The self inductance of a toroid is :
A varying current in a coil changes from 10 amp to zero in 0.5 sec. If average EMF is induced in
the coil is 220 volts, the self inductance of coil is :
How much length of a very thin wire is required to obtain a solenoid of length l0 and inductance L.
What is the mutual inductance of a two-loop system as shown with centre separation l ?

A coil resistance 20 Ω and inductance 5 H is connected with a 100 V battery. Energy stored in the coil will be :
Energy stored = 62.5 J
A current of \(2\text{ A}\) is increasing at a rate of \(4\text{ A/s}\) through a coil of inductance \(2\text{ H}\). The energy stored in the inductor per unit time in given instant is:
Formula for rate of change of energy in an inductor is \(\frac{dU}{dt} = LI\frac{dI}{dt}\). Given \(L = 2\text{ H}\), \(I = 2\text{ A}\), and \(\frac{dI}{dt} = 4\text{ A/s}\), we get \(\frac{dU}{dt} = 2 \times 2 \times 4 = 16\text{ J/s}\).
A coil of \(\text{Cu}\) wire (radius \(r\), self inductance \(L\)) is bent in two concentric turns each having radius \(\frac{r}{2}\). The self-inductance now is:
Self-inductance of a coil is \(L \propto N^2 r\). When bent into \(2\) turns of radius \(r/2\), new self-inductance is \(L' \propto (2)^2 (r/2) = 2 L\).
Assertion (A): The self inductance of a solenoid can be increased by decreasing length if number of turns are fixed.
Reason (R): Self inductance of a solenoid is directly proportional to current passing through it.
Self inductance of a solenoid is given by \(L = \frac{\mu_0 N^2 A}{l}\). So, Assertion (A) is true as \(L\) is inversely proportional to \(l\). Self inductance \(L\) is a property of the coil's geometry and material, not dependent on current. So, Reason (R) is false. Thus, (A) is true but (R) is false.