Motional EMF — Full Derivations
Two approaches: (A) using Faraday’s Law and (B) using charge motion (Lorentz force).
Key Result
For a straight conducting rod of length \(l\) moving at speed \(v\) perpendicular to a uniform magnetic field \(B\),
\[ \varepsilon = B\,l\,v \]
(Polarity/direction follows Lenz’s law.)
(A) Derivation using Faraday’s Law
Setup: A U-shaped conductor with a sliding rod PQ (length \(l\)) forms a rectangular loop in a uniform field \(B\) (into the page). The rod moves right with constant speed \(v\), so the loop area increases.
- Let the distance between the fixed rail and the rod be \(x(t)\). The loop area \(A=l\,x(t)\). Magnetic flux:
\[ \Phi_B = B\,A = B\,l\,x(t) \quad (\text{since } B \perp \text{area}) \] - By Faraday’s law:
\[ \varepsilon = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}\big(B\,l\,x(t)\big) = -B\,l\,\frac{dx}{dt} \]
Since \(v=\frac{dx}{dt}\),
\[ \varepsilon = -B\,l\,v \]
The negative sign indicates the sense of the induced emf (Lenz’s law). Magnitude:
\[ |\varepsilon|=B\,l\,v \]
Physical meaning: as area (hence flux) increases, the induced current sets up a magnetic field opposing the increase (Lenz’s law), ensuring energy conservation.
(B) Derivation using Charges (Lorentz Force)
Idea: Mobile charges in the rod move with the rod and feel a magnetic force \( \mathbf{F}=q\,\mathbf{v}\times\mathbf{B} \), separating along the rod and creating a potential difference.
- Magnitude of magnetic force on a carrier: \( F = q\,v\,B \) (with \( \mathbf{v}\perp\mathbf{B} \)).
- Charge separation sets up an electric field \(E\) along the rod. In steady state, \( qE = qvB \Rightarrow E=vB \).
- Potential difference (emf) across the rod of length \(l\):
\[ \varepsilon = \int \mathbf{E}\cdot d\mathbf{l} = E\,l = v\,B\,l \]
Direction/polarity follows Lenz’s law or the right-hand rule.
Diagram (Rod PQ on U-shaped Rails)
Assumptions & General Form
- Uniform \(B\); rod moves with speed \(v\); rod length \(l\) is perpendicular to both \(B\) and \(v\).
- General orientation: \( \varepsilon = B\,l\,v\,\sin\theta \), where \(\theta\) is the angle between \(\mathbf{v}\) and the component of \(\mathbf{B}\) perpendicular to the rod.
- No emf if motion is strictly parallel to \(B\) (no flux change, no \( \mathbf{v}\times\mathbf{B} \) component along the rod).
Energy View (Why Lenz’s Law)
To keep the rod moving at constant speed, an external agent must do work against the magnetic force on the induced current. The mechanical power supplied equals the electrical power dissipated (Joule heating) in the circuit, preserving energy conservation.
Worked Examples
Example 1 (Linear motion)
A 0.50 m rod slides at 8 m/s on rails in a uniform field \(B=0.30\,\text{T}\) (into the page). Find the induced emf.
\( \varepsilon = B\,l\,v = 0.30 \times 0.50 \times 8 = 1.2\,\text{V} \). Answer: \(1.2\,\text{V}\).
Example 2 (Rotating rod)
A rod of length \(R=1.0\,\text{m}\) rotates about one end with angular speed \(\omega=50\pi\,\text{rad s}^{-1}\) in a uniform field \(B=1.0\,\text{T}\) (field along the axis). Emf between the center and the rim it touches?
For an element at radius \(r\), \(d\varepsilon = B(\omega r)\,dr\). Hence
\[ \varepsilon = \int_{0}^{R} B\,\omega r\,dr = \tfrac{1}{2}B\,\omega R^2 = \tfrac{1}{2}\times 1 \times 50\pi \times 1^2 = 25\pi \approx 78.54\,\text{V}. \]
Answer: \(\approx 78.5\,\text{V}\).
Quick Reference
| Quantity | Symbol | Unit | Relation |
|---|---|---|---|
| Motional emf | \(\varepsilon\) | volt (V) | \(\varepsilon = B\,l\,v \) (max); polarity via Lenz’s law |
| Magnetic flux | \(\Phi_B\) | weber (Wb) | \(\Phi_B = B\,A\) (for \(B\perp A\)) |
| Faraday’s Law | — | — | \(\varepsilon = -\dfrac{d\Phi_B}{dt}\) |
| Lorentz force | \( \mathbf{F} \) | newton (N) | \( \mathbf{F}=q(\mathbf{v}\times \mathbf{B}) \) |