Q. In a coil rate of change of area is $5\ m^{2}/milli\ second$ and current between becomes 1 $amp$ from $2\ amp$ in $2 \times 10^{- 3}\ sec$. If magnitude of field is $1\ tesla$ inductance of the coil is
A) $2\ H$
B) $5\ H$
C) $20\ H$
D) $10\ H$
Explanation
(d) $N\phi = Li \Rightarrow \frac{Nd\phi}{dt} = \frac{Ldi}{dt} \Rightarrow NB\frac{dA}{dt} = \frac{Ldi}{dt}$
$\Rightarrow \frac{1 \times 1 \times 5}{10^{- 3}} = L \times \left( \frac{2 - 1}{2 \times 10^{- 3}} \right) \Rightarrow L = 10H$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.