Informative line

Mutual Inductance

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.

Mutual Inductance

  • 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, \(H\).

Illustration Questions

Which of the following orientation will have more mutual inductance?

A

B

C

D

×

As distance is increased, number of field lines passing through coil 2 decreases. Thus, flux is decreased.

So, mutual inductance \(M_{12}\) decreases.

Hence, option (A) is correct.

Which of the following orientation will have more mutual inductance?

A image
B image
C image
D image

Option A is Correct

Induced EMF Due to Mutual Inductance in Coil

  • 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 , N2 = 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 \(\mathcal{E}_2=-M\dfrac{dI_1}{dt}\)

 

Illustration Questions

Two closely wounded coils having mutual inductance \(M=2\,H\) and the current in second coil is a function of time \(I=3t+5\). Find the induced emf in coil 1.

A \(-6\,V\)

B \(-10\,V\)

C \(-5\,V\)

D \(-12\,V\)

×

Induced emf \((\mathcal{E}_1)\) in coil 1 due to coil 2 is given as

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

where \(M=\) Mutual Inductance

\(I_2=\) Current in coil 2

Given : \(M=2H,\;I=3t+5\)

\(\mathcal{E}_1=-2×\dfrac{d}{dt}(3t+5)\)

\(\mathcal{E}_1=-2×3\)

\(\mathcal{E}_1=-6\,V\)

Two closely wounded coils having mutual inductance \(M=2\,H\) and the current in second coil is a function of time \(I=3t+5\). Find the induced emf in coil 1.

A

\(-6\,V\)

.

B

\(-10\,V\)

C

\(-5\,V\)

D

\(-12\,V\)

Option A is Correct

Average Induced EMF

  • 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 , N2 = 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, \(|\mathcal{E}_2|=M\dfrac{\Delta I_1}{\Delta t}\)

 

Illustration Questions

Two closely wound coils each having mutual inductance \(M=2\,H\). The current in second coil changes from \(I_i=5\,A\) to \(I_f=10\,A\) in \(\Delta t=2\,sec\). Find the magnitude of average induced emf in coil 1.

A \(3\,V\)

B \(8\,V\)

C \(7\,V\)

D \(5\,V\)

×

Average induced emf in coil 1 due to coil 2 is given as

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

where \(I_i=\) Initial current

\(I_f=\) final current

\(\Delta t=\) time

\(M=\) Mutual inductance

Given : \(I_i=5\,A,\;I_f=10\,A,\;\Delta t=2\,sec,\,M=2\,H\)

\(|\mathcal{E}_1|=2\left(\dfrac{10-5}{2}\right)\)

\(|\mathcal{E}_1|=5\,V\)

 

Two closely wound coils each having mutual inductance \(M=2\,H\). The current in second coil changes from \(I_i=5\,A\) to \(I_f=10\,A\) in \(\Delta t=2\,sec\). Find the magnitude of average induced emf in coil 1.

A

\(3\,V\)

.

B

\(8\,V\)

C

\(7\,V\)

D

\(5\,V\)

Option D is Correct

Mutual Inductance Between Two Coaxial Solenoid

  • 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 \(\ell.\)
  • 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. 

  • 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.

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

 

Illustration Questions

Two co-axial solenoid of same cross-sectional area \(A\) and number of turns \(N_1\) and \(N_2\) respectively, are of same length and current \(I\) is flowing in inner solenoid. Choose the correct sequence to calculate mutual inductance between them. Given the magnetic field of inner and outer solenoid are \(B_1\) and \(B_2\), respectively.

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 \(B_1\to\,M=\dfrac{V_2B_2}{A_2}\)  

×

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.

image

Magnetic field in solenoid 1

\(B_1=\dfrac{\mu_0\,N_1I}{\ell}\)

image

Flux in coil 1 due to coil 2

\(\phi_{12}=\dfrac{\mu_0\,N_1IA}{\ell}\)

image

Thus, mutual inductance in coil 1 due to coil 2

\(M=N_2\left(\dfrac{\phi_{12}}{I}\right)\)

\(M=N_2\dfrac{\mu_0\,N_1IA}{\ell×I}\)

\(M=\dfrac{\mu_0\,N_1N_2A}{\ell}\)

image

Two co-axial solenoid of same cross-sectional area \(A\) and number of turns \(N_1\) and \(N_2\) respectively, are of same length and current \(I\) is flowing in inner solenoid. Choose the correct sequence to calculate mutual inductance between them. Given the magnetic field of inner and outer solenoid are \(B_1\) and \(B_2\), respectively.

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

 

Option A is Correct

Induced EMF in a Coil as a Function of Time

  • 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, \(H\).

Illustration Questions

Two closely wounded coils, having mutual inductance \(M = 2 \; H\) and current in the second coil is alternating current \(I_2 = 3 \; sin \; 2 \pi t\). Find the induced emf in coil 1 as a function of time. 

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

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

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

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

×

Average induced emf in coil 1 is given as 

\(\mathcal{E}_1= \dfrac{-MdI_2}{dt}\)

where M = mutual inductance 

\(I_2\) = current in coil 2

Given:- \(M = 2 \; H \; , I_2 = 3 \; sin 2 \pi t \)

\(\mathcal{E}_1 = -2 \dfrac{d}{dt} (3 sin \; 2 \pi t)\)

\(\mathcal{E}_1 = - 2 × 3 × 2 \pi \; cos \; 2\pi t \)

\(\mathcal{E} _1 = - 12 \pi cos \;2\pi t\)

Two closely wounded coils, having mutual inductance \(M = 2 \; H\) and current in the second coil is alternating current \(I_2 = 3 \; sin \; 2 \pi t\). Find the induced emf in coil 1 as a function of time. 

A

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

.

B

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

C

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

D

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

Option B is Correct

Mutual Inductance Between two Concentric Single Turn Rings 

  • Consider two single turn co–planar concentric circular coils of radii R1 and R2 respectively, such that  R>>> R2.

  • 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 B1

\(\phi_{21} = B_1A_2\)

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

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

  • 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: R1 should be very greater than R2 otherwise the field becomes variable.

Illustration Questions

The two concentric single turn rings are of radius \(R_1 = 12 \; m\) and \(R_2 = 6 \; cm\) respectively. Calculate the mutual inductance between them when current is flowing through outer coil.   

A \(3 \pi ^2 × 10 ^{-7}H\)

B \(2 \pi ^2 × 10 ^{-6} H\)

C \(6 \pi ^2 × 10 ^{-5} H\)

D \(6\pi ^2 × 10 ^{-11}H\)

×

Mutual inductance is given as 

\(M = \dfrac{\mu_0 \pi R_2^2}{2 R_1}\)

where  \(R_2 \) = radius of outer ring 

\(R_1\) = radius of inner ring  

Given:- \(R_ 1 = 12 \; m, R_2 = 6 \; cm\)

\(M = \dfrac{4 \pi × 10 ^{-7} × \pi (6\times10^{-2})^2}{ 2 × (12)}\)

\(M =6 \pi ^2 × 10 ^{-11}\; H\)

The two concentric single turn rings are of radius \(R_1 = 12 \; m\) and \(R_2 = 6 \; cm\) respectively. Calculate the mutual inductance between them when current is flowing through outer coil.   

A

\(3 \pi ^2 × 10 ^{-7}H\)

.

B

\(2 \pi ^2 × 10 ^{-6} H\)

C

\(6 \pi ^2 × 10 ^{-5} H\)

D

\(6\pi ^2 × 10 ^{-11}H\)

Option D is Correct

Practice Now