Motional Electromotive Force

A Rod Sweeping Flux

So far we changed $B$. Now let's change the area. Consider this setup: a U-shaped conducting frame in a uniform field $\vec{B}$ (perpendicular to its plane, into the page), with a rod PQ of length $l$ sliding on it with velocity $v$. If $x$ is the length of the enclosed loop,

$$\Phi_B = Blx$$

As PQ moves, $x$ changes, and Faraday's law gives

$$\varepsilon = -\frac{d\Phi_B}{dt} = -Bl\frac{dx}{dt} = Blv$$

This induced emf $\boxed{\varepsilon = Blv}$ is called the motional emf - produced by moving a conductor instead of varying the field.

Key Point: The formula assumes $\vec{B}$, the rod's length, and $\vec{v}$ are mutually perpendicular. In general it is the perpendicular components that count - if the rod moves parallel to the field, no flux is swept, and $\varepsilon = 0$.

The Lorentz-Force Explanation

The same result drops out of Chapter 4 physics — no Faraday's law needed. Each free charge $q$ inside the moving rod travels with velocity $v$ through field $B$, so it feels the Lorentz force

$$F = qvB$$

directed along the rod (towards Q for positive charges, by $q\vec{v}\times\vec{B}$). The charges pile up: one end of the rod becomes positive, the other negative, until the resulting electric field balances the magnetic push.

The work done in carrying a charge $q$ across the rod's length $l$ is $W = qvBl$, so the emf - work per unit charge - is

$$\varepsilon = \frac{W}{q} = Blv$$

exactly as before. The moving rod is a genuine seat of emf, like a battery with the positive terminal at the end the positive charges drift to.

And when the conductor is stationary? With $v = 0$ the magnetic part of $\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$ vanishes - yet experiments (Section 1!) show emf is still induced when $B$ changes. The inescapable conclusion, and the deepest line of the chapter:

A time-varying magnetic field generates an electric field. (This induced electric field differs in character from the electrostatic field of fixed charges.)

This is the fundamental significance of Faraday's discovery: electricity and magnetism are two faces of one subject.

The Rotating Rod

A rod of length $R$ rotates with angular speed $\omega$ about one end, perpendicular to a uniform field $B$ (think of a wheel spoke). A small element at distance $r$ moves with speed $v = \omega r$, contributing $d\varepsilon = Bv\,dr$. Integrating along the rod:

$$\varepsilon = \int_0^R B\omega r\,dr = \frac{1}{2}B\omega R^2$$

Check by the area method: the rod sweeps area at the rate $\frac{dA}{dt} = \frac{1}{2}R^2\frac{d\theta}{dt} = \frac{1}{2}R^2\omega$, so $\varepsilon = B \times \frac{1}{2}\omega R^2$ - identical.

With $\omega = 2\pi\nu$:

$$\varepsilon = \frac{1}{2}B(2\pi\nu)R^2 = \pi B \nu R^2$$

[NEET Important] For a wheel with many spokes, all spokes are emf sources in parallel between the axle and the rim - same emf each - so the number of spokes is immaterial: the emf stays $\frac{1}{2}B\omega R^2$.

[JEE Tip] Don't mix the two formulas: translating rod → $Blv$ (linear in length), rotating rod → $\frac{1}{2}B\omega R^2$ (quadratic in length). The factor $\frac{1}{2}$ is the give-away that an integration over the rod happened.

Solved Examples

Example 1: The rotating metre rod

A metallic rod of 1 m length rotates at 50 rev/s about an axis through one end, perpendicular to a uniform field of 1 T. Find the emf between the centre and the rim.

Solution:

1. Formula: $\varepsilon = \frac{1}{2}B\omega R^2$ with $\omega = 2\pi\nu = 2\pi \times 50$.
2. $\varepsilon = \frac{1}{2} \times 1.0 \times 2\pi \times 50 \times (1)^2 = 50\pi$.
3. Answer: $\varepsilon \approx 157$ V. (Physically: the Lorentz force drives free electrons along the rod until a steady charge separation — and emf — is reached.)

Example 2: The wheel with ten spokes

A wheel with 10 metallic spokes, each 0.5 m long, rotates at 120 rev/min in a plane normal to the earth's horizontal field $H_E = 0.4$ G. Find the emf between axle and rim. (1 G = $10^{-4}$ T.)

Solution:

1. $\nu = 120/60 = 2$ rev/s, so $\omega = 4\pi$ rad/s; $B = 0.4 \times 10^{-4}$ T.
2. $\varepsilon = \frac{1}{2}\omega B R^2 = \frac{1}{2} \times 4\pi \times 0.4 \times 10^{-4} \times (0.5)^2$.
3. Answer: $\varepsilon \approx 6.28 \times 10^{-5}$ V.
4. The spokes don't matter: all ten emfs sit in parallel between axle and rim — same emf, so the number of spokes is immaterial.

