JEE Mains Physics · Electromagnetic Induction

Motional EMF: Rods, Rails and Moving Loops

A conductor of length l moving at v across a field B develops an emf Blv; on rails the current it drives feels a force B²l²v/R that opposes the motion, and a loop crossing the edge of a field has an emf only while its flux is changing.

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Why this matters

Nineteen PYQs, twelve of them multiple choice, and three from 2026. Eight are a single rod, wire or wing cutting the field, often the earth's; six put a rod on rails or pull a loop out of a field and ask for the force, the work or a terminal speed; five follow a loop across the edge of a field or through a field that changes along its path. Ten of them come with a figure.

Concept 1 of 3: Motional emf of a moving rod

Charges inside a moving conductor are carried across the field, so each feels a magnetic force qvB along the conductor. They pile up at one end until an electric field stops them, and that leaves an emf Blv between the ends. Only the parts that are mutually perpendicular count: the length across the motion, and the field component across both. In the earth's field, decide which component, horizontal or vertical, the rod actually cuts.

Definition

$\varepsilon = Blv$ when B, l and v are mutually perpendicular; in general $\varepsilon = (\vec v \times \vec B)\cdot \vec l$ . A rod moving along its own length, or along B, has no emf.
Earth's field with total B and dip $\delta$ : $B_H = B\cos\delta$ , $B_V = B\sin\delta$ , so $B_V = B_H\tan\delta$ .
A horizontal rod moving horizontally (aircraft wings, a rod lying N–S moving east) cuts $B_V$ .
A horizontal E–W wire falling, or a vertical rod moving east or west, cuts $B_H$ . A wire falling from rest through h has $v = \sqrt{2gh}$ .
Rails meeting at a vertex at angle $\alpha$ , bar moving away at constant v: the length between the rails is $2vt\tan(\alpha/2)$ for a symmetric V starting at the vertex. Put that length into Blv before deciding how the emf grows with time.
A block with no circuit still develops a potential difference $vBd$ across the faces separated along $\vec v \times \vec B$ .
Convert km/h with $\times 5/18$ ; 1 gauss is $10^{-4}\ \text{T}$ .

Motional emf

\varepsilon = Blv \qquad B_V = B\sin\delta,\ B_H = B\cos\delta

Worked example

A horizontal rod 2 m long lies along the north–south line and moves east at

15\ \text{m/s}

. The earth's total field there is

5 \times 10^{-5}\ \text{T}

and the angle of dip is

37^{\circ}

. Find the emf across the rod. (Take

\sin 37^{\circ} = 0.6

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 26 · Q17Moderate

Example 1 · Electromagnetic Induction · Motional EMF: Rods, Rails and Moving Loops

An aeroplane, with its wings spread

10\text{ }m

, is flying at a speed of

180\text{ }km/h

in a horizontal direction. The total intensity of earth's field at that part is

2.5 \times10^{- 4}\text{ }Wb/m^{2}

and the angle of dip is

60^{\circ}

. The emf induced between the tips of the plane wings will be

Using the total field instead of the cut component

Aircraft wings and a horizontal rod moving horizontally cut only the vertical component, B sin δ. A falling horizontal wire cuts only the horizontal component, B cos δ.

Sine and cosine of the dip swapped

The dip is the angle the field makes with the HORIZONTAL. So the horizontal component is B cos δ and the vertical one is B sin δ.

Leaving the speed in km/h or the field in gauss

Blv gives volts only with tesla, metres and metres per second. 180 km/h is 50 m/s, and 0.5 gauss is 5 × 10⁻⁵ T.

Drill 7 more on motional emf of a moving rod

Concept 2 of 3: Force, power and terminal speed for a rod on rails

Once the moving rod closes a circuit, its emf drives a current, and that current in the field feels a force BIl that points against the motion. To keep the speed constant, something must push with exactly that force, and all the work it does turns into heat in the resistance. A rod falling on vertical rails speeds up until this magnetic drag equals its weight.

Definition

Current $I = \dfrac{Blv}{R}$ , where l is the length BETWEEN the rails and R the whole circuit's resistance.
Retarding force $F = BIl = \dfrac{B^{2}l^{2}v}{R}$ ; the force to keep a constant speed is equal to it.
Power $P = Fv = \dfrac{B^{2}l^{2}v^{2}}{R} = I^{2}R$ : the work done becomes heat.
Falling rod of mass m on smooth vertical rails: terminal speed when $mg = \dfrac{B^{2}l^{2}v_t}{R}$ , so $v_t = \dfrac{mgR}{B^{2}l^{2}}$ .
Pulling a square loop of side a out of a field slowly and uniformly in time t: $v = a/t$ , and the work is $\dfrac{(Bav)^{2}}{R}\,t$ , the same as F × a. With N turns the emf is N times larger and the work $N^{2}$ times.
A network on the rails: reduce it to one resistance, then add the rod's own resistance.

