Class 12 Physics Electromagnetic Induction | Back Emf |

**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 cm

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

Area of cross-section, A = 25cm^{2} =25x10^{-4}m^{2}

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

= μ_{0 }(NI/l_{ength})

Where,

μ_{0}= Permeability of free space =4nx10s^{-7}xTmA^{-1}

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

e= (μ_{0}N_{2}IA)/ (t l_{ength})

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

=6.5V

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

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