(a) What is meant by the statement that a solenoid has an inductance of 2.0 H?
A 2.0 H solenoid is connected in series with a resistor, so that the total resistance is 0.50 Ω, to a 2.0 V DC supply. Sketh the graph of current against time when the crrent is switched on.
(i) the final current,
(ii) the initial rate of change of current with time, and
(iii) the rate of change of current with time when the current is 2.0 A?
Explain why an EMF greatly in excess of 2.0 V will be produced when the current is switched off.
(b) A long air-cored solenoid has 1000 turns of wire per metre and a cross-sectional area of 8.0 cm2. A secondary coil, of 2000 turns, is wound around its centre and connected to a ballistic galvanometer, the total resistance of coil and galvanometer being 60 Ω. The sensitivity of the galvanometer is 2.0 divisions per microcoulomb. If a current of 4.0 A in the primary solenoid were switched off, what would be the deflection of the galvanometer?
Answers: (a) (i) 4.0 A, (ii) 1.0 A s-1, (iii) 0.5 A s-1, (b) 268 divisions.
2. The attempt at a solution
(a) (i) I = V / R = 2 / 0.5 = 4 A.
(a) (ii) E = L [d I / d t] → d I / d t = 2 / 2 = 1 A s-1.
(a) (iii) This one can't get right. E = L [d I / d t] → 2 = 2 [2 / d t] → 2 d t = 4 → d t = 2 s. d I / d t = 2 / 2 = 1 A s-1.
I also tried: P = I E = 2 * 2 = 4 W. P = L I [d I / d t] → 4 = 2 * 2 [d I / d t] → d I / d t = 1 A s-1.
Maybe it should be like: 2 A is half the 4 A, so 1 A s-1 / 2 = 0.5 A s-1?
(b) As I understand we need to find θ: θ = a N A B / R, where a is the division per microcoulomb, N = turns, A = cross-sectional area, B = magnetic field and R = resistance. If this formula is correct, I'm not sure how to find B. Maybe something like: B = μ0 n I = 4 π * 10-7 * 1000 * 4 = 5 * 10-3 T. Then plug into the formula: θ = 2 * 2000 * (8 * 10-4) * (5 * 10-3) / 60 = 2.68 * 10-4. It has the 2.68 in it as in the answer 268 divisions. But still I'm not sure whether I'm in the right direction.