The self inductance of a coil is L. Keeping the length and area same, the number of turns in the coil is increased to four times. The self inductance of the coil will now be 

1. $\frac{1}{4}L$

2. L

3. 4 L

4. 16 L

A coil has an inductance of 2.5 H and a resistance of 0.5 r. If the coil is suddenly connected across a 6.0 volt battery, then the time required for the current to rise 0.63 of its final value is 

1. 3.5 sec

2. 4.0 sec

3. 4.5 sec

4. 5.0 sec

Pure inductance of 3.0 H is connected as shown below. The equivalent inductance of the circuit is

1. 1 H

2. 2 H

3. 3 H

4. 9 H

If a current of 10 A flows in one second through a coil, and the induced e.m.f. is 10 V, then the self-inductance of the coil is 

1. $\frac{2}{5}H$

2. $\frac{4}{5}H$

3. $\frac{5}{4}H$

4. 1 H

The adjoining figure shows two bulbs \(B_1\) and \(B_2\) resistor \(R\) and an inductor \(L\). When the switch \(S\) is turned off 

An inductance L and a resistance R are first connected to a battery. After some time the battery is disconnected but L and R remain connected in a closed circuit. Then the current reduces to 37% of its initial value in time ?

1. RL sec

2. $\frac{R}{L}sec$

3. $\frac{L}{R}sec$

4. $\frac{1}{LR}sec$

In an LR-circuit, the time constant is that time in which current grows from zero to the value (where I₀ is the steady-state current) 

1. 0.63 I₀

2. 0.50 I₀

3. 0.37 I₀

4. I₀

In the figure magnetic energy stored in the coil is 

1. Zero

2. Infinite

3. 25 joules

4. None of the above

A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity ω. The induced e.m.f. between the two ends is 

1. $\frac{1}{2}B\omega l^2$

2. $\frac{3}{4}B\omega l^2$

3. $B\omega l^2$

4. $2B\omega l^2$

Two conducting circular loops of radii \(R_1\) and \(R_2\) are placed in the same plane with their centres coinciding. If \(R_1>>R_2\), the mutual inductance \(M\) between them will be directly proportional to: