Mutual Inductance MCQ - Practice Questions & Answers

Solve by difficulty

An electron moves on a straight line path XY as shown. The abcd is a coil adjacent to the path of electron. What will be the direction of current, if any, induced in the coil ?

adcb

The current will reverse its direction as the electron goes past the coil

No current induced

abcd

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Concepts Covered - 3

Mutual Inductance

Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighboring coil or circuit will also change. Hence an emf will be induced in the neighboring coil or circuit. This phenomenon is called ‘mutual induction’.

or The phenomenon of producing an induced emf in a coil due to the change in current in the other coil is known as mutual induction.

Coefficient of mutual induction (M)-

If two coils (P-primary coil or coil 1, S-Secondary Coil or coil 2) are arranged as shown in the below figure.

If we change the current through the coil P (i.e $i_1$ ) then flux passing through Coil S (i.e $\phi _2$ ) will change.

I.e $N_{2}\phi_{2} \, \alpha \, i_{1}\Rightarrow N_{2}\phi_{2} =M_{21}i_{1}=Mi_1$

where

$M_{21}$ = mutual induction of Coil 2 w.r. t Coil 1

$N_{1}=$ Number of turns in the primary coil

$N_{2}=$ Number of turns in the secondary coil

$i_{1}=$ current through the primary coil or coil 1

Similarly, if we exchange the position of Coil 1 and Coil 2

then

If we change the current through the coil S (i.e $i_2$ ) then flux passing through Coil P (i.e $\phi _1$ ) will change.

I.e $N_{1}\phi_{1} \, \alpha \, i_{2}\Rightarrow N_{1}\phi_{1} =M_{12}i_{2}=Mi_2$

where

$M_{12}$ = mutual induction of Coil 1 w.r. t Coil 2

$N_{1}=$ Number of turns in the primary coil

$N_{2}=$ Number of turns in the secondary coil

$i_{2}=$ current through the coil 2 or Coil S

As $N_{2}\phi_{2} = Mi_1$

$\text { If } i_1=1 \text { amp, } N_2=1 \text { then, } M=\phi_2$

I.e coefficient of mutual induction of two coils is numerically equal to the magnetic flux linked with one coil when unit current flows through the neighboring coil.

Using Faraday Second Law of Induction emf we get

$\varepsilon _{2}=-N_{2} \frac{d \phi_{2}}{d t} =-M \frac{d i_{1}}{d t}$

$\text { If } \frac{d i_1}{d t}=1 \frac{amp}{sec} \ and \ N_2=1 \ \text { then } \ |\varepsilon _2|=M$

I.e The coefficient of mutual induction of two coils is numerically equal to the emf induced in one coil when the rate of change of current through the other coil is unity.

Units and dimensional formula of ‘M’-

S.I. Unit - Henry (H)

And $1 H = \frac{1V.sec }{Amp}$

And its dimensional formula is $ML^{2}T^{-2}A^{-2}$

Dependence of mutual inductance

Number of turns (N₁, N₂) of both coils
Coefficient of self inductances (L₁, L₂) of both the coils

and the relation between $M , L_1, L_2$ is given as

$M= K\sqrt{L_{1}L_{2}}$

where K =coeffecient of coupling.

If L=0 then M = 0

If K = 0 i.e case of No coupling then M = 0.

Distance(d) between two coils (i.e As d increases then M decreases)
The magnetic permeability of medium between the coils ( $\mu _r$ )

Mutual Inductance for two coaxial long solenoids

Consider two long co-axial solenoids of the same length $l$ ..Let A₁ and A₂ be the area of cross-section of the solenoids with A₁ being greater than A₂ as shown in the below figure.

The turn density of these solenoids are n₁ and n₂ respectively are given as $n_1=\frac{N_1}{l} \ and \ n_2=\frac{N_2}{l}$

Let i₁ be the current flowing through solenoid 1, then the magnetic field produced inside it is given as

$B_{1}=\mu_{o} n_{1} i_{1}$

As the field lines of $\vec{B_{1} }$ are passing through the area A₂

So the magnetic flux linked with each turn of solenoid 2 due to solenoid 1 and is given by

$\begin{array}{l}{\Phi_{21}=\int_{A_{2}} \bar{B}_{1} \cdot d \vec{A}=B_{1} A_{2}=\left(\mu_{0} n_{1} i_{1}\right) A_{2}}\end{array}$

The total flux linkage of solenoid 2 with total turns N₂ is

$\begin{array}{c}{(\phi _{21})_{total} =N_{2} \Phi_{21}=\left(n_{2} l\right)\left(\mu_{0} n_{1} i_{1}\right) A_{2}} \\ { \Rightarrow (\phi _{21})_{total}=N_{2} \Phi_{21}=\left(\mu_{0} n_{1} n_{2} A_{2} l\right) i_{1}}\end{array}$

And Using $(\phi _{21})_{total}=N_{2} \Phi_{21}=M_{21} i_{1}$ we get

$M_{21}=\mu_{0} n_{1} n_{2} A_{2} l$

Where M₂₁ is the mutual inductance of the solenoid 2 with respect to solenoid 1.

Similarly M₁₂ =mutual inductance of solenoid 1 with respect to solenoid 2 is given as

$M_{12}=\mu_{0} n_{1} n_{2} A_{2} l$

Hence $M_{21}= M_{12}=M$

So, In general, the mutual inductance between two long co-axial solenoids is given by

$M=\mu_{0} n_{1} n_{2} A_{2} l$

If a dielectric medium of permeability $\mu$ is present inside the solenoids, then

$\begin{array}{l}{\mathrm{M}=\mu \mathrm{n}_{1} \mathrm{n}_{2} \mathrm{A}_{2} \mathrm{l}} \ \ {\text { or} \ \ \ \mathrm{M}=\mu _0\mu_{\mathrm{r}} \mathrm{n}_{1} \mathrm{n}_{2} \mathrm{A}_{2} \mathrm{l}}\end{array}$

Mutual Inductance for a pair of concentric coils

Consider two circular coils one of radius 'r₁' and the other of radius' r₂'placed coaxially with their centers coinciding as shown in the below figure.