JEE Mains Physics · Electromagnetic Induction

Rotating Coils, Rods and Discs

A coil spinning in a field has an emf NBAω sin ωt, largest when its plane lies along the field; a rod turning about one end sweeps out area and develops ½Bωl², and a disc does the same between its axle and its rim.

All Electromagnetic Induction notes JEE Mains Physics strategy

Why this matters

Fifteen PYQs, five of them multiple choice, and three from 2026. Seven are a coil spinning in a field, the AC generator, and ask for its peak or instantaneous emf; eight are a rod, disc, fan blade or pendulum wire turning about one end. Of the ten that want a number, six start by turning revolutions per minute or per second into radians per second.

Concept 1 of 2: Emf of a coil rotating in a magnetic field

As a coil turns at ω about an axis perpendicular to the field, the angle between its normal and the field is ωt, so its flux is NBA cos ωt. The emf is the rate of change of that, NBAω sin ωt. The flux changes fastest when it is passing through zero, so the emf is largest when the coil's plane lies along the field, and zero when the plane faces the field.

Definition

$\Phi = NBA\cos\omega t$ , $\varepsilon = NBA\omega\sin\omega t$ , peak $\varepsilon_0 = NBA\omega$ .
$\omega = 2\pi f$ ; from revolutions per minute, $\omega = 2\pi \times \text{rpm}/60$ . Half a revolution per second is $\pi\ \text{rad/s}$ .
Plane perpendicular to B: flux is greatest, emf zero. Plane parallel to B: flux zero, emf greatest.
Plane at angle $\alpha$ to B: the normal is at $90^{\circ} - \alpha$ , so $\varepsilon = \varepsilon_0\cos\alpha$ .
A circular coil of radius r: $A = \pi r^{2}$ . The rms value of the emf is $\varepsilon_0/\sqrt 2$ .

AC generator

\varepsilon = NBA\omega\sin\omega t \qquad \varepsilon_0 = NBA\omega

Worked example

A coil of 50 turns and area

0.04\ \text{m}^{2}

rotates at 600 rpm about an axis perpendicular to a field of

0.5\ \text{T}

. Find the peak emf, and the emf at the instant the plane of the coil makes

30^{\circ}

with the field.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q117Moderate

Example 1 · Electromagnetic Induction · Rotating Coils, Rods and Discs

A coil of 200 turns and area

0.20{\text{ }m}^{2}

is rotated at half a revolution per second and is placed in uniform magnetic field of

0.01\text{ }T

perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is

\frac{2\pi}{\beta}

volt. The value of

\beta

is______

Leaving rpm unconverted

NBAω needs ω in rad/s. Multiply rpm by 2π/60, and revolutions per second by 2π. Using the raw rpm gives an answer about ten times too big.

Zero flux does not mean zero emf

When the plane lies along the field, no lines pass through the coil, but the flux is changing fastest. That is the instant of PEAK emf.

Angle with the plane versus angle with the normal

With the normal at angle θ to B, the emf is ε₀ sin θ. With the plane at angle α to B, it is ε₀ cos α. Check which angle the question gives.

Drill 6 more on emf of a coil rotating in a magnetic field

Concept 2 of 2: Emf of a rod or disc rotating about one end

A rod turning about one end moves faster the farther out you go: a piece at distance r moves at ωr. Each piece adds B·ωr·dr, and summing from the pivot to the tip gives ½Bωl². Equivalently, the rod sweeps out ½l² of area per radian. A disc is a crowd of such rods side by side, all in parallel, so the emf between axle and rim is the same ½BωR².

Definition

Rod about one end, plane of rotation perpendicular to B: $\varepsilon = \tfrac12 B\omega l^{2}$ .
Disc of radius R about its axis along B: $\varepsilon = \tfrac12 B\omega R^{2}$ between axle and rim.
Pivot inside the rod: each part gives its own emf from the pivot outwards, with the same sign at both ends, so they partly cancel. For parts $l_1$ and $l_2$ : $\Delta V_{\text{ends}} = \tfrac12 B\omega(l_1^{2} - l_2^{2})$ .
Field that varies along the rod: $\varepsilon = \displaystyle\int_0^{l} B(r)\,\omega r\,dr$ .
Earth's field: a horizontal ceiling fan cuts $B_V$ ; a rod turning in a vertical plane cuts the horizontal component perpendicular to that plane. Fan blades are in parallel between hub and tips, so the emf is that of ONE blade.
A pendulum's wire is a rod pivoted at the suspension; its greatest ω comes from energy conservation.

Rotating rod or disc

\varepsilon = \frac12 B\omega l^{2}

Worked example

A rod

1.5\ \text{m}

long turns at

4\ \text{rad/s}

about one end, in a plane perpendicular to a field of

0.5\ \text{T}

. Find the emf between its ends.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 15 Apr 2023 · Q21Moderate

Example 2 · Electromagnetic Induction · Rotating Coils, Rods and Discs

20\text{ }cm

long metallic rod is rotated with

210rpm

about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field

0.2\text{ }T

parallel to the axis exists everywhere. The emf developed between the centre and the ring is___

mV

. Take

\pi =\frac{22}{7}

Forgetting the half

The pieces of a rotating rod move at speeds from zero to ωl, so the average speed is ωl/2. The emf is ½Bωl², not Bωl².

Adding the emfs of fan blades

Every blade runs from the hub to a tip, so the blades are connected in parallel. The emf between hub and tips is that of one blade, whatever the number of blades.

Total field for a horizontal fan

A ceiling fan turns in a horizontal plane, so only the vertical component B sin δ is perpendicular to that plane.

Drill 7 more on emf of a rod or disc rotating about one end

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 (2)

Emf of a coil rotating in a magnetic field
AC generator
$\varepsilon = NBA\omega\sin\omega t \qquad \varepsilon_0 = NBA\omega$
Emf of a rod or disc rotating about one end
Rotating rod or disc
$\varepsilon = \frac12 B\omega l^{2}$

Watch out for (6)

Leaving rpm unconverted
→ Emf of a coil rotating in a magnetic field
Zero flux does not mean zero emf
→ Emf of a coil rotating in a magnetic field
Angle with the plane versus angle with the normal
→ Emf of a coil rotating in a magnetic field
Forgetting the half
→ Emf of a rod or disc rotating about one end
Adding the emfs of fan blades
→ Emf of a rod or disc rotating about one end
Total field for a horizontal fan
→ Emf of a rod or disc rotating about one end

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