Question

In a magnetic field of $0.05T$ , area of a coil changes from $101c{m}^2$ to $100c{m}^2$ 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,
$q=\frac{B(\Delta A)}{R}$
$=\frac{\text{0.05}\times \left(101\times 10^{-4}-100\times 10^{-4}\right)}{2}$
$=\frac{\text{0.05}\times 1\times 10^{-4}}{2}$
$=\text{2.5}\times 10^{-6}C$