Question

In a magnetic field of $0.05\,T$ area of the coil changes from $101\,c{m^2}\,to\,100c{m^2}$ without changing the resistance which is $2\Omega $. The amount of charge flow during the period is
A. $2.5 \times {10^{ - 6}}C$
B. $2 \times {10^{ - 6}}C$
C. $10 \times {10^{ - 6}}C$
D. $8 \times {10^{ - 6}}C$

Answer 1

Hint:First we will calculate the change in area of the coil without changing the resistance. We know the magnetic field now applying change in flux formula we get the flux linked on the coil. Now the resistance is known to us. Now applying the change flux and resiatbec relation formula we can find the charge flow through it.

Formula used:
$\Delta \phi = B\Delta A$
Where, $B$ = Magnetic Field and $A$ = Area.

Complete step by step answer:
As per the given problem there is a magnetic field of $0.05\,T$ area of the coil changes from $101\,c{m^2}\,to\,100c{m^2}$ without changing the resistance which is $2\Omega $.We need to calculate the amount of charge flow during this period.As the area change from $101\,c{m^2}\,to\,100\,c{m^2}$,
$\Delta A = \left( {101 - 100} \right)c{m^2}$
$ \Rightarrow \Delta A = 1c{m^2}$
Where $\Delta A$ is the change in area of the coil.
We know,
$B = 0.05T$

Now applying change in flux formula we will get,
$\Delta \phi = B\Delta A$
Putting the given value in the flux formula we will get,
$\Delta \phi = 0.05T \times 1c{m^2}$
Convert the centimeter into meter so as to get the flux in its SI unit that is Wb.
$\Delta \phi = 0.05T \times {10^{ - 4}}{m^2}$
$ \Rightarrow \Delta \phi = 5 \times {10^{ - 6}}{\text{Wb}}$

Now applying change flow formula we will get,
$q = \dfrac{{\Delta \phi }}{R}$
We know,
$R = 2\Omega $
Now putting the known values in the change formula we will get,
$q = \dfrac{{5 \times {{10}^{ - 6}}{\text{Wb}}}}{{2\Omega }}$
$ \therefore q = 2.5 \times {10^{ - 6}}{\text{C}}$

Therefore the correct option is $\left( A \right)$.

Note:When a coil moves in a magnetic field it will induce current in the coil and as the number of magnetic field lines passes through the coil it will generate a flux around it.Also keep in mind magnetic flux linked in a coil can be measured using a magnetometer.