Question

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

• A. $2\cdot 5\times 10^{- 6}C$
• B. $2\times 10^{- 6}C$
• C. $10^{- 6}C$
• D. $8\times 10^{- 6}C$

Answer 1

Answer: (A)

$B=\text{0.05}T$
$A_{1}=101 \, \, \text{cm}^{2}=101\times 10^{- 4} \, \, \text{m}^{2}$
$A_{2}=100 \, \, \text{cm}^{2}=100\times 10^{- 4} \, \, \text{m}^{2}$
$R=2 \, \Omega$
Amount of charge flown is given by,
$q=\frac{B \left(\right. \Delta \, A \left.\right)}{R}$
$=\frac{\text{0.05} \times \left(1 0 1 \times 1 0^{- 4} - 1 0 0 \times 1 0^{- 4}\right)}{2}$
$=\frac{\text{0.05} \times 1 \times 1 0^{- 4}}{2}$
$=\text{2.5}\times 10^{- 6}C$