English

Draw a Labelled Diagram of a Moving Coil Galvanometer. Describe Briefly Its Principle and Working. - Physics

Advertisements
Advertisements

Question

Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working.

Solution

Principle:

Its working is based on the fact that when a current carrying coil is placed in a magnetic field, it experiences a torque.

Working:

Suppose the coil PQRS is suspended freely in the magnetic field.

Let l = length PQ or RS of the coil

b = breadth QR or SP of the coil

n = number of turns in the coil

Area of each turn of the coil, A = l × b

Let B = strength of the magnetic field in which the coil is suspended

I = current passing through the coil in the direction PQRS

At any instant, let α be the angle that the normal drawn on the plane of the coil makes with the direction of magnetic field.

The rectangular coil carrying current, when placed in the magnetic field, experiences a torque whose magnitude is given by:

τ = nIBA sinα

Due to the deflecting torque, the coil rotates and the suspension wire gets twisted. A restoring torque is set up in the suspension wire.

Let θ be the twist produced in the phosphor bronze strip due to the rotation of the coil and K be the restoring torque per unit twist of the phosphor bronze strip. Then, we have:

Total restoring torque produced = kθ

In equilibrium position of the coil, we have:

Deflecting torque = Restoring torque

NIBA = kθ

or 

`I=k/(NBA)theta " or "Gtheta`

Here, 

`K/(NBA)=G=a["constant for a galvanometer"]`

It is known as galvanometer constant.

shaalaa.com
  Is there an error in this question or solution?
2013-2014 (March) All India Set 2

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

 

Why does a galvanometer show a momentary deflection at the time of charging or discharging a capacitor? Write the necessary expression to explain this observation.


A circular coil of 250 turns and diameter 18 cm carries a current of 12A. What is the magnitude of magnetic moment associated with the coil?


Two moving coil meters, M1 and M2 have the following particulars:

R1 = 10 Ω, N1 = 30,

A1 = 3.6 × 10–3 m2, B1 = 0.25 T

R2 = 14 Ω, N2 = 42,

A2 = 1.8 × 10–3 m2, B2 = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of

  1. current sensitivity and
  2. voltage sensitivity of M2 and M1.

Explain how moving coil galvanometer is converted into a voltmeter. Derive the necessary formula.


Explain, giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter?


Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a given range ?


The deflection in a moving coil galvanometer is ______.


When a galvanometer is shunted with a 4 Ω resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2 Ω wire. The further reduction (find the ratio of decrease in current to the previous current) in the deflection will be (the main current remains the same)


A galvanometer coil bas 500 turns and each tum has an average area of 3 × 10-4 m2. If a torque of 1.5 Nm is required to keep this coil parallel to a magnetic field when a current of 0.5 A is flowing through it, the strength of the field (in T) is ______.


A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10-divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×