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Mutual Inductance MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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.

  • 18 Questions around this concept.

Solve by difficulty

The flux linked with the secondary coil due to current I1=2Amp in primary coil is 4×105 Wb No. of turns in primary and secondary coil is 200 and 400 respectively. find the coefficient of mutual induction

What will be mutual inductance if there is no current flowing in the primary coil(Open Circuit)?

What will be mutual inductance if current I is flowing in primary coil? (Coefficient of coupling K=0)

What will be the inductance of the secondary coil if mutual inductance is 0. (Coefficient of cooling is K).

Two coils, X and Y, are kept in close vicinity of each other. When a varying current, I(t), flows through coil X, the induced emf (V(t)) in coil Y, varies in the manner shown here. The variation of I(t), with time, can then be represented by the graph labeled as graph: 

A rectangular loop of sides ‘ a ‘ and ‘ b ‘ is placed in xy plane. A very long wire is also placed in xy plane such that side of length ‘ a ‘ of the loop is parallel to the wire. The distance between the wire and the nearest edge of the loop is ‘ d ‘. The mutual inductance of this system is proportional to :

A time varying current I(t)=2coswt is flowing in primary coil of 200 turns with frequency 40 Hz . The coefficient of mutual induction is 10 mH .

Find the emf induced (max) (in mV ) is secondary coil of 400 turns

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An e.m.f. of 5 millivolt is induced in a coil when in a nearby placed another coil, the current changes by 5 ampere in 0.1 second. The coefficient of mutual induction between the two coils will be :

The coefficient of mutual inductance of the two coils is 0.5 H. If the current is increased from 2A to 3 A in 0.01 sec. in one of them, then the induced e.m.f. in the second coil is :

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The coefficient of mutual inductance of the two coils is 0.5 H. If the current is increased from 2 to 3 A in 0.01 sec. in one of them, then the induced e.m.f. in the second coil 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 i1 ) then flux passing through Coil S (i.e ϕ2 ) will change.

 I.e N2ϕ2αi1N2ϕ2=M21i1=Mi1

where
M21= mutual induction of Coil 2 w.r. t Coil 1
N1= Number of turns in the primary coil
N2= Number of turns in the secondary coil

i1= 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 i2 ) then flux passing through Coil P (i.e ϕ1 ) will change.

 I.e N1ϕ1αi2N1ϕ1=M12i2=Mi2

where

M12= mutual induction of Coil 1 w.r. t Coil 2N1= Number of turns in the primary coil N2= Number of turns in the secondary coil 
 

i2= current through the coil 2 or Coil S
- As N2ϕ2=Mi1

If i1=1amp,N2=1 then, M=ϕ2
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's Second Law of Induction emf we get

ε2=N2dϕ2dt=Mdi1dt


If di1dt=1ampsec and N2=1 then |ε2|=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

1H=1VsecAmp


And its dimensional formula is ML2T2A2

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,L1,L2 is given as

    M=KL1L2

    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 dincreases then M decreases)
    - The magnetic permeability of medium between the coils (μr)

 

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 n1=N1l and n2=N2l
Let i1 be the current flowing through solenoid 1 , then the magnetic field produced inside it is given as

B1=μon1i1


As the field lines of B1 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

Φ21=A2B¯1dA=B1A2=(μ0n1i1)A2


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

(ϕ21)total =N2Φ21=(n2l)(μ0n1i1)A2(ϕ21)total =N2Φ21=(μ0n1n2A2l)i1


And Using (ϕ21)total =N2Φ21=M21i1 we get

M21=μ0n1n2A2l


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

M12=μ0n1n2A2l
 

Hence M21=M12=M
So, In general, the mutual inductance between two long co-axial solenoids is given by

M=μ0n1n2A2l


If a dielectric medium of permeability μ is present inside the solenoids, then

M=μn1n2 A2l or M=μ0μrn1n2 A2l
 

 

 

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

 

 

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

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