Question

Two coils have a mutual inductance of $0.005\,H$. The current changes in first coil according to the equation i= \[{{i}_{m}}\mathbf{\omega }\sin \mathbf{\omega }t\] where \[{{i}_{m}}=10\,A\], \[\mathbf{\omega }=100\pi rad/s\]. The maximum value of the emf induced in the second coil is ?

Answer 1

Hint: The magnitude of emf is the rate of change of magnetic flux with respect to time, it reaches its maximum value when the flux curve has the greatest slope.Mutual induction is a phenomenon by which an induced emf is produced in one coil due to change in strength of current flowing through another coil.

Complete step by step answer:
Mutual induction is the property of two coils by virtue of which each opposes any change in the strength of current flowing through the other by developing an induced emf. Faraday’s law of Induction states that the induced emf in a circuit is proportional to the time rate of change of magnetic flux through the circuit. i.e,
\[\mathbf{\varepsilon }=\dfrac{-d\varphi }{dt}\]
The negative sign signifies that the induced emf always opposes any change in magnetic flux associated with the circuit.

Now, \[\varphi =Mi\]
where $M$ is the coefficient of mutual induction and $i$ is the current flowing in the coil.
Also,
\[i={{i}_{m}}\sin \mathbf{\omega }t\],
where $\omega$ = the angular velocity and ${i_m}$ = amplitude of electric current.
Now, if we differentiate magnetic flux with respect to time, we get,
\[\varepsilon =\dfrac{-d\varphi }{dt} \\
\Rightarrow \varepsilon =-\dfrac{Md({{i}_{m}}\sin \omega t)}{dt} \\
\Rightarrow \varepsilon=-M{{i}_{m}}\omega \cos \omega t\]
Since, in the above formula only the value of \[\cos \omega t\] varies with time, and it’s value can vary from -1 to +1, for \[\varepsilon ={{\varepsilon }_{\max .}}\], we need to have the value of \[\cos \omega t\] to be -1. So that the overall value becomes positive.So,\[{{\varepsilon }_{\max .}}=M{{i}_{m}}\omega \] and from the question, we get \[M=5 \times {{10}^{-3}}H\], \[{{i}_{m}}=10A\], \[\omega =100\pi rad/s\].Thus
\[{{\mathbf{\varepsilon }}_{\max .}}=0.005 \times 10 \times 100\pi \]
\[\therefore {{\mathbf{\varepsilon }}_{\max .}}=5\pi \]

Hence, the maximum value of the emf induced in the second coil is $5\pi$.

Note: In order to get the highest value of a quantity that depends on a trigonometric function, look at the formula and try to use one of the known values of that trigonometric function. While using a graph, the maximum value of induced emf is obtained when the flux curve (in ‘magnetic flux vs emf’ graph) has the greatest slope.