Electromagnetic Induction

Motional Electromotive Force

Did you know that the Electromotive force is essential for an electronic circuit to drive currents through the circuit? The electromotive force also like a charge pump. Let us learn more about them.

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Motional Electromotive Force

An emf induced by the motion of the conductor across the magnetic field is a motional electromotive force. The equation is given by E = -vLB.  This equation is true as long as the velocity, field, and length are mutually perpendicular. The minus sign associated with the Lenz’s law.

Motional Electromotive Force(Source: Exam.com)

For us to understand the motional electromotive force, let us make a particular setup. Let us take a rectangular coil, a metal rod of length L, moving with velocity V, through a magnetic field B. There is a magnetic field at some location.

Length, velocity and magnetic field should always be at a right angle with each other. The direction of the magnetic field is going inside. Assume the metal rod is frictionless that means there is no loss of energy due to friction and we apply a uniform magnetic field. The conductor rod is moved with a constant velocity and placed in the magnetic field.

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ΦB = Blx

But ‘x’ changes with time,

E = – \( \frac{dΦ_B}{dt} \) = – \( \frac{d }{dt} \) (Blx) = -Bl \( \frac{dx}{dt} \)

E = Blv

The induced emf Blv is motion electromotive force. So we produce emf by moving a conductor inside the uniform magnetic field. The power required to move a conductor rod in a magnetic field is,

P =  \( \frac{B² l² v²}{R} \)


  • B is the magnetic field,
  • l is the length of the conductor
  • v is the velocity of the conductor
  • R is the resistance

The magnetic flux associated with the coil is given by Φ = BA cos θ. We know that cos θ = 0, so Φ = BA. The motion of electromotive force can be further explained by Lorentz force which acts on free charge carriers. The Lorentz force on charge is:

F = qVB

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Motional Electromotive Force


Solved Questions For You

Q1. A coil having n turns and area A is initially placed with its plane normal to the magnetic field B. It is then rotated through 180º in 0.2 sec. The emf induced at the ends of the coils is

  1. 0.1 nAB
  2.  nAB
  3. 5 nAB
  4. 10 nAB

Answer: D. Total change in flux = ΔΦ = 2 nAB
Total time of change = Δt = 0.2s
Emf induced = \( \frac{ΔΦ}{Δt} \) = 10nAB

Q2.  A straight line conductor of length 0. 4m is moved with a speed of 7ms-1 perpendicular to a magnetic field of an intensity of 0.9wbm-2  The induced emf across the conductor is:

  1. 25.2 V
  2. 5.24 V
  3. 2.52 V
  4. 1.26 V

Answer: C. The induced emf across the conductor E= Blv
= 0.98 × 0.4 × 7 = 2.52V

Q.3 Two conducting rings of radii r and 2r move in opposite directions with velocities 2v and v respectively on a conducting surface S. There is a uniform magnetic field of magnitude B perpendicular to the plane of the rings. The potential difference between the highest points of the two rings is:

  1. Zero
  2. 2rvB
  3. 4rvB
  4. 8rvB

Answer: D. Replace the emf in the rings by the cells.
E1= B2r(2V) = 4Brv
E= B(4r)v  = 4Brv
V–  V= 8Brv

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