Class 12 Physics Electromagnetic Induction | Self and Mutual Inductance |

**Self and Mutual Inductance**

- Consider a coil (1) which is connected to a battery and there is a key.
- There is another coil (2) which is connected to galvanometer.
- When the key is closed, the current flows through coil (1) there is a change in the flux of coil (1) because of which self-induced emf will be produced.
- Because of current flow in coil(1) there will be change in the magnetic flux of coil(2) because of which induced emf will be produced in coil(2).
- As a result mutual induction takes place in coil(2).
- This shows both self-inductance and mutual inductance takes place simultaneously.
- Therefore flux linked to coil 1:-N
_{1}Φ_{1}=L_{1}I_{1}+M_{12}I_{2} - Induced emf in coil (1)
**e= -L**_{1}(dI_{1}/dt) - M_{12}(dI_{2}/dt)- Where(-)ive sign shows it opposes the growth of current.

** Problem:- **Two coils have mutual inductance of 1.5 H. If the current in the primary circuit is raised to 5A in 1 millisecond after closing the circuit, what is the emf induced in the secondary?

__Answer:-__

M=1.5 H

I_{i}=0 A, I_{f}=5A, dt =1ms =10^{-3}sec

e_{2} =-M (dI_{i}/dt)

=1.5 x ((I_{f}-I_{i})/dt)

=1.5 x ((5-0)/10^{-3})

=7.5x103V.

** Problem:- **Two concentric circular coils, one of small radius r

arrangement.

__Answer:-__

Let a current I_{2} flow through the outer circular coil. The field at the centre of the coil is

B_{2} = μ_{0}I_{2} / 2r_{2}. Since the other co-axially placed coil has a very small radius, B_{2} may be considered constant over its cross-sectional area. Hence,

Φ_{1} = πr_{1}^{2} B_{2}

=I_{2 }(μ_{0} π r_{1}^{2})/ (2r_{2})

= M_{12}I_{2}

Thus, M_{12} = (μ_{0} π r_{1}^{2})/ (2r_{2})

Using equation M_{12}=M_{21}= (μ_{0} π r_{1}^{2})/ (2r_{2})

Note that we calculated M_{12} from an approximate value of Φ_{1}, assuming the magnetic field B_{2} to be uniform over the area π r_{1}^{2}. However, we can accept this value because r1 << r2.

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