Class 12 > Unit # 19:Electromagnetic Induction > Motional EMF


Motional EMF - Definition and Derivation - Talha's Physics Academy

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Electromagnetic Induction - Motional EMF

Define Motional EMF. Derive expression for motional EMF.

Definition

"Motional electromotive force (emf) is an electromotive force induced in a conductor when it moves through a magnetic field."

This concept stems directly from Faraday's law of electromagnetic induction and represents a fundamental principle of electromagnetism.

Example: The generation of electricity by an electrical generator. As a coil of wire rotates within a magnetic field, a motional emf is induced, producing an electric current.

Figure: A straight conducting wire $PQ$ of length $L$ moving uniformly with velocity $v$ in a uniform magnetic field $B$.

Conceptual Explanation

Consider a straight wire $PQ$ of length $L$ moving uniformly at a constant velocity $v$ within a rectangular loop $PQRS$ placed in a uniform magnetic field $B$ directed perpendicularly into the plane of the page.

As the conductor moves, each free electron inside the wire moves along with it and experiences a magnetic force given by: $$F = -e (\vec{v} \times \vec{B}) \quad \text{or} \quad F = e (\vec{B} \times \vec{v})$$ This force causes free electrons to accumulate at one end of the conductor (creating a negative charge), while leaving a deficiency of electrons at the opposite end (creating a positive charge). This charge separation establishes an electrostatic potential difference across the ends of the wire, which is known as motional emf.

Mathematical Derivation

Potential difference is defined as the work done per unit charge in moving a charge across the conductor:

$\text{EMF } (\mathcal{E}) = \frac{W}{q}$

Since work done is equal to the product of force ($F$) and displacement ($L$), and the magnetic force on a charge $q$ moving with velocity $v$ in magnetic field $B$ is $F = q v B$:

$W = F \cdot L = (q v B) L$

Substituting the work expression into the emf formula:

$\mathcal{E} = \frac{(q v B) L}{q}$

Canceling the charge $q$ from the numerator and denominator gives the final mathematical expression for motional emf:

$\mathcal{E} = v B L$

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