JEE Main Last 10 Years Question Papers with Solutions PDF 2024 to 2014

# Mutual Inductance - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Mutual Inductance, Mutual Inductance for two coaxial long solenoids, Mutual Inductance for a pair of concentric coils is considered one of the most asked concept.

• 38 Questions around this concept.

## Solve by difficulty

Two loops with sides L and $1(L>>1)$ are placed concentrically and coplanar as shown in figure. The mutual inductance of the system is proportional to

A small square loop of wire of side l is place inside a large square loop of wire of side
L(L > > l). The loops are coplanar and their centers coincide. The mutual inductance of the system is proportional to :

Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be:

(a)      (b)        (c)

A circular copper disc 10 cm in diameter rotates at 1800 rev per minute about an axis through its centre and at right angles to the disc. A uniform field of induction B of 1 Wb$\mathrm{m^{-2}}$ is perpendicular to the disc. What potential difference is developed between the axis of the disc and the rim?

A pair of adjacent coils has a mutual inductance of 2.5 H. If the current in one coil changes from 0 to 40 A in 0.8s, then the change in flux linked with the other coil is:

Two coils A and B having turns 300 and 600 respectively are placed near each other. On passing a current of 3.0 ampere in A, the flux linked with A is $\mathrm{1.2 \times 10^{-4} \mathrm{~Wb}}$ and with B it is $\mathrm{9.0 \times 10^{-5} \mathrm{~Wb}}$. The mutual inductance of the system is:

Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be:

##### Amity University, Noida B.Tech Admissions 2024

Asia's Only University with the Highest US & UK Accreditation

##### UPES B.Tech Admissions 2024

Ranked #52 among universities in India by NIRF | Highest CTC 50 LPA | 100% Placements

As shown in the figure, P and Q are two co-axial conducting loops separated by some distance. When the switch S is closed, a clockwise current $I_{P}$ flows in P (as seen by E) and an induced current $I_{Q1}$ flows in Q. The switch remains closed for a long time. When S is opened, a current $I_{Q3}$ flows in Q. Then, the directions of $I_{Q1}$ a $I_{Q2}$ (as seen by E) are

Two different coils have self-inductance $\mathrm{L_1=8 m H, L_2=2 m H .}$. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same rate. At a certain instant of time, the power given to the two coils is the same. At that time the current, the induced voltage, and the energy stored in the first coil are  respectively. Corresponding values for the second coil at the same instant are respectively. Then choose the incorrect option.

JEE Main Exam's High Scoring Chapters and Topics
This free eBook covers JEE Main important chapters & topics to study just 40% of the syllabus and score up to 100% marks in the examination.

A small square loop of wire of side l is placed inside a large square loop of wire of side L(>> l ). The loops are coplanar and their centres coincide. The mutual inductance of the system is:

## 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.

Since $r_1 >>> r_2$ so we can assume coil 2 is at the center of coil 1.

If

Suppose a current  $i_1$ flows through the outer circular coil.
Then Magnetic field at the center of the coil 1 is given as

$B_1=\frac{\mu _0N_1i_1}{2r_1}$

So the total flux passing through coil 2 will be given as

$( \phi _2)_{total}=N_2B_1A_2=\frac{\mu _0N_1N_2i_1A_2}{2r_1}$

And using $( \phi _2)_{total}= Mi_1$

we get $M=\frac{\mu _0N_1N_2 A_2}{2r_1}=\frac{\mu _0N_1N_2 (\pi r_2^2)}{2r_1}$

Where M=mutual inductance between two concentric coils

## Study it with Videos

Mutual Inductance
Mutual Inductance for two coaxial long solenoids
Mutual Inductance for a pair of concentric coils

"Stay in the loop. Receive exam news, study resources, and expert advice!"

## Books

### Reference Books

#### Mutual Inductance for a pair of concentric coils

Errorless Physics Vol 2

Page No. : 1232

Line : 37

Back to top