Question

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

Answer 1

Hint : The mutual inductance is similar to the concept of self-inductance, but it occurs in between two coils. It has the same unit as that of inductance.

Formula used: In this solution we will be using the following formula;
 $ M = - \dfrac{{{E_s}}}{{\dfrac{{d{I_p}}}{{dt}}}} $ where, $ M $ is the coefficient of mutual induction, $ {E_s} $ is the emf induced on a secondary coil, $ {I_p} $ is the instantaneous current in a primary coil, and $ t $ is time.

Complete step by step answer
In general, when a changing current (such as an alternating current), is flowing through a coil, the changing electric current causes a changing magnetic field around the coil. Now when another coil is in the path of these changing magnetic field lines, by Faraday’s law of induction, emf becomes induced in that other coil, and the value is said to be directly proportional to the rate of change of the changing current in the first coil. Mathematically,
 $ {E_p} \propto \dfrac{{d{I_p}}}{{dt}} $
 $ \Rightarrow {E_p} = - M\dfrac{{d{I_p}}}{{dt}} $
where $ M $ is a constant of proportionality, called the coefficient of mutual induction. or simply mutual inductance (the negative sign shows that the emf in the secondary coil produces a current in the reverse direction as the current in the primary coil)
From above, $ M = - \dfrac{{{E_s}}}{{\dfrac{{d{I_p}}}{{dt}}}} $ , hence can be defined as the ratio of the emf induced in a secondary coil to the rate of change of current in a primary coil.
It has the same unit as any other inductance. The Henry (H)
The dimensional formula can be given as
 $ \left[ M \right] = \dfrac{{\left[ V \right]}}{{\dfrac{A}{T}}} = \dfrac{{\left[ V \right] \times T}}{A} $ but $ \left[ V \right] = \dfrac{{\left[ {BA} \right]}}{T} $
Hence, by substituting and cancelling $ T $ , we have
 $ \left[ M \right] = \dfrac{{\left[ {BA} \right]}}{A} $
The dimension of flux can be given as
 $ \left[ {BA} \right] = M{L^2}{T^{ - 2}}{A^{ - 1}} $
Then,
 $ \left[ M \right] = \dfrac{{M{L^2}{T^{ - 2}}{A^{ - 1}}}}{A} = M{L^2}{T^{ - 2}}{A^{ - 2}} $.

Note
for clarity, we can derive the dimension of flux from the formula,
 $ F = BIL $
 $ \Rightarrow B = \dfrac{F}{{IL}} $
Dimensionally,
 $ \left[ B \right] = \dfrac{{ML{T^{ - 2}}}}{{AL}} $
 $ \left[ B \right] = M{T^{ - 2}}{A^{ - 1}} $
Hence, $ \left[ {BA} \right] = M{T^{ - 2}}{A^{ - 1}} \times {L^2} = M{T^{ - 2}}{A^{ - 1}}{L^2} $