Back Emf

• Self-induced Emf is also known as back Emf.
• Back Emf tries to oppose change in the current.It tries to bring back the current.
• This implies the current needs to do work against back Emf.
• The work done by the current is stored as magnetic potential energy.
  • In a coil there is increase in the current as a result there is change in the magnetic flux because of that there is induced Emf.
  • This induced Emftry to oppose the change in the current.
  • The current will do some work to oppose the back Emf.
  • This work done is stored as magnetic potential energy.
• Mathematically:-
  • Work done = Potential Energy

• Note: - Self-inductance acts like inertia.

Problem:- An air-cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10^–3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Answer:- Length of the solenoid, l = 30 cm = 0.3 m

Area of cross-section, A = 25cm² =25x10^-4m²

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Current flows for time, t = 10^-3 s

Average back emf, e= (dΦ/dt)   … (i)

Where,dΦ = Change in flux

= NAB … (2)

Where,

B = Magnetic field strength

= μ₀ (NI/l_ength)

Where,

μ₀= Permeability of free space =4nx10s^-7xTmA^-1

Using equations (2) and (3) in equation (1), we get

e= (μ₀N₂IA)/ (t l_ength)

= (4x3.14x10^-7x (500)²x2.5x25x10^-4)/ (3.0x10^-3)

=6.5V

Hence, the average back emf induced in the solenoid is 6.5 V.