Rod on rails

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

Worked example

A rod slides on rails

0.5\ \text{m}

apart at a steady

2\ \text{m/s}

, across a field of

0.4\ \text{T}

. The total resistance of the circuit is

0.2\ \Omega

. Find the current, the force needed to keep the speed and the power supplied.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q101Moderate

Example 2 · Electromagnetic Induction · Motional EMF: Rods, Rails and Moving Loops

A square loop of area

25{\text{ }cm}^{2}

has a resistance of

10\Omega

. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0sec, will be

Using the rod's length instead of the rail gap

Only the part of the rod between the rails carries current. A rod longer than the gap still has l equal to the gap in Blv and in BIl.

Forgetting that the force goes as v

The magnetic drag B²l²v/R grows with speed. That is why a falling rod reaches a terminal speed instead of accelerating at g for ever.

Work done is not the stored energy

Pulling a loop out at constant speed stores nothing. All the work appears as heat, I²R t, in the loop's resistance.

Drill 5 more on force, power and terminal speed for a rod on rails

Concept 3 of 3: Loop crossing a field boundary or a non-uniform field

A moving loop has an emf only while the flux through it is changing. Entering a field, one side cuts the field and the emf is Blv; once the loop is fully inside a uniform field, both sides cut it equally and the emfs cancel. In a field that changes from place to place, the two sides sit in different fields, so the net emf is the difference between them.

Definition

Partly inside a uniform field: $\varepsilon = Blv$ , with l the side lying across the boundary. Constant at constant speed.
Fully inside, or fully outside: $\varepsilon = 0$ . Track the front and back edges with $x = vt$ .
A ring crossing a straight boundary: the effective length is the chord on the boundary. When the centre is on the boundary, the chord is the diameter, so $\varepsilon = B(2r)v$ .
Field varying along x: $\varepsilon = (B_{\text{front}} - B_{\text{back}})\,lv$ , with l the side across the motion. For $B = kx$ and a loop of length a along x, $\varepsilon = kalv$ .
If the field also changes in time, add $A\,\dfrac{\partial B}{\partial t}$ to the motional part.

Loop in a non-uniform field

\varepsilon = (B_{\text{front}} - B_{\text{back}})\,lv

Worked example

A square loop of side

10\ \text{cm}

moves at

2\ \text{cm/s}

into a region

30\ \text{cm}

wide with a uniform field of

0.5\ \text{T}

. Its front edge enters at

t = 0

. Find the emf at

t = 3\ \text{s}

8\ \text{s}

and

18\ \text{s}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 6 · Q17Moderate

Example 3 · Electromagnetic Induction · Motional EMF: Rods, Rails and Moving Loops

The magnetic field in a region is given by

\overrightarrow{B}=B_{0}\left( \frac{x}{a} \right)\widehat{k}

. A square loop of side

d

is placed with its edges along the

x

and

y

axes. The loop is moved with a constant velocity

\overrightarrow{v}=v_{0}\widehat{i}

. The emf induced in the loop is:

An emf while the loop is fully inside

Inside a uniform field the flux through a moving loop is constant, so the emf is zero, however fast it moves. The emf appears only while an edge is crossing the boundary.

Adding the two sides in a non-uniform field

The front and back sides drive current in opposite senses round the loop. Their emfs subtract; the net is the field difference times lv.

Using the arc for a ring at an edge

For a ring crossing a straight boundary, the effective length is the straight chord along the boundary, not the arc of the ring inside the field.

Drill 4 more on loop crossing a field boundary or a non-uniform field

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 emf of a moving rod
Motional emf
$\varepsilon = Blv \qquad B_V = B\sin\delta,\ B_H = B\cos\delta$
Force, power and terminal speed for a rod on rails
Rod on rails
$I = \frac{Blv}{R} \qquad F = \frac{B^{2}l^{2}v}{R} \qquad v_t = \frac{mgR}{B^{2}l^{2}}$
Loop crossing a field boundary or a non-uniform field
Loop in a non-uniform field
$\varepsilon = (B_{\text{front}} - B_{\text{back}})\,lv$

Watch out for (9)

Using the total field instead of the cut component
→ Motional emf of a moving rod
Sine and cosine of the dip swapped
→ Motional emf of a moving rod
Leaving the speed in km/h or the field in gauss
→ Motional emf of a moving rod
Using the rod's length instead of the rail gap
→ Force, power and terminal speed for a rod on rails
Forgetting that the force goes as v
→ Force, power and terminal speed for a rod on rails
Work done is not the stored energy
→ Force, power and terminal speed for a rod on rails
An emf while the loop is fully inside
→ Loop crossing a field boundary or a non-uniform field
Adding the two sides in a non-uniform field
→ Loop crossing a field boundary or a non-uniform field
Using the arc for a ring at an edge
→ Loop crossing a field boundary or a non-uniform field

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