1 having \(N_1\) turns creates a magnetic field.
- Some of the magnetic field lines pass through coil 2 having \(N_2\) turns.
- The magnetic flux caused by the current in coil 1 passing through coil 2 is represented by \(\phi_{21}\).
- Magnetic flux linked with coil 2 is proportional to the current in coil 1.

\(N_2\phi_{21}\propto I_1\)

\(N_2\dfrac{d\phi_{21}}{dt}\propto \dfrac{dI_1}{dt}\)

\(N_2\dfrac{d\phi_{21}}{dt}=M_{21}\dfrac{dI_1}{dt}\)

where \(M_{21}\) is a proportionality constant and is expressed as

\(M_{21}=\dfrac{N_2\,\phi_{21}}{I_1}\)

Mutual inductance (M) can be defined as flux through coil when current in second coil is 1 ampere.

**Note : **Mutual inductance \(M_{21}\) depends only on the geometrical properties of the two coils such as their orientation with respect to each other. Unit of mutual inductance is Henry, \(H\).

- Consider two closely wound coils of wire, as shown in figure.

- The current \(I_1\) in coil 1 having \(N_1\) turns creates a magnetic field.
- Some of the magnetic field lines pass through coil 2 having \(N_2\) turns.
- The magnetic flux caused by the current in coil 1 passing through coil 2 is represented by \(\phi_{21}\).
- Magnetic flux linked with coil 2 is proportional to the current in coil 1.

\(N_2\phi_{21}\propto I_1\)

\(N_2\dfrac{d\phi_{21}}{dt}\propto \dfrac{dI_1}{dt}\)

\(N_2\dfrac{d\phi_{21}}{dt}=M_{21}\dfrac{dI_1}{dt}\)

where \(M_{21}\) is a proportionality constant and is expressed as

\(M_{21}=\dfrac{N_2\,\phi_{21}}{I_1}\)

Mutual inductance (M) can be defined as flux through coil when current in second coil is 1 ampere.

**Note : **Mutual inductance \(M_{21}\) depends only on the geometrical properties of the two coils such as their orientation with respect to each other. Unit of mutual inductance is henry, \(H\).

- Let current \(I_1\) be a function of time, then from Faraday's law, an emf in induced by coil 1 in coil 2.

\(\mathcal{E}_2=-N_2\dfrac{d\phi_{21}}{dt}\)

\(\Rightarrow\;\mathcal{E}_2=-N_2\dfrac{d}{dt}\left(\dfrac{M_{21}\,I_1}{N_2}\right)\)

\(\mathcal{E}_2=-M_{21}\dfrac{dI_1}{dt}\) [\(\because\) for single loop , N_{2 = }1]

- Let us assume current \(I_2\) is flowing in coil 2. Due to variation in current \(I_2\) with time, an emf is induced in coil 1 by coil 2.

\(\mathcal{E}_1=-M_{12}\dfrac{dI_2}{dt}\)

- Since, \(M_{12}=M_{21}=M\) [\(M\) is a constant]

So, \(\mathcal{E}_1=-M\dfrac{dI_2}{dt}\)

and \(\mathcal{E}_2=-M\dfrac{dI_1}{dt}\)

A \(-6\,V\)

B \(-10\,V\)

C \(-5\,V\)

D \(-12\,V\)

- Consider two closely wound coils of wire, as shown in figure.

- The current \(I_1\) in coil 1 having \(N_1\) turns creates a magnetic field.
- Some of the magnetic field lines pass through coil 2 having \(N_2\) turns.
- The magnetic flux caused by the current in coil 1 passing through coil 2 is represented by \(\phi_{12}\).
- Magnetic flux linked with coil 2 is proportional to the current in coil 1.

\(N_2\phi_2\propto I_1\)

\(N_2\dfrac{d\phi_{12}}{dt}\propto \dfrac{dI_1}{dt}\)

\(N_2\dfrac{d\phi_{12}}{dt}=M_{12}\dfrac{dI_1}{dt}\)

where \(M_{12}\) is a proportionality constant and is expressed as

\(M_{12}=\dfrac{N_2\,\phi_{12}}{I_1}\)

Mutual inductance (M) can be defined as flux through coil when current in second coil is 1 ampere.

**Note : **Mutual inductance \(M_{12}\) depends only on the geometrical properties of the two coils such as their orientation with respect to each other. Unit of mutual inductance is henry, \(H\).

A \(6 \pi cos \; 2 \pi t\)

B \(- 12 \pi cos \; 2 \pi t\)

C \(- 3 \pi cos \; 2 \pi t\)

D \(4.8 \pi cos \; 2 \pi t\)

- Consider two single turn co–planar concentric circular coils of radii R
_{1}and R_{2}respectively, such that R_{1 }>>> R_{2}.

- Since the calculation of flux from solenoid \(S_1\) to solenoid \(S_2\) is complicated. So, only calculation of flux from solenoid \(S_2\) to solenoid \(S_1\) is done for calculation of mutual inductance between them.

