Question 41:
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 42:
easy42. 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: (2021)
Let current $I_1$ pass through the outer loop of radius $R_1$. The magnetic field at its centre is $B_1 = \frac{\mu_0 I_1}{2 R_1}$.
Since $R_1 >> R_2$, this field is nearly uniform over the inner loop of area $A_2 = \pi R_2^2$.
Flux linked is $\Phi_2 = B_1 A_2 = \frac{\mu_0 I_1 \pi R_2^2}{2 R_1} = M I_1$, hence $M \propto \frac{R_2^2}{R_1}$.
Question 43:
easy43. A long solenoid of diameter $0.1 m$ has $2 \times 10^4$ turns per metre. At the centre of solenoid, a coil of 100 turns and radius $0.01 m$ is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to $0 A$ from $4 A$ in $0.05 s$. If the resistance of the coil is $10 \pi^2 \Omega$, the total charge flowing through the coil during this time is: (2017-Delhi)
Total charge flowing is $q = \frac{\Delta \Phi}{R} = \frac{N_{coil} (\Delta B) A_{coil}}{R}$.
Change in magnetic field is $\Delta B = \mu_0 n \Delta I = (4\pi \times 10^{-7}) \times (2 \times 10^4) \times 4 = 32\pi \times 10^{-3} T$.
$q = \frac{100 \times (32\pi \times 10^{-3}) \times [\pi (0.01)^2]}{10 \pi^2} = \frac{100 \times 32\pi^2 \times 10^{-7}}{10 \pi^2} = 32 \times 10^{-6} C = 32 \mu C$.
Question 44:
easy44. Two coils of self-inductances $2 mH$ and $8 mH$ are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is: (2006)
For complete coupling between two coils, the coupling coefficient is $k = 1$.
The mutual inductance is given by $M = k \sqrt{L_1 L_2}$.
$M = 1 \times \sqrt{2 mH \times 8 mH} = \sqrt{16} = 4 mH$.
Question 45:
easy45. Two coil have a mutual inductance $0.005 H$. The current changes in first coil according to equation $I = I_0 \sin \omega t$ where $I_0 = 2 A$ and $\omega = 100pi rad/sec$. The maximum value of emf in second coil is: (1998)
Induced emf in the second coil is $e = -M \frac{dI}{dt} = -M \frac{d}{dt}(I_0 \sin \omega t) = -M I_0 \omega \cos \omega t$.
The maximum (peak) value of induced emf is $e_{max} = M I_0 \omega$.
$e_{max} = 0.005 \times 2 \times 100\pi = 0.01 \times 100\pi = \pi V$.
Question 46:
easy32. A long solenoid has 1000 turns. When a current of $4 A$ flows through it, the magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} Wb$. The self inductance of the solenoid is: (2016 – I)
Total magnetic flux linked with the solenoid is $N \Phi = L I$.
Given $N = 1000$, $\Phi = 4 \times 10^{-3} Wb$, and $I = 4 A$.
$L = \frac{N \Phi}{I} = \frac{1000 \times 4 \times 10^{-3}}{4} = 1 H$.
Question 47:
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 48:
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 49:
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 50:
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.