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Maharashtra State BoardHSC Science (Electronics) 12th Standard Board Exam
प्रश्न पत्र२५४
पाठ्यपुस्तक उत्तर१०४७७
MCQ ऑनलाइन अभ्यास परीक्षा६०
महत्वपूर्ण प्रश्न एवं उत्तर७८४१
अवधारणा स्पष्टीकरण और वीडियो७९३
समय सारणी३५
पाठ्यक्रम

Motional Electromotive Force (e.m.f.)

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Estimated time: 13 minutes
CBSE: Class 12
Maharashtra State Board: Class 11

Definition: Motional emf

The emf induced across the ends of a conductor due to its motion in a magnetic field is called motional emf.

CBSE: Class 12

Formula: Motional EMF

e = Blv

  • B = magnetic field
  • l = length of conductor
  • v = velocity
CBSE: Class 12

Derivation of Motional EMF

Method 1: Using Faraday's Law

Let the rod move a distance dx in time dt:

dΦ = B ⋅ l ⋅ dx
e = \[-\frac{d\Phi}{dt}=-B\cdot l\cdot\frac{dx}{dt}=-Blv\]

The magnitude of the induced EMF:

e = Blv   ...(1)

Method 2: Using Lorentz Force

Each free electron in the moving rod experiences the Lorentz force:

F = qvB (\[\text{since } \vec{v} \perp \vec{B}\])

This force acts as the "EMF agent" that does work W in moving charge q from one end to another across length l:

e = ​\[\frac{W}{q}=\frac{F\cdot l}{q}=\frac{qvB\cdot l}{q}\] = Blv

Both methods give the same result, confirming Eq. (1).

CBSE: Class 12

Conditions for Motional EMF

  • The conductor must be moving relative to the magnetic field (or the field must be changing)
  • The conductor's velocity must have a component perpendicular to both \[\vec{B}\] and \[\vec{l}\]
  • If \[\vec v\] || \[\vec B\] → No EMF (no force on charges perpendicular to motion)
  • If \[\vec v\] || \[\vec l\] → No EMF (no effective length component)
  • For maximum EMF: \[\vec{v} \perp \vec{B} \perp \vec{l}\] all mutually perpendicular
CBSE: Class 12

Example 1

Setup: A metallic rod of length 1 m is hinged at the centre of a circular metallic ring of radius 1 m. It rotates at 50 rev/s in a uniform magnetic field B = 1 T directed parallel to the axis of rotation (i.e., perpendicular to the plane of the rod).

What's being asked: EMF between the centre (hinge) and the rim (ring).

Core idea: As the rod spins, each small element dr at distance r from the centre moves with velocity v = ωr. The motional EMF across that tiny element is dε = Bv dr = Bωr dr. Integrating from 0 to R:

ε = \[\int_0^RB\omega rdr=\frac{1}{2}B\omega R^2\]

Substituting values:

ω = 2π × 50 = 100π rad/s
ε = \[\frac {1}{2}\] × 1 × 100π × (1)2 = 50π ≈ 157 V
CBSE: Class 12

Example 2

Setup: A wheel has 10 metallic spokes, each 0.5 m long. It rotates at 120 rev/min in a plane normal to Earth's horizontal magnetic field HE = 0.4 G.

What's being asked: Induced EMF between the axle (centre) and the rim.

Step 1 — Convert units:

B = 0.4 G = 0.4 × 10−4 T
ω = \[\frac {120×2π}{60}\] = 4π rad/s

Step 2 — Apply the rotating rod formula:

ε = \[\frac{1}{2}B\omega R^2=\frac{1}{2}\times0.4\times10^{-4}\times4\pi\times(0.5)^2\]
ε = \[\frac{1}{2}\times0.4\times10^{-4}\times4\pi\times0.25\approx{6.28\times10^{-5}\mathrm{V}}\]

The critical insight about the 10 spokes: Each spoke generates the same EMF. But because they are all connected in parallel (all share the same axle and rim), the total EMF equals the EMF of just one spoke — not 10 times. This is a very common exam trap!

Shaalaa.com | Electromagnetic Induction Part 3

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Related QuestionsVIEW ALL [21]

An aircraft of wing span of 50 m flies horizontally in the Earth's magnetic field of 6 x 10-5 T at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.

A 20 cm long conducting rod is set into pure translation with a uniform velocity of 10 cm s−1 perpendicular to its length. A uniform magnetic field of magnitude 0.10 T exists in a direction perpendicular to the plane of motion. (a) Find the average magnetic force on the free electrons of the rod. (b) For what electric field inside the rod, the electric force on a free elctron will balance the magnetic force? How is this electric field created? (c) Find the motional emf between the ends of the rod.

A magnetic field B = Bo sin ( ωt )`hatk` wire AB slides smoothly over two parallel conductors separated by a distance d (Figure). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity?

Find the current in the wire for the configuration shown in figure. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. θ is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d.

An e.m.f is produced in a coil, which is not connected to an external voltage source. This can be due to ______.

  1. the coil being in a time varying magnetic field.
  2. the coil moving in a time varying magnetic field.
  3. the coil moving in a constant magnetic field.
  4. the coil is stationary in external spatially varying magnetic field, which does not change with time.

A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because ______.

  1. the magnetic field is constant.
  2. the magnetic field is in the same plane as the circular coil and it may or may not vary.
  3. the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably.
  4. there is a constant magnetic field in the perpendicular (to the plane of the coil) direction.

Find the current in the sliding rod AB (resistance = R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance. v is constant. Switch S is closed at time t = 0.

A wire of length 50 cm moves with a velocity of 300 m/min, perpendicular to a magnetic field. If the e.m.f. induced in the wire is 2 V, the magnitude of the field in tesla is ______.

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