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Question: Define the coefficient of mutual inductance. Give its unit and dimensional formula....

Define the coefficient of mutual inductance. Give its unit and dimensional formula.

Explanation

Solution

Hint A magnetic field will be generated when current flows through a conductor. This magnetic field will produce an electromotive force in the second coil that happens to be in the vicinity of the magnetic field of the first coil. This is related to the coefficient of mutual inductance.

Complete Step by step solution
Let us consider two coils C1{C_1} and C2{C_2} . Let the magnetic flux in the second coil be ϕ2{\phi _2} . Let the current in the first coil C1{C_1} be I1{I_1}.
The magnetic flux in the coil C2{C_2} due to the current in C1{C_1} will be
ϕ2I1{\phi _2} \propto {I_1}
Introducing a constant of proportionality into the above equation gives us
ϕ2=MI1{\phi _2} = M{I_1}
Here MM is the coefficient of mutual inductance. This constant of proportionality will depend on the number of turns of the secondary coil, area of the secondary coil, and the medium of propagation of the magnetic field.
If the current in the coil C1{C_1} changes with time, then the magnetic flux on the second coil C2{C_2} will change with time and according to Faraday’s law, the induced emf in the coil C2{C_2} will be equal to the rate of change of magnetic flux. Mathematically,
ε2=dϕ2dt{\varepsilon _2} = - \dfrac{{d{\phi _2}}}{{dt}}
But, we have already written an equation for the magnetic flux in the second coil. Substituting that into the above equation, we get
ε2=MdI1dt{\varepsilon _2} = - M\dfrac{{d{I_1}}}{{dt}}
Therefore,
M=ε2dI1dtM = - \dfrac{{{\varepsilon _2}}}{{\dfrac{{d{I_1}}}{{dt}}}}
Thus, we can define the mutual inductance as the induced emf when the rate of change of current in the primary coil is unity.
The unit of mutual inductance is Weber or Henry.
Unit of M=volt×secampereM = \dfrac{{volt \times \sec }}{{ampere}}
Potential is defined as the energy per unit charge. Therefore,
Dimension of volt=ML2T2IT = \dfrac{{M{L^2}{T^{ - 2}}}}{{IT}}
By simplifying, we get
Dimension of volt=ML2T3I1 = M{L^2}{T^{ - 3}}{I^{ - 1}}
\therefore dimension of M=MLT3I1T1IM = \dfrac{{ML{T^{ - 3}}{I^{ - 1}}{T^1}}}{I}
Hence dimension of M=MLT2I2M = ML{T^{ - 2}}{I^{ - 2}}

Note
The negative sign in the equation for mutual inductance is explained using Lenz’s law. The induced electromotive force in the second coil due to the changing magnetic field in the first coil will oppose the initial changing magnetic field.