Class 12 Physics Electromagnetic Induction | Mutual Inductance |

**Mutual Inductance**

In mutual inductance there are 2 coils, current is passed in one coil, as current increases there is change in the flux, and as aresult current is induced in the second coil.

Consider coil 1 connected to battery and coil 2 is connected to the galvanometer.

- When the key is pressed attached to the coil 1 the current starts flowing, when the current starts increasing flux linked also starts increasing.
- Because of the increase in the flux linked with the coil1, the flux of coil 2 also increases.
- There is change in the flux of the coil 2 as a result emf is induced in the coil 2.
- Because of the induced emf induced current will be there in coil2.
- This induced current opposes the increase of the current in coil 1.

Mathematically:-

- Φ
_{(2)}∝ I_{(1)} - =>Φ
_{(2)}= MI_{(1)}where M = constant of proportionality known as Mutual Inductance. - Induced emf in coil 2 e=-(dΦ
_{(2)}/dt) - =>e =-d/dt(MI
_{(1)}) where I current flowing in coil (1). - Therefore
**e =-d/dt (M****I**_{(1)})

- M (mutual inductance) depends on:-

- Geometry of both coils.
- Distance between coils.
- Orientation of coils.

__Mutual Inductance between long co-axial solenoids__

- Co-axial solenoids means the centres of both the solenoids are same.
- Radius of the smaller solenoid (1) =r
_{1}. Number of turns in smaller solenoid= N_{1}. - Radius of the bigger solenoid (2) = r
_{2}.Number of turns in bigger solenoid= N_{2}.

** Case 1**:-

- Current flowing in the bigger solenoid = I
_{2}, as a result magnetic flux Φ_{1}will be induced in smaller solenoid. - Therefore N
_{1}Φ_{1}∝I_{2} - =>N
_{1}Φ_{1}=M_{12}I_{2}equation(1)- where M
_{12}= mutual inductance of 1 w.r.t 2

- where M
- Magnetic field due to I
_{2}in (bigger solenoid 2) B =μ_{0}n_{2}I_{2 } - => B =(μ
_{0}n_{2}I_{2})/l_{ength} - Total Flux N
_{1}Φ_{1}=N_{1}BA_{1} - => =(N
_{1}A_{1}μ_{0}N_{2}I_{2})/l_{ength}equation(2) - From equation(1) and (2)
- M
_{12}I_{2}= (N_{1}A_{1}μ_{0}N_{2}I_{2})/l_{ength} - =>
**M**equation(a)_{12}= (μ_{0}N_{1}N_{2}A)/ l_{ength}

__Case 2:-__

- Current I
_{1}flowing through solenoid (1) this will result in flux solenoid (2) - Total flux N
_{2}Φ_{2}= M_{21}I_{1 }equation(3) - Also total flux N
_{2}Φ_{2}= N_{2}B_{1}A_{1}- where B =magnetic field due to smaller solenoid;
- B
_{1}=μ_{0}n_{1}I_{1}; - =(μ
_{0}N_{1}I_{1})/l_{ength}

- =>N
_{2}Φ_{2}= N_{2}((μ_{0}N_{1}I_{1})/l_{ength})A_{1}equation(4) - Comparing (3) and (4)
- M
_{21}I_{1}= N_{2}((μ_{0}N_{1}I_{1})/l_{ength})A_{1} **M**equation(b)_{21}= (μ_{0}N_{2}N_{1}A_{1})/ l_{ength}- Comparing (a) and (b)
- Therefore
**M**_{12}= M_{21}

** Problem:- **A solenoid of length 50cm with 20 turns per cm and area of cross-section 40cm

** Answer:- **Let the outer solenoid be 1 and inner solenoid be 2.

Length l_{1} =50cm=50x10^{-2}m

n_{1} =2000 turns /m

A_{1} = 40 x10^{-4} m^{2}

Length l_{2}=50x10^{-2}m

n_{2} =2500 turns/m

A_{2}=25x10^{-4} m^{2}

M_{12}=M_{21}

M_{12}= (μ_{0}N_{1}N_{2}A_{2})/l_{ength} =μ_{0}n_{1}n_{2}A_{2}l_{ength}

=4x3.14x10^{-7}x2000x2500x50x10^{-2}x25x10^{-4}

=7.85x10^{-3} H

** Problem:- **(a) A toroidal solenoid with an air core has an average radius of 15cm, area of cross-section 2cm

(b) A second coil of 300 turns is wound closely on the toroid above. If the current in the primary coil is increased from 0 to 2A in 0.05s.Obtain the induced emf in the second coil.

** Answer:- **(a) Radius r

Area A_{1}=2x10^{-4}m^{2}

N_{1}=1200

B=μ_{0}nI= (μ_{0}N_{1}I)/ (2πr_{1})

Total flux: -N_{1}xBxA_{1}=N_{1}x(μ_{0}xN_{1}I/2πr_{1}) x A_{1}

Also Total flux =LI

LI =μ_{0}xA_{1}x N_{1}^{2}I/ (2πr_{1})

= (4πx10^{-7}x (1200)^{2}x 2x10^{-4})/ (2πx15x10^{-2})

=2.304x10^{-4} H

The self-inductance of the toroid is 2.304x10^{-4} H.

(b) A_{1}=2x10^{-4}m^{2}

r_{1}=15x10^{-2}m

N_{1}=1200

N_{2} =300

(dI_{1}/ dt) = (I_{f} –I_{i})/dt = (2-0)/ (0.05) =40A/s

Total flux in 2:- Φ_{2} =N_{2}xB_{1}xA_{1}

=N_{2}(μ_{0}n_{1}I_{1})/A_{1} equation (1)

Also,Φ_{2} =MI_{1} equation (2)

=> M=N_{2}xμ_{0}xn_{1}xA_{1}

e_{2} =M (dI_{1}/dt) =μ_{0}N_{2}n_{1}A_{1}x40

=4πx10^{-7}x 300x (N_{1}/2πr_{1}) x2x10^{-4}x40

=0.023V

The emf induced in the second coil is 0.023V.

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