Question

A solenoid has 2000 turns wound over a length of 0.3m. Its cross sectional area is equal to $1.2\times {{10}^{-3}}{{m}^{2}}$. Around its central cross section, a coil of 300 turns is wound. If an initial current of 2A flowing in the solenoid is reversed in 0.25s, the emf induced in the coil is
a) 0.6mV
b) 60mV
c) 40.2mV
d) 48mV

Answer 1

It is given in the question that along the central cross section of the solenoid a coil is kept. Since this coil is placed coaxially it will experience the magnetic field due to the solenoid. If the current across the solenoid changes it will induce the emf in a coil placed coaxially as per the law of electromagnetic induction. Hence we will use the expression of emf produced in the coil, when the current changes in the solenoid.

Formula used:
$E=M\dfrac{di}{dt}$

Complete answer:
If two coils are placed coaxially and the current in one of the coils changes then an emf is induced in another coil by law of electromagnetic induction. This emf induced in the coil placed coaxially into the central region of the solenoid is given by, $E=M\dfrac{di}{dt}...(1)$ where M is the coefficient of mutual induction and $\dfrac{di}{dt}$ represents the rate of change of current in the circuit. The coefficient of mutual induction M is numerically equal to, $M=\dfrac{{{\mu }_{\circ }}nNA}{l}$ where ${{\mu }_{\circ }}=4\pi \times {{10}^{-7}}$ is the permeability of free space, n is the number of turns of the coil coaxially placed, N is the number of turns in the solenoid, A is the area of cross section of the solenoid and l is the length of the solenoid. Substituting this in equation 1 we get,
$E=\dfrac{{{\mu }_{\circ }}nNA}{l}\dfrac{di}{dt}$. It is given in the question that the current is reversed from 2A in 0.25s. Hence the final value of current will be -2A. hence the emf induced in the coil is,
$\begin{aligned} & E=\dfrac{{{\mu }_{\circ }}nNA}{l}\dfrac{di}{dt} \\\ & E=\dfrac{4\pi \times {{10}^{-7}}\times 300\times 2000\times 1.2\times {{10}^{-3}}}{0.3}\dfrac{2-(-2)}{0.25} \\\ & E=\dfrac{90.432\times 4\times {{10}^{-5}}}{0.075}=4823.04\times {{10}^{-5}}V \\\ & E=4.8\times {{10}^{-3}}V=E=4.8mV \\\ \end{aligned}$

So, the correct answer is “Option D”.

Note:
It is to be noted that there is 100% of flux of the solenoid to the coil which is placed coaxially. If this condition is not satisfied then the above expression does hold its validity. It is to be noted that the induced emf in the coil can be increased by increasing the number of turns of the solenoid as well as the coil.