Motional EMF - EMF in Moving Conductors

Movie Hook: The Electromagnetic Railgun

In sci-fi movies and military tech demonstrations, railguns launch projectiles at hypersonic speeds using electromagnetic forces. When massive currents flow through conducting rails, the Lorentz force accelerates a sliding projectile. The same principle works in reverse: when a conductor moves through a magnetic field, it generates EMF. This is motional EMF - the basis of every electric generator, from massive hydroelectric dams to the alternator in your car!

What is Motional EMF?

When a conductor moves through a magnetic field (or a magnetic field moves relative to a conductor), an EMF is induced even though no coil or loop is involved.

Both are manifestations of the same phenomenon, just different viewpoints!

The Physics: Lorentz Force on Charges

When a conductor moves with velocity $\vec{v}$ in a magnetic field $\vec{B}$:

1. Free electrons in the conductor also move with velocity $\vec{v}$
2. These moving charges experience Lorentz force: $\vec{F} = q(\vec{v} \times \vec{B})$
3. Force pushes charges to one end of conductor
4. Charge accumulation creates potential difference (EMF)
5. Equilibrium when electric force = magnetic force

Motional EMF Formula

For a straight conductor of length $\ell$ moving perpendicular to magnetic field $B$ with velocity $v$:

Vector Form

For a conductor element $d\vec{\ell}$ moving with velocity $\vec{v}$:

$$d\mathcal{E} = (\vec{v} \times \vec{B}) \cdot d\vec{\ell}$$

Total EMF:

$$\mathcal{E} = \int (\vec{v} \times \vec{B}) \cdot d\vec{\ell}$$

Derivation: Two Perspectives

Method 1: From Lorentz Force (Microscopic View)

Consider a conductor of length $\ell$ moving with velocity $v$ in field $B$:

1. Force on electron: $F = evB$ (perpendicular to $v$ and $B$)
2. Electrons accumulate at one end until electric field $E$ develops
3. At equilibrium: $eE = evB$ → $E = vB$
4. Potential difference: $\mathcal{E} = E\ell = vB\ell = B\ell v$ ✓

Method 2: From Faraday’s Law (Macroscopic View)

Consider a rod sliding on rails forming a closed loop:

1. Area swept in time $dt$: $dA = \ell \cdot (v \cdot dt)$
2. Change in flux: $d\Phi_B = B \cdot dA = B\ell v \cdot dt$
3. Induced EMF: $\mathcal{E} = \frac{d\Phi_B}{dt} = B\ell v$ ✓

Both methods give same result! Choose based on problem context.

The Sliding Rod Problem (Classic JEE)

Setup: A conducting rod of length $\ell$ slides on two parallel conducting rails in a perpendicular magnetic field $B$. The rails are connected by resistor $R$.

Case 1: Rod Pulled with Constant Velocity

Given: Rod pulled with constant velocity $v$ to the right.

Find: Induced EMF, current, force required, power dissipated.

Solution:

Induced EMF:

$$\mathcal{E} = B\ell v$$

Induced current:

$$I = \frac{\mathcal{E}}{R} = \frac{B\ell v}{R}$$

Direction (by Lenz’s law): Opposes increase in flux → Current flows counterclockwise

Magnetic force on rod:

$$F_B = BI\ell = B \cdot \frac{B\ell v}{R} \cdot \ell = \frac{B^2\ell^2 v}{R}$$

Direction: Opposes motion (to the left, by Lenz’s law)

Applied force (to maintain constant velocity):

$$F_{app} = F_B = \frac{B^2\ell^2 v}{R}$$

(to the right)

Power dissipated in resistor:

$$P = I^2 R = \frac{(B\ell v)^2}{R}$$

Power delivered by applied force:

$$P = F_{app} \cdot v = \frac{B^2\ell^2 v}{R} \cdot v = \frac{B^2\ell^2 v^2}{R}$$

Energy conservation: ✓ Both powers are equal!

Case 2: Rod Released from Rest (No Applied Force)

Given: Rod given initial velocity $v_0$, then released.

Find: Velocity as function of time.

