AP Physics C - Multiple Choice Questions on Inductance

1. An air cored coil of self inductance L has N turns of fine insulated copper wire wound on a former of cross section area A. If the area and number of turns are doubled and the core is a medium of relative permeability 1000, the self inductance of the coil will be

(a) 8000 L (b) 4000 L (c) 8×10^–3 L (d) 4×10^–3 L (e) L

Self inductance of a coil is directly proportional to the area of cross section, relative permeability of the core and the square of the number of turns. The answer therefore is 8000 L.

2. A straight air cored solenoid has length 1 m, area of cross section 10 cm² and total number of turns 2000. If a current of 1 A flowing in it is reversed in 0.1 s, the average emf induced in it will be nearly

(a) 100V (b) 10 V (c) 1 V (d) 0.5 V (e) 0.1 V

The emf induced is given by

ε = –L(dI/dt),

where dI is the change in current in the time dt and L = μ₀n²Aℓ where ‘ℓ’ is the length of the solenoid, ‘A’ is its cross section area, ‘n’ is the number of turns per metre and μ₀ is the permeability of air (or free space) which you must remember as 4π×10^–7.

Ignoring the negative sign which is because of Lenz’s law, we have

ε = μ₀n²Aℓ (dI/dt)

Substituting, ε = 4π×10^–7×(2000)²×(10×10^–4)×1×[1– (–1)]/(0.1) volt.

Note that the change in the current is 2 A ( from 1 A to –1 A) since the current is reversed.

Therefore, ε = 0.1 (nearly).

3. In the circuit shown, the switch S is kept closed so that the current flowing in the circuit is the final steady value. When the switch S is opened, the time constant for the decay of current in the inductance is

(a) L/R

(b) L/5R

(c) 5L/4R

(d) 4L/5R

(e) L/3R

The current decays through the resistors R and 4R in series with L and therefore the time constant is L/(R+4R) = L/5R.

4. In the above question, suppose the resistance 4R is disconnected and is then connected between the inductance L and the negative terminal of the battery. The switch S is kept closed. If the switch is opened, the time constant for the decay of current in L is

(a) L/R

(b) infinity

(c) zero

(d) L/5R

(e) 5L/4R

This is a simple case which may however generate confusion in your mind. Note that when you open the switch, the resistance in series with L is infinite and the time constant will be L/∞ = 0.

Physically, the current becomes zero abruptly since there is no conducting path for the current to flow.

[Discharging of a charged capacitor through a resistor is an easy thing since the capacitor can retain the charge during the small time during which the capacitor is disconnected from the charging battery and then connected to the resistor, using a charge-discharge key. In the case of the LR circuit, for studying the current decay pattern, after establishing the steady final current through the circuit, you have to short circuit the leads connected to the battery and immediately disconnect the battery so that the battery is not damaged].

5. A small plane circular coil of radius ‘r’ having ‘n’ turns is placed at a distance ‘d’ (d >> r) from a straight vertical conductor. The coil and the straight conductor are contained in the same vertical plane. The mutual inductance between the coil and the straight conductor is

(a) μ₀ nr²/2πd

(b) π μ₀ nr²/2d

(c) μ₀ nr²/2

(d) μ₀ nr²/2π²d

(e) μ₀ nr²/2d

The mutual inductance is given by

M = Ф/I where I is the current in the straight conductor and Ф is the magnetic flux through the coil.

Since the coil is small, the magnetic field produced at the coil by the current in the straight conductor can be assumed to be constant over the entire area of the coil so that

Ф =nBA = n(μ₀I/2πd)(πr²)

Therefore, M = μ₀ nr²/2d

You will find similar multiple choice questions at physicsplus: Multiple Choice Questions (MCQ) involving Inductance