Question

A conducting circular loop made of thin wire has an area $3.5 \times {10^{ - 3}}{m^2}$ and resistance $10\Omega $. It is placed perpendicular to a time dependent magnetic field $B(t) = 0.4T\sin (50\pi t)$. The field is uniform in space. Then, the net charge flowing through the loop during $t = 0$ s and $t = 10$ ms is close to:
(A) $0.14$mC
(B) $0.21$mC
(C) 6mC
(D) 7mC

Answer 1

Hint
The application of magnetic fields in and around an electric circuit generates an electromotive force by the process of electromagnetic induction. This force is responsible for the flow of charges in the circuit.
Formula used: $V = \dfrac{{d\Phi }}{{dt}}$, where V is the induced voltage, $\Phi $ is the magnetic flux and t is the time.

Complete step by step answer
Electromagnetic induction is the phenomenon by which an electromotive force across an electrical conductor is produced in the presence of a time-varying magnetic field. This induced emf or voltage is given by Faraday's Law. This law states that the induced voltage in a circuit is proportional to the rate of change of the magnetic flux through that circuit with time.
In this question, we are provided with the following information:
Area of the wire $A = 3.5 \times {10^{ - 3}}{m^2}$
Resistance of the wire $R = 10\Omega $
Time dependent magnetic field $B(t) = 0.4T\sin (50\pi t)$
On comparing this field with the standard form of $B = {B_0}\sin (\omega t)$, we get:
$\Rightarrow {B_0} = 0.4T$
$\Rightarrow \omega = 50\pi $
We know that the Faraday’s Law states that:
$\Rightarrow V = \dfrac{{d\Phi }}{{dt}}$ [Eq. 1]
The magnetic flux is given as:
$\Rightarrow \Phi = B.A = {B_0}\sin (\omega t).A$
Putting this in Eq. 1, we get:
$\Rightarrow V = \dfrac{d}{{dt}}({B_0}\sin (\omega t).A) = A{B_0}\dfrac{d}{{dt}}\sin \omega t$
Solving the differential, we get:
$\Rightarrow V = A{B_0}\omega \cos \omega t$
Since, the voltage is directly proportional to charge flow, we can write it as:
$\Rightarrow \dfrac{{dQ}}{{dt}}R = V = A{B_0}\omega \cos \omega t$ [As V=IR]
$\Rightarrow \dfrac{{dQ}}{{dt}} = \dfrac{{A{B_0}}}{R}\omega \cos \omega t$
To get the amount of charge flowing, we integrate this equation:
$\Rightarrow \int {dQ} = \int\limits_0^{10ms} {\dfrac{{A{B_0}\omega }}{R}\cos \omega t} dt$
Taking the constant outside of the integral and solving gives us:
$\Rightarrow Q = \dfrac{{A{B_0}\omega }}{R}\int\limits_0^{10ms} {\cos \omega t} dt = \dfrac{{A{B_0}\omega }}{R}\left| {\dfrac{{\sin \omega t}}{\omega }} \right|_0^{10ms}$
We know that $1ms = {10^{ - 3}}s$, hence:
$\Rightarrow Q = \dfrac{{A{B_0}}}{R}\left| {\sin \omega t} \right|_0^{10ms} = \dfrac{{A{B_0}}}{R}[\sin \omega \times 10 \times {10^{ - 3}} - \sin \omega \times 0]$
On substituting the values, we get:
$\Rightarrow Q = \dfrac{{3.5 \times {{10}^{ - 3}} \times 0.4}}{{10}}[\sin 50\pi \times 10 \times {10^{ - 3}}]$
$\Rightarrow Q = 0.14 \times {10^{ - 3}}\sin \dfrac{{50\pi }}{{100}} = 0.14 \times {10^{ - 3}}\sin \dfrac{\pi }{2}$
This gives us the charge as:
$\Rightarrow Q = 0.14 \times {10^{ - 3}}C = 0.14mC$
Hence, the correct answer is option (A).

Note
This property of electromagnetic induction can be used to generate DC and/or AC power by eliminating the need of an external electrical energy source. This property is widely used in electric motors, trains, and transformers along with non-contact position sensors.