Solution:

Equation of motion:

$$m\frac{dv}{dt} = -F_B = -\frac{B^2\ell^2 v}{R}$$ $$\frac{dv}{v} = -\frac{B^2\ell^2}{mR}dt$$

Let $\alpha = \frac{B^2\ell^2}{mR}$

$$\int_{v_0}^{v}\frac{dv}{v} = -\alpha\int_0^t dt$$ $$\ln\frac{v}{v_0} = -\alpha t$$

Time constant: $\tau = \frac{mR}{B^2\ell^2}$

Distance traveled before stopping:

$$s = \int_0^{\infty} v \, dt = v_0\int_0^{\infty} e^{-\alpha t}dt = \frac{v_0}{\alpha} = \frac{v_0 mR}{B^2\ell^2}$$

Memory Trick: “BLV”

B-field × Length × Velocity = Motional EMF

Just remember “BLV” for $\mathcal{E} = B\ell v$ (when perpendicular)

Rotating Rod in Magnetic Field

Setup: A rod of length $L$ rotates about one end with angular velocity $\omega$ in perpendicular field $B$.

Key insight: Different parts of rod have different velocities!

At distance $r$ from pivot: $v = \omega r$

Motional EMF in small element $dr$:

$$d\mathcal{E} = B \cdot v \cdot dr = B\omega r \, dr$$

Total EMF:

$$\mathcal{E} = \int_0^L B\omega r \, dr = B\omega \int_0^L r \, dr = B\omega \frac{L^2}{2}$$

Physical interpretation:

• Average velocity: $v_{avg} = \frac{v_{tip}}{2} = \frac{\omega L}{2}$
• Average EMF: $\mathcal{E} = B \cdot v_{avg} \cdot L = \frac{1}{2}B\omega L^2$ ✓

Rotating Disc in Magnetic Field

Setup: Conducting disc of radius $R$ rotates about center with angular velocity $\omega$ in perpendicular field $B$.

EMF between center and rim:

Using the same logic as rotating rod:

$$\mathcal{E} = \frac{1}{2}B\omega R^2$$

This is a “disc generator” - used in some practical generators.

If connected in circuit:

• Current flows radially (from center to rim or vice versa)
• Magnetic force creates retarding torque (Lenz’s law)

Common JEE Problem Types

Type 1: Rod on Rails (Variable Force)

Problem: Rod starts from rest. Find velocity as function of time if constant force $F_0$ is applied.

Solution:

$$F_0 - \frac{B^2\ell^2 v}{R} = m\frac{dv}{dt}$$

Terminal velocity (when $\frac{dv}{dt} = 0$):

$$v_{terminal} = \frac{F_0 R}{B^2\ell^2}$$

Complete solution requires solving differential equation.

Type 2: Inclined Rails

Problem: Rod slides down frictionless inclined rails (angle θ) in perpendicular field.

Forces:

1. $mg\sin\theta$ (down the incline)
2. $\frac{B^2\ell^2 v}{R}$ (up the incline, electromagnetic braking)

Terminal velocity:

$$mg\sin\theta = \frac{B^2\ell^2 v_t}{R}$$ $$v_t = \frac{mgR\sin\theta}{B^2\ell^2}$$

Type 3: Rectangular Loop Entering/Exiting Field

Problem: Rectangular loop (width $w$, length $L$) moves into magnetic field region with velocity $v$.

While entering (distance $x$ inside):

• Flux: $\Phi_B = Bwx$
• EMF: $\mathcal{E} = \frac{d\Phi_B}{dt} = Bw\frac{dx}{dt} = Bwv$
• Same as motional EMF formula!

Completely inside: No change in flux → No EMF

While exiting: EMF again = $Bwv$ (opposite direction)

Direction of Induced Current

Use Fleming’s Right-Hand Rule (Generator rule):

• Thumb: Motion of conductor
• First finger: Field direction
• Second finger: Induced current direction

Or use Lenz’s law (always reliable):

1. Determine if flux is increasing/decreasing
2. Induced current opposes this change

Common Mistakes Students Make

❌ Mistake 1: Using $B\ell v$ when vectors aren’t perpendicular ✅ Reality: Use $B\ell v\sin\theta$ or vector form

❌ Mistake 2: Forgetting that different parts of rotating rod have different velocities ✅ Reality: Must integrate: $\mathcal{E} = \int_0^L B\omega r \, dr$

❌ Mistake 3: Not considering the retarding force (Lenz’s law) ✅ Reality: Moving conductor always experiences opposing force

❌ Mistake 4: Confusing EMF and potential difference ✅ Reality:

• EMF = $B\ell v$ (open circuit)
• PD across resistor = $IR$ (closed circuit)

❌ Mistake 5: Wrong direction of force or current ✅ Reality: Always use Lenz’s law or Fleming’s rules systematically

Practice Problems

Level 1: JEE Main (Basics)

Problem 1.1: A rod of length 0.5 m moves perpendicular to a magnetic field of 0.4 T with velocity 10 m/s. Calculate the induced EMF.

