Class 12 Physics
Moving Charges and Magnetism
Moving Coil Galvanometer:
- Moving coil galvanometer is an electromagnetic device that can measure small values of current. It is also known as Weston galvanometer.
- It works on the principle that when a current loop is placed in an external magnetic field, it experiences torque, and the value of torque can be changed by changing the current in the loop
- Moving coil galvanometer consists of permanent horse-shoe magnets, coil, soft iron core, pivoted spring, non-metallic frame, scale and pointer
- We know that a current loop having N number of turns,and the cross sectional area A, carrying current i, when placed in and along the direction of external magnetic field B, experiences a torque given by:
ԏ = NiAB
The pivoted spring of spring constant k would oppose the above torque with restoring torque C given by: C = kΦ
Here, Φ is the angular deflection of spring
- Both, the torque, and the restoring torque would be equal:
kΦ = NiAB
Φ = NiAB/k
- In the above equation, except for current, every other quantity on the right hand side is constant for a galvanometer, hence:
Φ ∝ i
- So, the angular deflection Φ produced in the pointer could be measured in terms of current in the scale calibrated on the basis of above equations.
- To use galvanometer as an ammeter (to measure higher values of current), we need to connect a shunt wire, with very small resistance(Rs), in parallel with the galvanometer (which have very low resistance of Rg
Equivalent resistance R of ammeter will be:
R = RgRs/(Rg + Rs)
Rg ˃˃ Rs
∴ R = RgRs/Rg = Rs
- So, the equivalent resistance of ammeter is very less, which is a must for sensitivity of ammeter to be higher. Also, most of the current will pass through the shunt, thus protecting the galvanometer from any damage.
- Ammeter is connected in series with the circuit where current is to be measured
- Current sensitivity(deflection per unit current) of galvanometer is given by:
Φ/i = NAB/k
- To use galvanometer as a voltmeter, we need to connect a wire, with very high resistance(Rw˃˃Rg), in series with the galvanometer to ensure that our voltmeter equivalent resistance is high and so that it will draw a very small current. Equivalent resistance will be given by:
R = Rg + Rw = Rw
Voltmeter is connected in parallel with the circuit where voltage is to be measured
- Voltage sensitivity (deflection per unit voltage) of galvanometer is given by:
Φ/V = NAB/(kR)
Numerical Problem:
1) 2 moving coil galvanometers M1 and M2(having the same spring constants), have the following specifications: R1 = 10Ω, N1 = 30, A1 = 3.6×10-3m2, B1 = 0.25T
R2 = 14Ω, N2 = 42, A2 = 1.8×10-3m2, B2 = 0.5T
Find the ratios of current and voltage sensitivities between the 2 galvanometers.
Solution: current sensitivity is given by:
Φ/i = NAB/k
(Φ/i)1 = (30×3.6×10-3×0.25)/k
(Φ/i)2 = (42×1.8×10-3×0.5)/k
∴(Φ/i)1/(Φ/i)2 = (30×3.6×10-3×0.25)/(42×1.8×10-3×0.5) = 5/7 (ans)
Voltage sensitivity is given by:
Φ/V = NAB/(KR)
(Φ/V)1 = (30×3.6×10-3×0.25)/(k×10)
(Φ/V)2 = (42×1.8×10-3×0.5)/(k×14)
∴ (Φ/V)1/(Φ/V)2 = (30×3.6×10-3×0.25×14)/(42×1.8×10-3×0.5×10) = 1 (ans)
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