Practice free Electromagnetic Induction (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
According to the context, what does 'L' represent primarily in the energy storage of an inductor?
The NCERT text explains this clearly: 'L is analogous to m (i.e., L is electrical inertia and opposes growth and decay of current in the circuit).' This highlights L's role as a measure of electrical inertia.
For a long solenoid, the magnetic energy stored per unit volume ($u_B$) depends on which of the following?
The formula derived is $u_B = \frac{B^2}{2\mu_0}$. This implies that the magnetic energy per unit volume depends solely on the magnetic field B and the permeability of free space $\mu_0$ (assuming a vacuum core or air core solenoid). Although B itself depends on current and solenoid properties, the question asks what $u_B$ depends on directly from the given formula.
The expression $W = \frac{1}{2} L I^2$ for energy stored in an inductor is valid under what condition?
The NCERT text explicitly states: 'If we ignore the resistive losses and consider only inductive effect, then using Eq. (6.14), $\frac{dW}{dt} = I L \frac{dI}{dt}$.' The subsequent integration to find W leads to $W = \frac{1}{2} L I^2$. Therefore, this expression assumes negligible resistive losses.
The magnetic energy stored in an inductor of inductance $4\ \mu\text{H}$ carrying a current of $2\ \text{A}$ is:
$U = \tfrac{1}{2} L I^2 = \tfrac{1}{2} (4 \times 10^{-6})(2)^2 = 8\ \mu\text{J}$.
In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:
Lenz's law: solenoid-1 sees N-flux decreasing (current $A \to B$); solenoid-2 sees S-flux increasing (current $D \to C$).
AB is a part of an electrical circuit (see figure). The potential difference $V_A - V_B$, at the instant when current $i = 2$ A and is increasing at a rate of 1 amp/second is:
Traversing A→B in the current direction the drops add: $V_A - V_B = L\dfrac{di}{dt} + \varepsilon + iR = (1)(1) + 5 + (2)(2) = 10\ \text{V}$.
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