- The magnetic field in coil 1 due to current \(I_1\)

\(B_1 = \dfrac{\mu _0 I_1}{2 R_1}\)

- Flux through coil 2 (inner) due to B
_{1}

\(\phi_{21} = B_1A_2\)

\(\phi_{21} = \dfrac{\mu_0 I_1}{2 R_1} × \pi R_2^2\)

\(\phi_{21} = \dfrac{\mu_0I_1\; \pi R_2^2}{2 R_1}\)

- So, the mutual inductance is given as

\(M = \dfrac{\phi _{21}}{I_1}\)

\(M = \dfrac{\mu _0 \pi R_2^2}{2 R_1}\)

Note: R_{1} should be very greater than R_{2} otherwise the field becomes variable.

A \(3 \pi ^2 × 10 ^{-7}H\)

B \(2 \pi ^2 × 10 ^{-6} H\)

C \(6 \pi ^2 × 10 ^{-5} H\)

D \(6\pi ^2 × 10 ^{-11}H\)

- Consider two closely wound coils of wire as shown in figure.

- The current \(I_1\) in coil 1 having \(N_1\) turns creates a magnetic field.
- Some of the magnetic field lines pass through coil 2 having \(N_2\) turns.
- The magnetic flux caused by the current in coil 1 passing through coil 2 is represented by \(\phi_{21}\).
- Magnetic flux linked with coil 2 is proportional to the current in coil 1.

\(N_2\phi_{21}\propto I_1\)

\(N_2\dfrac{d\phi_{21}}{dt}\propto \dfrac{dI_1}{dt}\)

\(N_2\dfrac{d\phi_{21}}{dt}=M_{21}\dfrac{dI_1}{dt}\)

where \(M_{21}\) is a proportionality constant and is expressed as

\(M_{21}=\dfrac{N_2\,\phi_{21}}{I_1}\)

Mutual inductance (M) can be defined as flux through coil when current in second coil is 1 ampere.

**Note : **Mutual inductance \(M_{21}\) depends only on the geometrical properties of the two coils such as their orientation with respect to each other. Unit of mutual inductance is henry, \(H\).

- Let current \(I_1\) be a function of time, then from Faraday's law, an emf in induced by coil 1 in coil 2.

\(\mathcal{E}_2=-N_2\dfrac{d\phi_{21}}{dt}\)

\(\Rightarrow\;\mathcal{E}_2=-N_2\dfrac{d}{dt}\left(\dfrac{M_{21}\,I_1}{N_2}\right)\)

\(\mathcal{E}_2=-M_{21}\dfrac{dI_1}{dt}\) [\(\because\) for single loop , N_{2 = }1]

- Let us assume current \(I_2\) is flowing in coil 2. Due to variation in current \(I_2\) with time, an emf is induced in coil 1 by coil 2.

\(\mathcal{E}_1=-M_{12}\dfrac{dI_2}{dt}\)

- Since, \(M_{12}=M_{21}=M\) [\(M\) is a constant]

So, \(\mathcal{E}_1=-M\dfrac{dI_2}{dt}\)

and \(\mathcal{E}_2=-M\dfrac{dI_1}{dt}\)

- Average induced emf in coil 1 due to coil 2

\(\mathcal{E}_1=-M\dfrac{\Delta I_2}{\Delta t}\)

or, \(|\mathcal{E}_1|=M\dfrac{\Delta I_2}{\Delta t}\)

- Similarly, average induced emf in coil 2 due to coil 1

\(\mathcal{E}_2=-M\dfrac{\Delta I_1}{\Delta t}\)

or, \(|\mathcal{E}_2|=M\dfrac{\Delta I_1}{\Delta t}\)

- Consider a solenoid \(S_1\), placed inside another solenoid \(S_2\), as shown in figure, such that the radii of both inner and outer solenoid are \(r_1\) and \(r_2\) respectively and number of turns per unit length are \(n_1\) and \(n_2\) respectively.
- The length of each solenoid is \(\ell.\)

- Suppose a current \(i\) is passed through the inner solenoid \(S_1\).
- A magnetic field \(B=\mu_0n_1i\) is produced inside \(S_1\) whereas the field outside it, is zero.
- The flux through each turn of \(S_2\) is

\(\phi_{12}=B\pi\, r^2_1\)

\(\phi_{12}=\mu _0n_1i\,\pi\, r^2_1\)

- The total flux through each turn through length \(\ell\) of \(S_2\) is

\(\phi_{12}=(\mu_0n_1i\pi\, r^2_1)×n_2\,\ell\)

\(\phi_{12}=(\mu_0n_1n_2\,\pi\, r^2_1\,\ell)i\)

- Thus, mutual inductance

\(M_{12}=\mu_0n_1n_2\pi\, r^2_1\,\ell\)

- The mutual inductance of both the coils is same since the amount of area through which field lines pass, is same.

\(M_{12}=M_{21}\)

A Calculate \(B_1\to\,\phi_{12}=B_1A\to\,M=\dfrac{N_2\phi_{12}}{I}\)

B Calculate \(B_2\to\,\phi_{12}=B_2A\to\,M=\dfrac{N_2\phi_{21}}{I}\)

C Calculate \(B_2\to\,\phi_{21}=B_2A\to\,M=N_2\,\phi_{12}\)

D Calculate \(B_1\to\,M=\dfrac{V_2B_2}{A_2}\)