Solveeit Logo
Login

Question

Question: In a magnetic field of \(0.05\,T\) area of the coil changes from \(101\,c{m^2}\,to\,100c{m^2}\) with...

In a magnetic field of 0.05 T0.05\,T area of the coil changes from 101 cm2 to 100cm2101\,c{m^2}\,to\,100c{m^2} without changing the resistance which is 2Ω2\Omega . The amount of charge flow during the period is
A. 2.5×10−6C2.5 \times {10^{ - 6}}C
B. 2×10−6C2 \times {10^{ - 6}}C
C. 10×10−6C10 \times {10^{ - 6}}C
D. 8×10−6C8 \times {10^{ - 6}}C

Explanation

Solution

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:
Δϕ=BΔA\Delta \phi = B\Delta A
Where, BB = Magnetic Field and AA = Area.

Complete step by step answer:
As per the given problem there is a magnetic field of 0.05 T0.05\,T area of the coil changes from 101 cm2 to 100cm2101\,c{m^2}\,to\,100c{m^2} without changing the resistance which is 2Ω2\Omega .We need to calculate the amount of charge flow during this period.As the area change from 101 cm2 to 100 cm2101\,c{m^2}\,to\,100\,c{m^2},
ΔA=(101−100)cm2\Delta A = \left( {101 - 100} \right)c{m^2}
⇒ΔA=1cm2\Rightarrow \Delta A = 1c{m^2}
Where ΔA\Delta A is the change in area of the coil.
We know,
B=0.05TB = 0.05T

Now applying change in flux formula we will get,
Δϕ=BΔA\Delta \phi = B\Delta A
Putting the given value in the flux formula we will get,
Δϕ=0.05T×1cm2\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.
Δϕ=0.05T×10−4m2\Delta \phi = 0.05T \times {10^{ - 4}}{m^2}
⇒Δϕ=5×10−6Wb\Rightarrow \Delta \phi = 5 \times {10^{ - 6}}{\text{Wb}}

Now applying change flow formula we will get,
q=ΔϕRq = \dfrac{{\Delta \phi }}{R}
We know,
R=2ΩR = 2\Omega
Now putting the known values in the change formula we will get,
q=5×10−6Wb2Ωq = \dfrac{{5 \times {{10}^{ - 6}}{\text{Wb}}}}{{2\Omega }}
∴q=2.5×10−6C\therefore q = 2.5 \times {10^{ - 6}}{\text{C}}

Therefore the correct option is (A)\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.