Example 3: An aircraft's wings as a rod

A jet with a wing span of 25 m flies horizontally at 1800 km/h where the vertical component of the earth's field is $5 \times 10^{-4}$ T. Find the emf between the wing tips.

Solution:

1. Convert: $v = 1800$ km/h $= 500$ m/s.
2. Formula: $\varepsilon = B_v l v = 5 \times 10^{-4} \times 25 \times 500$.
3. Answer: $\varepsilon = 6.25$ V — the flying wings are a 25 m motional-emf rod!

Example 4: Full circuit bookkeeping on rails

A rod of length 0.2 m slides at 2 m/s on rails in a perpendicular field of 0.5 T; the circuit resistance is 0.5 $\Omega$. Find (a) the emf, (b) the current, (c) the force needed to keep the rod moving, (d) verify the power balance.

Solution:

1. (a) $\varepsilon = Blv = 0.5 \times 0.2 \times 2 = 0.2$ V.
2. (b) $I = \varepsilon/R = 0.2/0.5 = 0.4$ A.
3. (c) The field opposes the motion (Lenz) with force $F = BIl = 0.5 \times 0.4 \times 0.2 = 0.04$ N; an equal external force is needed.
4. (d) Mechanical power in: $Fv = 0.04 \times 2 = 0.08$ W. Heat out: $I^2R = (0.4)^2 \times 0.5 = 0.08$ W. They match — Lenz's law balancing the books.

Example 5: Which end is positive?

A rod PQ moves to the right with velocity $v$ in a field pointing into the page (rod vertical, P at top). Which end is at higher potential?

Solution:

1. Force on positive charges: $\vec{F} = q\vec{v}\times\vec{B}$. With $\vec{v}$ to the right ($+x$) and $\vec{B}$ into the page ($-z$), $\vec{v}\times\vec{B}$ points along $+y$ — towards P (up).
2. Positive charge accumulates at P: the top end is the positive terminal.
3. The rod acts as a battery of emf $Blv$ with P as its + terminal.

Example 6: Motion parallel to the field

The same rod now moves parallel to $\vec{B}$ instead. What emf is induced?

Solution:

1. Moving along the field gives $\vec{v}\times\vec{B} = 0$ for parallel vectors.
2. The Lorentz force on the charges vanishes; no charge separation occurs.
3. $\varepsilon = 0$. Only the mutually perpendicular geometry generates motional emf.

Example 7: A fresh rotating-rod numerical

A 0.5 m rod rotates at 100 rad/s about one end in a perpendicular field of 0.2 T. Find the emf across it.

Solution:

1. $\varepsilon = \frac{1}{2}B\omega R^2 = \frac{1}{2} \times 0.2 \times 100 \times (0.5)^2$.
2. $= \frac{1}{2} \times 0.2 \times 100 \times 0.25 = 2.5$ V.
3. Answer: 2.5 V, with the outer end's polarity set by $q\vec{v}\times\vec{B}$.

Example 8: Two methods, one answer

For the rotating rod, show the 'swept-area' method reproduces the integration result.

Solution:

1. In time $dt$ the rod turns by $d\theta$, sweeping area $dA = \frac{1}{2}R^2 d\theta$.
2. $\varepsilon = B\frac{dA}{dt} = B \cdot \frac{1}{2}R^2\omega$.
3. Identical to $\int_0^R B\omega r\,dr = \frac{1}{2}B\omega R^2$ — integrate the Lorentz emf or differentiate the flux: same physics, same answer.

Example 9: Charge through the circuit

In the rails setup of Example 4 (B = 0.5 T, l = 0.2 m, R = 0.5 $\Omega$), the rod moves 0.1 m. How much charge flows?

Solution:

1. Flux change: $\Delta\Phi = Bl\,\Delta x = 0.5 \times 0.2 \times 0.1 = 10^{-2}$ Wb.
2. Charge: $q = \Delta\Phi/R = 10^{-2}/0.5 = 2 \times 10^{-2}$ C.
3. Independent of the rod's speed — the charge counts the flux change, not the rate.

Example 10: The stationary-conductor puzzle

A stationary loop sits in a changing magnetic field, and a current is induced. The magnetic force $q\vec{v}\times\vec{B}$ on its stationary charges is zero. What drives the current?

Solution:

1. With $v = 0$, the force is $\vec{F} = q\vec{E}$ — only an electric field can move the charges.
2. So we must conclude: a time-varying magnetic field generates an electric field, which drives the induced current.
3. This induced $E$ differs from the electrostatic field of charges (its field lines close on themselves) — the fundamental significance of Faraday's discovery.