Question

A magnetic field $B = 2t + 4{t^2}$(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 S.I., the electric charge that flow through the circular wire during $t = 0s $ to $t = 2s$ is:
A) $\dfrac{{6\pi {r^2}}}{R}$
B) $\dfrac{{20\pi {r^2}}}{R}$
C) $\dfrac{{32\pi {r^2}}}{R}$
D) $\dfrac{{48\pi {r^2}}}{R}$

Answer 1

Hint: Recall that the magnetic field is defined as the area around the magnetic material in which its effect can be felt. The moving charges or magnet produce a force of magnetism. It is important to remember that the magnetic field and the electric field occur simultaneously. Magnetic field is a vector quantity and has both magnitude and direction.

Complete step by step solution:
Given $B = 2t + 4{t^2}$---(i)
At $t = 0,{B_1} = 2 \times 0 + 4{(0)^2}$
The value of magnetic field becomes,
$ \Rightarrow {B_1} = 0$
Now at $t = 2s,{B_2} = 2 \times 2 + 4{(2)^2}$
The value of magnetic field becomes,
$ \Rightarrow {B_2} = 4 + 16 = 20Wb/{m^2}$
Suppose that the magnetic flux through the circuit with resistance R is changing by an amount of $\Delta \phi $in a time of $\Delta t$. The total quantity of electric charge passing through the circuit in time $\Delta t$ is given by
$\Delta Q = \dfrac{{\Delta \phi }}{R}$---(ii)
Where $\Delta \phi $ is the magnetic flux
R is the resistance of the circular loop
Substitute the formula of magnetic flux which is given by $\phi = BA\cos \theta $
$\Delta Q = \dfrac{{BA\cos \theta }}{R}$---(iii)
Where B is the magnetic field
A is the area of the circular loop$ = \pi {r^2}$
$\theta $ is the angle between perpendicular vector and magnetic field. In this case, the angle is 0 because they are in the same direction.
So the equation (iii) becomes,
$\Delta Q = \dfrac{{\pi {r^2}({B_2} - {B_1})}}{R}$
$\Delta Q = \dfrac{{\pi {r^2}(20 - 0)}}{R}$
$\Delta Q = \dfrac{{20\pi {r^2}}}{R}$

Option B is the right answer.

Note: It is important to remember that the direction of the magnetic field can be known by using Fleming’s right hand rule. As per this rule, wrap the fingers of the right hand around the conductor. The thumb will point in the direction of current. The fingers will curl in the direction of the magnetic field.