Electromotive Force: Overview, Questions, Preparation

When we talk about electromagnetic induction, electromotive force or emf plays a crucial role. The emf is energy per unit electric charge, which is produced from an energy source.

An electromotive force, denoted by ‘ε’, is the electric potential produced by changing the magnetic field or using an electrochemical cell. Through this force, one terminal becomes negatively charged while the other gets positively charged. So, through this method, an emf is produced. 

ε= V + Ir 

ε = electromotive force

V = voltage of the cell

I = current across the circuit

r = Internal resistance of the cell

The emf is represented as the ratio of total work done on a unit charge. 

emf = Joules/ Coulombs 

So, the dimension is measured as M1 L2 T-3 I-1. 

An electromotive force can also be negative. When an inductor generates the emf that opposes the incoming power, the emf produced is negative as the direction of the emf is opposite to the incoming power. 

The chapter ‘Electromagnetic Induction’ will introduce you to the experiments conducted by Faraday and Henry, the magnetic flux, Faraday’s law of induction, Lenz’s law and conservation of energy, motional electromotive force, eddy currents, Inductance, and AC generator.

1. A solenoid with 15 turns/cm has a small loop of area 2 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid charges steadily from 2 A to 4 A in 0.1sec, what is the induced emf in the loop while the current is charging?

Ans-  no.of turns = 1500 turns/m

small loop area of solenoid A = 2 cm2

Change in current = di = 4 -2 = 2A

= dt = 0.1 s

Induced emf as per Faraday’s law

= e= dϕ/ dt  — (1)

where ϕ = induced flux through the small loop =BA — (2)

= B = μ0ni — (3)

= μ0 = 4π x 10-7 H/m

Reducing (1)

= e = d/dt (BA)

= e = Aμ0n x (di/dt)

= e = 2 x 10-2 x 4π x 10-7 x 1500 x (2/0.1)

= e = 7.54 x 10-6 V

2. A pair of adjacent coils have a mutual inductance of 1.5H. If the current in one coil changes from 0 to 20A in 0.5 secs, what is the change of flux linkage with the other coil?

Ans- 

= e= dϕ/ dt  = M e= di/ dt  

= dϕ = Mdi = 1.5 x (20-0)

= dϕ = 30 Wb

3. The current in a circuit falls from 5 A to 0 A in 0.1 secs. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.

Ans-

Change in current, ΔI = 0 - 5 = -5 A

Rate of change of current = ΔI /Δt

   = (-5/0.1) = -50 A/s

Avg. induced emf, ε = 200 V

= ε = -L x(ΔI /Δt)

= L = 4 H

Q: What is an electromotive force?

Q: What is the formula of electromotive force?

Q: Can an electromotive force produce a negative charge?

Q: Name the devices that measure emf and terminal voltage?

Q: State one difference between emf and terminal voltage.