When a current of $2 \ A$ is passed through a coil of $100$ turns,the magnetic flux associated with it is $5 \times 10^{-5} \ Wb$. Find the self-inductance of the coil.

Self-induction of a solenoid cannot be increased by

$A$ coil of $N = 100$ turns carries a current $I = 5 \, A$ and creates a magnetic flux $\phi = 10^{-5} \, Wb$ per turn. The value of its inductance $L$ will be ...... $mH$.

Calculate the self-inductance of a very long solenoid.

$A$ copper wire is wound on a wooden frame,whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of $3$,keeping the number of turns of the coil per unit length of the frame the same,then the self-inductance of the coil:

$A$ circuit has self-inductance $L$ and carries a current $I$. To prevent sparking when the circuit is switched off,a capacitor which can withstand potential difference $V$ is used. The least capacitance is