Homework Help: Two moving-coil galvanometers

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1. Oct 13, 2016

moenste

1. The problem statement, all variables and given/known data
(a) A moving-coil galvanometer consists of a rectangular coil of N turns each of area A suspended in a radial magnetic field of flux density B. Derive an expression for the torque on the coil when a current I passes through it. (You may assume the expression for the force on a current-carrying conductor in a magnetic field.)

(b) If the coil is suspended by a torsion wire for which the couple per unit twist is C, show that the instrumet will have a linear scale.

(c) How may the current sensitivity of the instrument be made as large as possible? What practical considerations limit the current sensitivity?

(d) Two galvanometers, which are otherwise identical, are fitted with different coils. One has a coil of 50 turns and a resistance of 10 ohms while the other has 500 turns and a resistance of 600 ohms. What is the ratio of the deflections when each is connected in turn to a cell of EMF 2.5 V and internal resistance 50 ohms?

2. The attempt at a solution
(a) T = B I A N, where T = torque, B = field of flux density, I = current, N = number of turns.

(b) T = C θ, where C is the torsion constant of the suspension. B I A N = C θ → θ = B I A N / C. The deflection of the coil, θ, is proportional to the current through it and therefore the instrument can be calibrated with a linear scale.

(c) Current sensitivity SI = θ / I and SI = B A N / C. So, to have large current sensitivity B, A and N must be large and C must be small.
• B must be large. Achieved by using a narrow air gap and a strong permanent magnet, also the instrument should is not be influenced by external magnetic fields like the Earth field.
• N must be large. There is an upper limit to N because the coil must be narrow enough to fit in the air gap.
• A must be large. There is an upper limit to A because it must not be so large that the instrument is unwieldy.
• C must be small. Weak suspension is required, but not too weak, because it would cause the coil to swing about its equilibrium position for long time.
(d) (B I1 A N1 / C) * (C / B I2 A N2) = I1 N1 / I2 N2.

I1 = V / (r + R1) = 2.5 / (50 + 10) = 0.042 A.
I2 = V / (r + R2) = 2.5 / (50 + 600) = 3.8 * 10-3 A.

I1 N1 / I2 N2 = (0.042 * 50) / [(3.8 * 10-3) * 500] = 1.083.

So, 1.083 : 1.

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Are there any mistakes here?

2. Oct 14, 2016

andrevdh

I get a different answer for (d) : I don't think you should add the turns to the internal resistances - just use the internal r's for the current.

3. Oct 14, 2016

moenste

We'll get I = V / r = 2.5 / 50 = 0.05 A.

I N1 / I N2 = (0.05 * 50) / (0.05 * 500) = 50 / 500 = 0.1.

Don't quite see how it fits the answer 1.08 : 1...

4. Oct 14, 2016

andrevdh

??? in your post the one is 10 ohm and the other 600 ohm ???

5. Oct 14, 2016

andrevdh

Aaah, sorry I forgot to add the internal resistance of the cell!

6. Oct 14, 2016

moenste

I just re-calculated (my calculations) and still got the same answer.

I assumed that the galvanometers and the battery are each connected in series, so I summed the resistance and calculated each current for the two situations.

Galvanometer 1: N1 = 50 turns, R1 = 10 Ω.
Galvanometer 2: N2 = 500 turns, R2 = 600 Ω.
Battery 1: E1 = 2.5 V, r1 = 50 Ω.
Battery 2: E2 = 2.5 V, r2 = 50 Ω.

7. Oct 14, 2016

andrevdh

Yes, you are (still) right 65/60