Question

A conducting loop of area $2\,{{\text{m}}^2}$ is placed to normal to uniform magnetic induction $2\,{\text{wb/}}{{\text{m}}^2}$. If the magnetic induction is uniformly reduced to $1\,{\text{wb/}}{{\text{m}}^2}$ in a time of 2 seconds. Calculate the e.m.f induced in the coil.

Answer 1

According to Faraday’s law of electromagnetic induction, the changing magnetic flux induces emf in the coil. The magnitude of the induced emf is equal to rate of change of magnetic flux around the coil. Recall the expression for the magnetic flux and determine the change in the magnetic flux around the coil.

Formula used:
Faraday’s law, $e = - \dfrac{{d\phi }}{{dt}}$
Here, $\phi$ is the magnetic flux and t is the time.
Magnetic flux, $\phi = BA\cos \theta$
Here, B is the magnetic field, A is the area and $\theta$ is the angle between the magnetic field and area vector of the coil.

Complete step by step answer:
We have given that the magnetic field is reduced from ${B_1} = 2\,{\text{wb/}}{{\text{m}}^2}$ to ${B_2} = 1\,{\text{wb/}}{{\text{m}}^2}$ in time $t = 2\,{\text{s}}$. We know that the changing magnetic field changes the magnetic flux. As we know, according to Faraday’s law of electromagnetic induction, the changing magnetic flux induces emf in the coil.

Let us express the emf induced in the coil as follows,
$e = - \dfrac{{d\phi }}{{dt}}$ …… (1)
Here, $\phi$ is the magnetic flux and t is the time.
We have the expression for the magnetic flux in the coil of surface area A,
$\phi = BA\cos \theta$
Here, B is the magnetic field, A is the area and $\theta$ is the angle between the magnetic field and area vector of the coil.

Since the magnetic field is applied normal to the plane of the coil, the area vector of the coil is parallel to the magnetic field. Therefore, the angle between the area vector and magnetic field is $0^\circ$. Therefore, the above equation becomes,
$\phi = BA\cos \left( 0 \right)$
$\Rightarrow \phi = BA$
Substituting the above equation in equation (1), we get,
$e = - \dfrac{{d\left( {BA} \right)}}{{dt}}$
$\Rightarrow e = - A\dfrac{{dB}}{{dt}}$ ……. (2)

Let us calculate the changing magnetic field as follows,
$\dfrac{{dB}}{{dt}} = \dfrac{{{B_2} - {B_1}}}{{\Delta t}}$
$\Rightarrow \dfrac{{dB}}{{dt}} = \dfrac{{1 - 2}}{2}$
$\Rightarrow \dfrac{{dB}}{{dt}} = - \dfrac{1}{2}\,{\text{wb/}}{{\text{m}}^2}$
Substituting $A = 2\,{{\text{m}}^2}$ and $\dfrac{{dB}}{{dt}} = - \dfrac{1}{2}\,{\text{wb/}}{{\text{m}}^2}$ in equation (2), we get,
$e = - \left( 2 \right)\left( { - \dfrac{1}{2}} \right)$
$\therefore e = 1\,{\text{V}}$

Thus, the induced emf in the coil is 1 V.

Note: Do not forget to write the negative sign for the changing magnetic field for the induced emf in the coil. The negative sign representing the emf induced in the coil opposes the change in the magnetic flux. The emf induces in the coil only if there is change in the magnetic flux or change in the magnetic field with respect to time. Note that if the applied magnetic field is along the plane of the coil, the angle between the area vector and magnetic field becomes 90 degrees. Therefore, the change in magnetic flux becomes zero.