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Electromagnetic Induction - Motional EMF

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Motional EMF is the electromotive force induced across the ends of a conductor when it is moved through a magnetic field. It is a consequence of the Lorentz force acting on the free charges within the conductor.

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For a straight conducting rod of length ll moving with a constant velocity vv perpendicular to a uniform magnetic field BB, the induced EMF ε\varepsilon is given by ε=Blv\varepsilon = Blv.

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The direction of the induced current is determined by Fleming's Right Hand Rule: If the thumb points in the direction of motion and the forefinger in the direction of the magnetic field, the middle finger indicates the direction of the induced current.

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When a conducting rod of length ll rotates with a constant angular velocity ω\omega in a uniform magnetic field BB (perpendicular to the plane of rotation), the induced EMF between the center and the endpoint is ε=12Bωl2\varepsilon = \frac{1}{2} B \omega l^2.

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Energy Conservation: To maintain a constant velocity vv for a conductor in a closed circuit of resistance RR, an external force F=B2l2vRF = \frac{B^2 l^2 v}{R} must be applied. The mechanical power Pmech=FvP_{mech} = Fv is exactly equal to the electrical power dissipated as heat Pelec=I2RP_{elec} = I^2R.

📐Formulae

ε=Blv\varepsilon = Blv

ε=∮(v⃗×B⃗)⋅dl⃗\varepsilon = \oint (\vec{v} \times \vec{B}) \cdot d\vec{l}

ε=12Bωl2\varepsilon = \frac{1}{2} B \omega l^2

I=εR=BlvRI = \frac{\varepsilon}{R} = \frac{Blv}{R}

Fmagnetic=IlB=B2l2vRF_{magnetic} = IlB = \frac{B^2 l^2 v}{R}

P=B2l2v2RP = \frac{B^2 l^2 v^2}{R}

💡Examples

Problem 1:

A horizontal straight wire 10 m10 \text{ m} long extending from east to west is falling with a speed of 5.0 m/s5.0 \text{ m/s} at right angles to the horizontal component of the Earth's magnetic field, 0.30×10−4 Wb/m20.30 \times 10^{-4} \text{ Wb/m}^2. What is the instantaneous value of the EMF induced in the wire?

Solution:

Given: l=10 ml = 10 \text{ m}, v=5.0 m/sv = 5.0 \text{ m/s}, and BH=0.30×10−4 TB_H = 0.30 \times 10^{-4} \text{ T}. Using the formula for motional EMF: ε=BHlv\varepsilon = B_H l v. Substituting the values: ε=(0.30×10−4)×10×5.0=1.5×10−3 V\varepsilon = (0.30 \times 10^{-4}) \times 10 \times 5.0 = 1.5 \times 10^{-3} \text{ V}.

Explanation:

The wire moves perpendicular to the horizontal component of the Earth's magnetic field, thereby cutting the magnetic flux lines and inducing an EMF.

Problem 2:

A metallic rod of 1 m1 \text{ m} length is rotated with a frequency of 50 rev/s50 \text{ rev/s} with one end hinged at the center and the other end at the circumference of a circular metallic ring of radius 1 m1 \text{ m}, about an axis passing through the center and perpendicular to the plane of the ring. A constant and uniform magnetic field of 1 T1 \text{ T} parallel to the axis exists everywhere. What is the EMF between the center and the metallic ring?

Solution:

Given: l=1 ml = 1 \text{ m}, f=50 Hzf = 50 \text{ Hz}, B=1 TB = 1 \text{ T}. First, find angular velocity ω=2πf=2×π×50=100π rad/s\omega = 2\pi f = 2 \times \pi \times 50 = 100\pi \text{ rad/s}. The induced EMF is ε=12Bωl2\varepsilon = \frac{1}{2} B \omega l^2. Substituting the values: ε=12×1×100π×(1)2=50π≈157 V\varepsilon = \frac{1}{2} \times 1 \times 100\pi \times (1)^2 = 50\pi \approx 157 \text{ V}.

Explanation:

As the rod rotates, its different segments move with different linear velocities (v=rωv = r\omega). Integrating these velocities over the length of the rod results in the rotational EMF formula.

Motional EMF - Revision Notes & Key Formulas | CBSE Class 12 Physics