An inductor carries a steady 2 A current. What is the back-EMF induced across it?
ALI — the inductance multiplied by the current
BZero, because the current is not changing
C½LI², converted to a voltage
Dμ₀N²A/ℓ multiplied by the current
The back-EMF is ε = −L(dI/dt). If the current is steady, dI/dt = 0, so ε = 0. An inductor does not oppose current itself — it opposes changes in current. A common misconception is that LI gives the back-EMF (it gives the total flux linkage), but voltage depends on the rate of change. Options A and C both confuse the energy/flux relationship with the EMF relationship.
Question 2 Multiple Choice
If you double the number of turns N in a solenoid while keeping length, cross-sectional area, and current constant, what happens to the inductance?
AIt doubles, since inductance is proportional to N
BIt triples
CIt quadruples, since inductance is proportional to N²
DIt remains unchanged, because inductance depends only on the physical geometry, not turn count
From L = μ₀N²A/ℓ, inductance scales as N². Doubling N gives L → μ₀(2N)²A/ℓ = 4μ₀N²A/ℓ — four times the original inductance. The N² dependence arises because each turn both produces more flux and links more of that flux: doubling turns doubles the flux and doubles how many turns 'see' it, giving a factor of 4. The misconception in option D is partly correct (geometry matters) but wrong to exclude turn count — N is part of the coil geometry.
Question 3 True / False
An inductor opposes the flow of current through it, similar to how a resistor limits current.
TTrue
FFalse
Answer: False
An inductor opposes changes in current (rate of change dI/dt), not current itself. An ideal inductor with a steady current flowing through it presents zero voltage drop — it behaves like a wire in DC steady state. A resistor, by contrast, always drops voltage proportional to the instantaneous current. The distinction matters: after a long time in a DC circuit, an inductor is a short circuit; a resistor never is.
Question 4 True / False
The energy stored in an inductor is proportional to the square of the current flowing through it.
TTrue
FFalse
Answer: True
The energy stored in an inductor is U = ½LI², which is quadratic in current I. This is analogous to the energy stored in a capacitor (U = ½CV²), which is quadratic in voltage. The energy is stored in the magnetic field distributed throughout the coil — not in the wire itself.
Question 5 Short Answer
When a switch in a highly inductive circuit is suddenly opened, the current is interrupted abruptly. Why does this produce a large voltage spike, and what physically causes it?
Think about your answer, then reveal below.
Model answer: Opening the switch forces the current to drop to zero nearly instantaneously, making dI/dt extremely large in magnitude. Since ε = −L(dI/dt), a large |dI/dt| produces a large back-EMF. The inductor 'insists' on maintaining its current and drives whatever voltage is necessary — potentially thousands of volts — across the switch gap, causing an arc.
This is Lenz's law in action: the induced EMF opposes the change (here, the decrease in current). The energy stored as ½LI² must go somewhere — it drives the arc discharge. This phenomenon explains why inductive loads (motors, solenoids, relay coils) require flyback diodes or snubber circuits to protect switching electronics from voltage spikes.