Thursday, December 20, 2007

Multiple Choice Questions (MCQ) involving Inductance

(1) A long straight solenoid of cross section area10 cm2 has 10 turns per cm. A short 50 turn coaxial coil of cross section area 2 cm2 is fixed inside the solenoid at the middle. The mutual inductance of the solenoid and the coil is (in micro henry)

(a) 400 (b) 40 (c) 4π (d) π (e) ) 0.4π

Mutual inductance between an infinitely long straight solenoid and a short secondary coil placed coaxially inside at the middle is given by

M = μ0nNA where μ0 is the magnetic permeability of free space (or air), n is the number of turns per metre of the solenoid, N is the total number of turns in the secondary coil and A is the cross section area of the secondary coil. (If the secondary coil is outside the solenoid, the cross section area of the solenoid will appear in place of the area of the secondary coil).

[Note that the mutual inductance as given by the above expression is the magnetic flux (linked with the secondary coil) per unit current in the solenoid: M = NBA where B is the magnetic field (μ0n×1) produced by unit current in the solenoid]

Substituting for the known quantities in the expression for mutual inductance, we have

M = 4π×10–7×1000 ×50 ×(2×10–4) = 4π×10–6 H = μH.

(2) The secondary coil in the above question is moved with uniform velocity coaxially over a distance of 5 cm in 2 seconds when a current of 2 ampere flows through the solenoid. The emf induced in the coil during the motion is

(a) 200π μV (b) 100π μV (c) 20π μV (d) 10π μV (e) zero

No emf will be induced in the coil since the same flux is linked with the coil throughout its motion and there is no flux change.

(3) In question No.1, suppose the solenoid carries a current of 4 A. If this current is switched off in 100 ms, the emf induced in the secondary coil will be

(a) 0.2π μV (b) 0.4π μV (c) π mV (d) 0.08π mV (e) 0.16π mV

We have ε = –M(dI/dt) = = 4π×10–6 [(4 – 0)/(100×10–3)] = 160π×10–6 volt = 0.16π mV.

4. A battery of emf V volt is connected in series with a coil of inductance L and resistance R at the instant t = 0. The current in the circuit when t = 2τ where τ is the time constant of the circuit is (base of natural logarithm = e)

(a) (V/R)[1 (1/e2)]

(b) (V/R (1/e2)

(c) (V/R)[1 (1/e)]

(d) (V/R)(1/e)]

(e) 2e(V/R)

The expression for the growth of current (with time) in an LR circuit is

I = I0[1– e–Rt/L]

where I0 is the final steady (maximum possible) current which is equal to V/R.

The time constant of the LR circuit is L/R. After 2 time constants, the current will be

I = I0[1– e–R× 2(L/R)/L] = (V/R) [1– e– 2] = (V/R)[1– (1/e2)]

You will find more questions (with solution) at AP Physics Resources: AP Physics C - Multiple Choice Questions on Inductance

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