MHT-CET Physics · Electromagnetic Induction

Motional e.m.f., Magnetic Braking and Rotating Rods

A straight conductor of length l moving at speed v across a field B has an e.m.f. Blv between its ends; if it closes a circuit, the current it drives feels a force that opposes the motion, and a rod or disc spinning about one end sweeps out area and develops ½Bωl².

All Electromagnetic Induction notes MHT-CET Physics strategy

Why this matters

19 PYQs, 4 HARD. Six are Blv itself — an aircraft's wings, a boat's mast, a wire falling from a height, a pendulum's bob; seven are the force and power that follow — the heat produced when a loop is pulled through a field, the work to pull it out, the speed at which a falling loop or rod stops accelerating; six are rotation — a rod about one end, a metal disc, a bicycle wheel, and a coil spinning in a field. Three cards.

Concept 1 of 3: Motional e.m.f. Blv

Charges in a conductor moving across a field feel a magnetic force along the conductor, which piles them at one end: an e.m.f. e = Blv, where l is the length perpendicular to both B and v. Only the component of B across the motion counts — for a horizontal wire falling or a vertical mast moving east it is the horizontal field; for an aircraft's wings it is the vertical one. For a coil leaving a field, l is the side crossing the edge, so a longer side there means a larger e.m.f.

Definition

$e = Blv$ with B, l, v mutually perpendicular. Field given as H in A/m: first $B = \mu_0 H$ .
Wing tips: $5 \times 10^{-5} \times 40 \times 500 = 1$ V. Boat's vertical rod moving east in a northward horizontal field: $Blv$ .
Wire falling from height h: $v = \sqrt{2gh}$ at the ground, $e = B_H l\sqrt{2gh}$ .
Pendulum of length L through angle θ: the bob's fastest speed is $2\sqrt{gL}\sin\frac{\theta}{2}$ , so $e_{\max} = 2BL\sqrt{gL}\sin\frac{\theta}{2}$ .
Coil leaving a field: the vertical side does the cutting — rotate the coil so its shorter side is vertical and the e.m.f. falls.

Motional e.m.f.

e = Blv

Worked example

A 3 m vertical rod on a car moving north at 20 m/s, where the horizontal field is 4 × 10⁻⁵ T pointing north. E.m.f. across the rod?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 1 · Q12Easy

Example 1 · Electromagnetic Induction · Motional EMF and Rotating Conductors

A boat is moving due east in a region where the earth's magnetic field is

3.6 \times 10^{-5}

N/Am due north and horizontal. The boat carries a vertical conducting rod 2 m long. If the speed of the boat is 2.00 m/s, the magnitude of the induced e.m.f. in the rod is

Using the total field instead of the cutting component

Only the part of B perpendicular to both the rod and its velocity induces an e.m.f. A rod moving along the field lines induces nothing.

Drill 5 more on motional e.m.f. blv

Concept 2 of 3: Magnetic Braking: Force, Heat and Terminal Speed

Close the circuit and the e.m.f. Blv drives I = Blv/R. That current sits in the field, so it feels F = BIl = B²l²v/R, opposing the motion. Pulling at steady speed means supplying exactly that force, so the heat produced per second is (Blv)²/R. A loop or rod falling through a field speeds up until the braking force equals its weight: mg = B²l²v/R, so v = mgR/(B²l²). Pulling a loop out slowly in time t: e = BA/t, and the work done is e²t/R.

Definition

Current $I = \dfrac{Blv}{R}$ ; braking force $F = \dfrac{B^2l^2v}{R}$ .
Heat per second (= power to pull at steady v): $P = \dfrac{B^2l^2v^2}{R}$ .
Terminal speed of a falling loop or rod of mass m: $v = \dfrac{mgR}{B^2l^2}$ .
Work to pull a loop out in time t: $e = \dfrac{BA}{t}$ , $W = \dfrac{e^2}{R}t$ (25 cm², 40 T, 10 Ω, 1 s ⇒ $10^{-3}$ J).
A current loop held by a field: $BIL = Mg$ with $I = \dfrac{V}{R}$ gives $V = \dfrac{MgR}{BL}$ .

Braking force and terminal speed

F = \frac{B^2l^2v}{R}, \qquad v_t = \frac{mgR}{B^2l^2}

Worked example

A 20 cm rod of mass 10 g and resistance 0.5 Ω slides down vertical rails in a 0.5 T horizontal field (g = 10). Terminal speed?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q17Moderate

Example 2 · Electromagnetic Induction · Motional EMF and Rotating Conductors

A horizontal wire of mass '

m

', length '

l

' and resistance '

R

' is sliding on the vertical rails on which uniform magnetic field '

B

' is directed perpendicular. The terminal speed of the wire as it falls under the force of gravity is (

g=

acceleration due to gravity)

Forgetting that the braking force grows with speed

F = B²l²v/R rises as the rod speeds up, which is why a falling rod reaches a terminal speed instead of accelerating at g forever.

Drill 6 more on magnetic braking: force, heat and terminal speed

Concept 3 of 3: Rotating Rods, Discs, Wheels and Coils

A rod spinning about one end moves faster further out; its average speed is ωl/2, so e = B·l·(ωl/2) = ½Bωl². A disc is a crowd of such radii in parallel — rim to axle it is ½BωR² too; so is a bicycle wheel, however many spokes it has, since they are in parallel. A whole coil turning in a field is different: its flux varies as cos ωt, so its e.m.f. peaks at NBAω and the average power into a resistor is (NBAω)²/(2R).

Definition

Rod about one end: $e = \tfrac{1}{2}B\omega l^2$ ; with n revolutions per second, $\omega = 2\pi n$ ⇒ $n = \dfrac{e}{\pi Bl^2}$ .
Disc or wheel, rim to axle: $\tfrac{1}{2}B\omega R^2$ , independent of the number of spokes. The paper's bicycle-wheel answer takes $\omega = 2\pi F$ , giving $B\pi FR^2$ .
Coil rotating: $e_0 = NBA\omega$ ; average power $\dfrac{N^2A^2B^2\omega^2}{2R}$ .

Rotating rod or disc

e = \tfrac{1}{2}B\omega l^2

Worked example

A 0.5 m rod rotates about one end at 20 rad/s in a 0.4 T field perpendicular to its plane of rotation. E.m.f.?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q5Moderate

Example 3 · Electromagnetic Induction · Motional EMF and Rotating Conductors

A metal rod of length

l

rotates about one of its ends in a plane perpendicular to a magnetic field of induction

B

. If the e.m.f. induced between the ends of the rod is

e

, then the number of revolutions made by the rod per second is

Multiplying by the number of spokes

The spokes of a wheel are in parallel between axle and rim, so the e.m.f. is that of one spoke, ½BωR². More spokes carry more current, not more voltage.

Drill 5 more on rotating rods, discs, wheels and coils

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Motional e.m.f. Blv
Motional e.m.f.
$e = Blv$
Magnetic Braking: Force, Heat and Terminal Speed
Braking force and terminal speed
$F = \frac{B^2l^2v}{R}, \qquad v_t = \frac{mgR}{B^2l^2}$
Rotating Rods, Discs, Wheels and Coils
Rotating rod or disc
$e = \tfrac{1}{2}B\omega l^2$

Watch out for (3)

Using the total field instead of the cutting component
→ Motional e.m.f. Blv
Forgetting that the braking force grows with speed
→ Magnetic Braking: Force, Heat and Terminal Speed
Multiplying by the number of spokes
→ Rotating Rods, Discs, Wheels and Coils

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Related notes

Mutual Inductance and Generators — the rotating coil in an a.c. generator →