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, $$H$$.

#### 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 B C D 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 $$\mathcal{E}_2=-M\dfrac{dI_1}{dt}$$

#### 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

# 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, $$H$$.

#### 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.

#### 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

# 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, $$|\mathcal{E}_2|=M\dfrac{\Delta I_1}{\Delta t}$$

#### 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.

$$M_{12}=M_{21}$$

#### 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. Magnetic field in solenoid 1

$$B_1=\dfrac{\mu_0\,N_1I}{\ell}$$ Flux in coil 1 due to coil 2

$$\phi_{12}=\dfrac{\mu_0\,N_1IA}{\ell}$$ Thus, mutual inductance in coil 1 due to coil 2

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

$$M=N_2\dfrac{\mu_0\,N_1IA}{\ell×I}$$

$$M=\dfrac{\mu_0\,N_1N_2A}{\ell}$$ ### 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 $$B_1\to\,M=\dfrac{V_2B_2}{A_2}$$

Option A is Correct