Solution

Given: $\ell = 0.5$ m, $B = 0.4$ T, $v = 10$ m/s

All three vectors are perpendicular, so:

$$\mathcal{E} = B\ell v = 0.4 \times 0.5 \times 10 = 2 \text{ V}$$

Answer: 2 V

Problem 1.2: A conducting rod rotates about one end at 120 rpm in a perpendicular magnetic field of 0.5 T. The rod length is 0.4 m. Find EMF between ends.

Solution

Given: $f = 120$ rpm $= 2$ rev/s, $B = 0.5$ T, $L = 0.4$ m

Angular velocity: $\omega = 2\pi f = 2\pi \times 2 = 4\pi$ rad/s

EMF in rotating rod:

$$\mathcal{E} = \frac{1}{2}B\omega L^2 = \frac{1}{2} \times 0.5 \times 4\pi \times (0.4)^2$$ $$\mathcal{E} = 0.5 \times 0.5 \times 4\pi \times 0.16 = 0.16\pi \approx 0.503 \text{ V}$$

Answer: 0.5 V

Level 2: JEE Advanced (Application)

Problem 2.1: A rod of length 1 m and mass 100 g slides on frictionless horizontal rails connected by a resistor of 10 Ω. The magnetic field is 0.5 T perpendicular to the plane. If the rod is given initial velocity 20 m/s, find: (a) Initial induced EMF (b) Initial acceleration (c) Time constant of velocity decay

Solution

Given: $\ell = 1$ m, $m = 0.1$ kg, $R = 10$ Ω, $B = 0.5$ T, $v_0 = 20$ m/s

(a) Initial EMF:

$$\mathcal{E}_0 = B\ell v_0 = 0.5 \times 1 \times 20 = 10 \text{ V}$$

(b) Initial acceleration: Initial current: $I_0 = \frac{\mathcal{E}_0}{R} = \frac{10}{10} = 1$ A

Initial force: $F_0 = BI_0\ell = 0.5 \times 1 \times 1 = 0.5$ N (opposing motion)

Initial acceleration: $a_0 = -\frac{F_0}{m} = -\frac{0.5}{0.1} = -5$ m/s² (negative = deceleration)

(c) Time constant:

$$\tau = \frac{mR}{B^2\ell^2} = \frac{0.1 \times 10}{(0.5)^2 \times 1^2} = \frac{1}{0.25} = 4 \text{ s}$$

Velocity equation: $v(t) = 20e^{-t/4}$ m/s

Answers: (a) 10 V, (b) -5 m/s², (c) 4 s

Problem 2.2: A rectangular loop (20 cm × 10 cm) enters a magnetic field region of 0.8 T with velocity 5 m/s. The field is perpendicular to the plane and the 10 cm side enters first. If the loop has resistance 2 Ω, find the power dissipated while entering.

Solution

Given: $w = 0.1$ m (entering side), $\ell = 0.2$ m, $B = 0.8$ T, $v = 5$ m/s, $R = 2$ Ω

While entering: EMF induced: $\mathcal{E} = Bwv = 0.8 \times 0.1 \times 5 = 0.4$ V

Current: $I = \frac{\mathcal{E}}{R} = \frac{0.4}{2} = 0.2$ A

Power dissipated:

$$P = I^2R = (0.2)^2 \times 2 = 0.08 \text{ W}$$

Alternative: $P = \frac{\mathcal{E}^2}{R} = \frac{(0.4)^2}{2} = 0.08$ W ✓

Answer: 0.08 W = 80 mW

Level 3: JEE Advanced (Challenging)

Problem 3.1: A metal disc of radius R rotates about its axis with angular velocity ω in a perpendicular magnetic field B. A circuit is formed by connecting center to rim through external resistance R. Find: (a) Current through resistor (b) Torque required to maintain constant ω (c) Power delivered by external agent

