Question 131:
easy36. If the number of turns per unit length of a coil of solenoid is doubled, the self-inductance of the solenoid will: (1991)
Self-inductance of a solenoid is $L = \mu_0 n^2 A l$, where $n$ is the number of turns per unit length.
Thus, self-inductance is directly proportional to the square of turns per unit length ($L \propto n^2$).
When $n$ is doubled, $L' = (2n)^2 = 4L$, so it becomes four times.
Question 132:
easy37. The current in self inductance $L = 40 mH$ is to be increased uniformly from 1 amp to 11 amp in 4 milliseconds. The e.m.f. induced in inductor during process is (1990)
Induced e.m.f. is given by $e = L \frac{\Delta I}{\Delta t}$.
Given $L = 40 mH = 40 \times 10^{-3} H$, $\Delta I = 11 - 1 = 10 A$, and $\Delta t = 4 ms = 4 \times 10^{-3} s$.
$e = 40 \times 10^{-3} \times \frac{10}{4 \times 10^{-3}} = 100 volt$.
Question 133:
easy38. The magnetic potential energy stored in a certain inductor is $25 mJ$, when the current in the inductor is $60 mA$. This inductor is of inductance: (2018)
The magnetic energy stored is $U = \frac{1}{2} L I^2$.
Given $U = 25 mJ = 25 \times 10^{-3} J$ and $I = 60 mA = 60 \times 10^{-3} A$.
$L = \frac{2U}{I^2} = \frac{2 \times 25 \times 10^{-3}}{(60 \times 10^{-3})^2} = \frac{50 \times 10^{-3}}{3600 \times 10^{-6}} = \frac{50000}{3600} \approx 13.89 H$.
Question 134:
easy39. For a inductor coil $L = 0.04 H$, then work done by source to establish a current of $5 A$ in it is: (1999)
Work done to establish a current in an inductor is stored as magnetic energy: $W = \frac{1}{2} L I^2$.
Given $L = 0.04 H$ and $I = 5 A$.
$W = \frac{1}{2} \times 0.04 \times 5^2 = 0.02 \times 25 = 0.5 J$.
Question 135:
easy40. A 100 millihenry coil carries a current of $1 A$. Energy stored in its magnetic field is (1991)
Energy stored in the magnetic field of an inductor is $U = \frac{1}{2} L I^2$.
Here, $L = 100 mH = 0.1 H$ and $I = 1 A$.
$U = \frac{1}{2} \times 0.1 \times 1^2 = 0.05 J$.
Question 136:
easy41. An inductor may store energy in (1990)
When a current flows through an inductor, a magnetic field is established around and within the coil.
The work done in establishing this current is stored in the magnetic field as magnetic potential energy ($U = \frac{1}{2} L I^2$).
Therefore, an inductor stores energy in its magnetic field.
Question 137:
easyA transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are:
(2014)
Input power is $P_{in} = 3000 W$, so primary current $I_p = \frac{P_{in}}{V_p} = \frac{3000}{200} = 15 A$.
Transformer efficiency is $\eta = \frac{P_{out}}{P_{in}} = \frac{V_s I_s}{P_{in}}$.
Substitute values: $0.90 = \frac{V_s \times 6}{3000}$, which gives secondary voltage $V_s = \frac{2700}{6} = 450 V$.