Question

The magnetic field \[B = 2{t^2} + 4{t^2}\] (where \[t\]\[ = \] time) is applied perpendicular to the plane of a circular wire of radius \[r\] and resistance\[R\]. If all the units are in SI the electric charge that flows through the circular wire during \[t = 0s\]to \[t = 2s\] is
A. \[\dfrac{{6\pi {r^2}}}{R}\]
B. \[\dfrac{{24\pi {r^2}}}{R}\]
C. \[\dfrac{{32\pi {r^2}}}{R}\]
D. \[\dfrac{{48\pi {r^2}}}{R}\]

Answer 1

Hint: Here, the magnetic field given is time dependent that is, it is varying with time. This time varying magnetic field induces an emf in the wire and from Faraday’s law, we have the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

Complete step by step answer:
Given, magnetic field, \[B = 2{t^2} + 4{t^2}\]
Radius of the circular wire \[ = r\]
And resistance of the circular wire\[ = R\]
We need to find the electric charge that flows through the wire during \[t = 0s\] to \[t = 2s\].
At \[t = 0s\], \[{B_1} = 2{(0)^2} + 4{(0)^2} \Rightarrow {B_1} = 0\]......................(1)
\[t = 2s\],\[{B_2} = 2{(2)^2} + 4{(2)^2} \Rightarrow {B_2} = 24\,{\text{Wb}}\,{{\text{m}}^{{\text{ - 2}}}}\]...................................(2)
As the magnetic field and the plane of the circular wire are perpendicular to each other so, there will be an induced emf in the wire due to varying magnetic field. The induced emf is given by,
\[\left| \varepsilon \right| = \dfrac{{\Delta {\phi _B}}}{{\Delta t}}\]..............................(3)
Where \[\Delta {\phi _B}\] is the change in the magnetic flux during the time interval \[\Delta t\].
The induced emf can also be written as,
\[\left| \varepsilon \right| = IR\].............................................(4)
where \[I\]is the current through the wire and \[R\]is the resistance of the wire.

Let \[\Delta Q\] be the charge flowing through the wire in the time interval\[\Delta t\]. So the current through the wire is, \[I = \dfrac{{\Delta Q}}{{\Delta t}}\]. Putting this value of \[I\]in equation (4), we get,
\[\left| \varepsilon \right| = \dfrac{{\Delta Q}}{{\Delta t}}R\].........................................(5)
Now, equating equations (3) and (5), we have,

\[\dfrac{{\Delta {\phi _B}}}{{\Delta t}} = \dfrac{{\Delta Q}}{{\Delta t}}R\]
\[ \Rightarrow \Delta {\phi _B} = \Delta QR\]
\[ \Rightarrow \Delta Q = \dfrac{{\Delta {\phi _B}}}{R}\]...........................................(6)
We have the formula for magnetic flux as, \[{\phi _B} = BA\cos \theta \] where \[B\] is the magnetic field and \[A\] is the area and \[\theta \]is the angle between the magnetic field vector and area vector of the wire. Here,\[\cos \theta = \cos 90 = 1\]and area of the circle of radius \[r\] is \[\pi {r^2}\]. Therefore, magnetic flux is
\[{\phi _B} = BA = B\pi {r^2}\]
Magnetic flux at time \[t = 0s\]is \[{\phi _{B1}} = {B_1}\pi {r^2} = 0\] [using equation (1)]
Magnetic flux at time \[t = 2s\]is \[{\phi _{B2}} = {B_2}\pi {r^2} = 24\pi {r^2}\,\] [using equation (2)]
Change in magnetic flux,\[\Delta {\phi _B} = {\phi _{B2}} - {\phi _{B1}} = 24\pi {r^2}\]........................... (7)
Now, putting the value of \[\Delta {\phi _B}\] from equation (7) in equation (6), we get
$\\Delta Q = \dfrac{{\Delta {\phi _B}}}{R} \\$
$ \Rightarrow \Delta Q = \dfrac{{24\pi {r^2}}}{R} \\$
Therefore charge flowing through the circular wire during the interval \[t = 0s\] to \[t = 2s\] is \[\Delta Q = \dfrac{{24\pi {r^2}}}{R}\]

So, the correct answer is “Option B”.

Note:
Here, the magnetic field is changing with time, that is the magnetic flux is changing, so there is an induced emf but if the magnetic field is constant then there will be no change in the magnetic field and there will be no induced emf. So while solving such problems, one should check whether the magnetic field is varying with time or is constant.