Solution

Given: Disc radius $R$, angular velocity $\omega$, field $B$, external resistance $R$

Let disc resistance = $r$ (usually negligible, assume $r \ll R$)

(a) Current: EMF: $\mathcal{E} = \frac{1}{2}B\omega R^2$

Current: $I = \frac{\mathcal{E}}{R+r} \approx \frac{B\omega R^2}{2R}$ (if $r \ll R$)

(b) Torque required: Current flows radially. Consider annular ring at radius $r$ of width $dr$:

• Current through ring: $I$ (same throughout)
• Force on ring: $dF = BI \cdot dr$
• Torque by ring: $d\tau = r \cdot dF = BIr \, dr$

Total torque:

$$\tau = \int_0^R BIr \, dr = BI\int_0^R r \, dr = BI \frac{R^2}{2}$$

Substituting $I = \frac{B\omega R^2}{2R}$:

$$\tau = B \cdot \frac{B\omega R^2}{2R} \cdot \frac{R^2}{2} = \frac{B^2\omega R^3}{4R}$$

(c) Power delivered:

$$P = \tau \omega = \frac{B^2\omega R^3}{4R} \cdot \omega = \frac{B^2\omega^2 R^3}{4R}$$

This equals $I^2R$ (check!):

$$I^2R = \left(\frac{B\omega R^2}{2R}\right)^2 R = \frac{B^2\omega^2 R^4}{4R^2} \cdot R = \frac{B^2\omega^2 R^3}{4R}$$

✓

Answers: (a) $I = \frac{B\omega R^2}{2R}$, (b) $\tau = \frac{B^2\omega R^3}{4R}$, (c) $P = \frac{B^2\omega^2 R^3}{4R}$

Problem 3.2: Two parallel conducting rails are placed on an inclined plane (angle 30°) in a region where magnetic field B = 1 T is perpendicular to the plane. A rod of mass 0.2 kg and length 0.5 m can slide on the rails. The rails are connected at top by resistance R = 5 Ω. Find the terminal velocity of the rod as it slides down.

Solution

Given: $\theta = 30°$, $B = 1$ T, $m = 0.2$ kg, $\ell = 0.5$ m, $R = 5$ Ω, $g = 10$ m/s²

At terminal velocity, net force = 0:

Forces:

1. Component of weight down the incline: $F_g = mg\sin\theta = 0.2 \times 10 \times 0.5 = 1$ N
2. Magnetic force up the incline: $F_B = BI\ell = B \cdot \frac{B\ell v_t}{R} \cdot \ell = \frac{B^2\ell^2 v_t}{R}$

At equilibrium:

$$mg\sin\theta = \frac{B^2\ell^2 v_t}{R}$$ $$v_t = \frac{mgR\sin\theta}{B^2\ell^2} = \frac{0.2 \times 10 \times 5 \times 0.5}{1^2 \times (0.5)^2}$$ $$v_t = \frac{5}{0.25} = 20 \text{ m/s}$$

Physical insight: Steeper incline or larger resistance → higher terminal velocity. Stronger field or longer rod → lower terminal velocity.

Answer: 20 m/s

Applications of Motional EMF

1. Electric Generators: Rotating coils in magnetic field (power plants)
2. Speedometers: Moving magnet induces current proportional to speed
3. Electromagnetic Railguns: Accelerate projectiles using Lorentz force
4. Electromagnetic Braking: Trains, elevators (eddy current braking)
5. Tachometers: Measure rotation speed
6. Velocity Sensors: Non-contact speed measurement
7. Electromagnetic Flowmeters: Measure flow of conducting fluids

Connection to Other Chapters

• Faraday’s Law: Motional EMF is special case
• Lenz’s Law: Determines direction of current and force
• Magnetic Force: Force on current-carrying conductor
• AC Generators: Rotating coil is continuous motional EMF

Key Takeaways

✓ Motional EMF: $\mathcal{E} = B\ell v$ (when perpendicular) ✓ Caused by Lorentz force on charge carriers ✓ Moving conductor experiences retarding force (Lenz’s law) ✓ Rotating rod: $\mathcal{E} = \frac{1}{2}B\omega L^2$ (must integrate) ✓ Energy conservation: Mechanical work = Electrical energy ✓ Terminal velocity when driving force = magnetic retarding force

Formula Summary

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Next Topic: Self Inductance - When a coil induces EMF in itself