Question 31:
easy33. A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} Tm^2$. The self inductance of the solenoid is: (2008)
Total flux linked is given by $N \Phi = L I$.
Here, $N = 500$, $\Phi = 4 \times 10^{-3} Wb$ (or $T m^2$), and $I = 2 A$.
$L = \frac{N \Phi}{I} = \frac{500 \times 4 \times 10^{-3}}{2} = 1.0 henry$.
Question 32:
easy34. If N is the number of turns in a coil, the value of self inductance varies as (1993)
Self-inductance of a coil/solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$.
This relation clearly shows that self-inductance is directly proportional to the square of the number of turns.
Therefore, $L \propto N^2$.
Question 33:
easy35. What is the self-inductance of a coil which produces $5 V$ when the current changes from 3 ampere to 2 ampere in one millisecond? (1993)
Induced emf is given by $e = L \left|\frac{\Delta I}{\Delta t}\right|$.
Here $e = 5 V$, $|\Delta I| = |2 - 3| = 1 A$, and $\Delta t = 1 ms = 10^{-3} s$.
$L = \frac{e}{|\Delta I / \Delta t|} = \frac{5}{1 / 10^{-3}} = 5 \times 10^{-3} H = 5 mili-henry$.
Question 34:
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 35:
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 36:
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 37:
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 38:
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 $V_s = \frac{2700}{6} = 450 V$.