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# Electromagnetic Field – Galvanometer, Inductance

The electromagnetic field is a combination of the electric field and the magnetic field. The electric  field component is produced by the voltage on a current carrying conductor while the magnetic  field is produced by the flow of electric current. The two fields permeate all the space around a  current-carrying conductor.

## Galvanometer

When current flows through a current carrying conductor placed in a magnetic field, it experiences  a force. This principle is applied in the design of galvanometer. The force is:

• directly proportional to the current flowing in the conductor
• directly proportional to the magnetic field strength of the magnetic
• directly proportional to the length of the conductor

The magnitude of the force is given as

F=BILsinθ ——– (1)

Where;

B- is the strength of the magnetic field of the magnet

I -is the current flowing in the conductor

L -is the length of the conductor

ϴ- is the angle the conductor makes with the direction of the magnetic field

A moving coil galvanometer is an instrument used for detection of very small current. It consist of

• magnet
• a coil
• a soft iron core. This is place in the centre of the coil to concentrate the magnetic line of
• Pointer and a scale

When current flows through the moving coil galvanometer, its coil turns to give the pointer a deflection which is directly proportional to the current flowing in. The torque experienced by the coil is given as:

Γ=BANI ——– (2)

Where Γ is the torque, B is the magnetic field strength,A is area of the coil, N is the number of  turns in the coil and I is the current.

The galvanometer can be adapted to measure large current like ammeter or to measure voltage like voltmeter

### Conversion of Galvanometer to Ammeter

To convert a galvanometer to an ammeter, resistor of low resistance (called a shunt) is  connected in parallel to the galvanometer.

The voltage drop across the shunt is equal to the voltage drop across the galvanometer

Vs=Vg(I−Ig)Rg=IsRs

### Conversion of Galvanometer to Voltmeter

A galvanometer can be converted to voltmeter by connecting a resistor of high resistance in  series to it.

The same amount of current flows through the shunt and the galvanometer.

VgRg=V−VgRm

## Transformer

Transformers are used in the transmission of electricity.

Principle: Whenever there is change in the magnetic flux linking a conductor or there is relative  motion between a conductor and a magnet, electric current is induced.

The transformer consists of two coils of wire wound on opposite ends of a soft iron. When an

alternating current is flowing in one of the coil, current will be induced in the other. The  alternating current in the (primary ) coil generates a changing magnetic flux. In accordance to the Faraday’s law of electromagnetic induction,

Whenever there is a change in the magnetic flux linked with a circuit, an electromotive force is induced, the strength of which is proportional to the rate of change of the magnetic flux linking the circuit.

E=−∂Ø∂t

The magnitude of the induced emf E is

• Directly proportional to the rate of change of the flux – Directly proportional to the number of turns in the coil
• Cross sectional area of the coil.

The direction of the emf is given by the Lenz’s law:

This states that the direction of the induced emf in such that it tends to oppose the change that is causing it.

### Types of Inductance

#### 1. Self inductance

This occurs when there is a change in the current flowing in a coil. This alters the flux in the  coil thereby inducing an emf in the coil. This type of induction is called self induction. The  emf generated is given as:

E=LδIδt

The energy stored in the inductor is given as:

energy=12LI2

Varying the resistor R causes change in  the current in the circuit. Consequently, the flux in the coil C is altered and EMF is generated. This is called self inductance

1. Mutual inductance

This is when the change in current in a coil induced emf in another coil close by:

The induced emf is given as

E2=MδI1δt

M is the mutual inductance and it is given as:

M=μ0AN1N2I

For a transformer, the alternating current flowing in primary coil generates a changing flux link to the secondary coil. This induces a emf in the secondary coil. Below is the ideal (100% efficiency) transformer equation.

npns=IsIpnpns=VpVsnpns=VpVsIsIp

The coil that brings current flow into the transformer is called the primary coil. The coil through  which current flow out of the transformer is called the secondary coil

In practice, the efficiency of the transformer is never 100%because energy is lost in the  transformer due to:

1. Heating in the coil: this is as a result of the resistance in the primary and secondary coil of the transformer. It can be reduced by thick wire or wire with low resistance for the coils
2. Eddy current losses in the coil: eddy current are produced by the varying flux cutting the iron They consume power from the primary coil. It can be reduced by laminating the core
3. Hysteresis loss: This is wasted energy due to reversing of magnetization of core. It can be reduced by using a special alloy in the core of the primary coil or by using soft iron core.
4. Flux leakage: This arises because not all the lines of induction due to current in the primary coil pass entirely through the iron core of the secondary coil. It can be reduced by using  special form of coil winding

All these tend to reduce the efficiency of the transformer.

Transformer formula:

Efficiency of a transformer:

ɳ=power output at secondary coilpower input at primary coil×100ɳ=P0PI×100ɳ=IsVsIpVp×100

## Solved Examples

1. A step down transformer operates a 2000 volt line and supplies a current of 60 A. the turns ratio of the transformer is 1/24. Calculate

(i) the secondary voltage

(ii) the primary current

(iii) the power output if the transformer is 100% efficient

Solution:

Primary voltage, V= 2000 V,

Secondary current I= 60 A

Turn ratio for step down transformer nsnp=124

(Since it is a step down transformer, the nis bigger)

(i) nsnp=VsVp124=Vs2000Vs=242000=0.012V

(ii) nsnp=IpIs124=Ip60Ip=6024=2.5A

(iii) Power output P0=IsVsP0=60×0.012=0.72W

1. A transformer is designed to convert a 25 V supply to an output of 240 V. If the transformer is  90% efficient, calculate the current in the primary winding when the output terminal are  connected to a 240 V 80 W lamp.

Solution:

Primary voltage V= 25 V

Secondary voltage Vs = 240 V

Power output P= 80 W

Efficiency η =0.9

P0Pi=η80PI=0.9PI=800.9=88.9W

But PI=IPVP88.9=IP×25IP=88.925=3.6A

EVALUATION (POST YOUR ANSWERS USING THE QUESTION BOX BELOW FOR EVALUATION AND DISCUSSION):

1. Differentiate between self inductance and mutual inductance
2. List the factors that reduces the efficiency of a transformer
3. State the Faraday’s law of electromagnetic induction
4. Describe the process by which a galvanometer is converted to an ammeter.

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