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Physics Question on Magnetic Field

In a magnetic field of 0.05 T0.05\, T area of coil changes from 101cm2101 cm ^{2} to 100 cm2100 \,cm ^{2} without changing the resistance which is 2 Ω2 \,\Omega. The amount of charge that flow during this period is

A

2.5×10−6C2.5 \times 10^{-6} C

B

2×10−6C2 \times 10^{-6} C

C

10−6C10^{-6} C

D

8×10−6C8 \times 10^{-6} C

Answer

2.5×10−6C2.5 \times 10^{-6} C

Explanation

Solution

Given, B=0.5TB=0.5 T A1=101 cm2=101×10−4 m2A_{1}=101\, cm ^{2}=101 \times 10^{-4}\, m ^{2} A2=100 cm2=100×10−4 m2A_{2}=100\, cm ^{2}=100 \times 10^{-4} \,m ^{2} R=2ΩR=2 \Omega Amount of charge q=BΔARq=\frac{B \Delta A}{R} =0.05×(101×10−4−100×10−4)2=\frac{0.05 \times\left(101 \times 10^{-4}-100 \times 10^{-4}\right)}{2} =0.05×1×10−42=2.5×10−6 C=\frac{0.05 \times 1 \times 10^{-4}}{2}=2.5 \times 10^{-6} \,C