Motional EMF in Rotating and Sliding Systems under Gravity: Calculating the terminal velocity, induced current, and rate of heat dissipation for a conducting rod sliding down parallel vertical rails under gravity in a uniform magnetic field.

1. Introduction

When a conducting rod moves through a magnetic field, the free charges inside the rod experience a magnetic force. This separation of charges produces an induced emf called motional EMF.

Basic idea:
Moving conductor + Magnetic field → Motional EMF → Current (if circuit is closed)

A very important application is a conducting rod sliding on two parallel conducting rails under gravity in a uniform magnetic field.

2. Motional EMF

For a rod of length L moving with velocity v perpendicular to a uniform magnetic field B:

ε = BLv

Where:

• B = magnetic field
• L = length of conducting rod
• v = velocity of rod
• ε = induced EMF

The faster the rod moves, the larger the induced EMF.

3. Why is EMF Produced?

A charge q moving with velocity v in magnetic field B experiences the magnetic Lorentz force:

F = qvB

This force separates positive and negative charges along the rod. Consequently, a potential difference develops between its ends.

ε = BLv

4. Conducting Rod Sliding Down Rails

Consider a conducting rod of mass m and length L placed on two parallel vertical rails. The rod is free to slide downward under gravity.

A uniform magnetic field B is perpendicular to the plane of the rails. The circuit has total resistance R.

The important quantities are:

• Weight = mg
• Motional EMF = BLv
• Induced current = BLv/R
• Magnetic force = BIL

5. Induced Current

Using Ohm's law:

I = ε/R

Since:

ε = BLv

Therefore:

I = BLv/R

Thus induced current increases with velocity.

6. Magnetic Force on the Rod

A current-carrying conductor in a magnetic field experiences force:

F = BIL

Substituting:

I = BLv/R

We get:

F_B = B²L²v/R

According to Lenz's law, this magnetic force opposes the downward motion.

Magnetic force acts upward when the rod is sliding downward.

7. Equation of Motion

The downward gravitational force is:

mg

The upward magnetic force is:

B²L²v/R

Therefore:

m dv/dt = mg - B²L²v/R

or:

m dv/dt = mg - (B²L²/R)v

8. Terminal Velocity

At terminal velocity, acceleration becomes zero:

dv/dt = 0

Therefore:

mg = B²L²v_t/R

Solving:

v_t = mgR/(B²L²)

Important:
Terminal velocity increases with mass m and resistance R, but decreases with B² and L².

9. Current at Terminal Velocity

The current is:

I = BLv/R

Putting:

v = mgR/(B²L²)

we get:

I_t = BL/R × mgR/(B²L²)

Therefore:

I_t = mg/(BL)

This result is important for competitive examinations.

10. Rate of Heat Dissipation

Electrical power converted into heat in the circuit is:

P = I²R

Since:

I = BLv/R

Therefore:

P = B²L²v²/R

At terminal velocity:

P_t = m²g²R/(B²L²)

11. Energy Conservation

When the rod falls through a vertical distance, gravitational potential energy is converted into:

• Kinetic energy
• Electrical energy
• Heat energy

At terminal velocity, kinetic energy does not increase because velocity remains constant.

Therefore:

mgv_t = I_t²R

This confirms that gravitational power is completely converted into electrical heating at terminal velocity.

12. Power Balance at Terminal Velocity

Mechanical power supplied by gravity:

P_gravity = mgv_t

Electrical heating:

P_heat = I_t²R

Since:

I_t=mg/(BL)

and:

v_t=mgR/(B²L²)

Both expressions give:

P_gravity = P_heat

13. Velocity as a Function of Time

The equation is:

m dv/dt = mg - (B²L²/R)v

Define:

k = B²L²/(mR)

Then:

dv/dt = g-kv

If the rod starts from rest:

v(t) = v_t(1-e^-kt)

where:

v_t=mgR/(B²L²)

and:

k = B²L²/(mR)

14. Time Constant

The time constant of the motion is:

τ = mR/(B²L²)

The velocity can be written as:

v = v_t(1-e^-t/τ)

After a few time constants, the rod approaches terminal velocity.

15. Rotating Conducting Rod

For a conducting rod of length L rotating with angular velocity ω in a uniform magnetic field perpendicular to the plane of rotation, different parts of the rod have different linear velocities.

At a distance r from the axis:

v = ωr

The small EMF generated in length dr is:

dε = Bv dr

Therefore:

dε = Bωr dr

Integrating from r=0 to r=L:

ε = ∫₀ᴸ Bωr dr

Hence:

ε = ½BωL²

16. Sliding Rod vs Rotating Rod

System	Velocity	Motional EMF
Sliding rod	v	BLv
Rotating rod	ωr	½BωL²

17. Solved Example — Terminal Velocity

A conducting rod of mass 0.5 kg and length 0.5 m slides vertically on conducting rails. The magnetic field is 1 T and total resistance is 2 Ω. Find terminal velocity. Take g=10 m/s².

v_t=mgR/(B²L²)

v_t = (0.5)(10)(2) / (1²)(0.5²)

v_t=40 m/s

18. Solved Example — Terminal Current

For a rod of mass 2 kg, length 1 m in a magnetic field of 0.5 T, find the terminal current. Take g=10 m/s².

I_t=mg/(BL)

I_t = (2)(10)/(0.5)(1)

I_t=40 A

19. Solved Example — Heat Dissipation

A rod moves with velocity 5 m/s in a magnetic field of 2 T. Its length is 0.4 m and circuit resistance is 4 Ω. Find the rate of heat production.

P=B²L²v²/R

P = (2²)(0.4²)(5²)/4

P=4 W

20. Solved Example — Rotating Rod

A conducting rod of length 0.4 m rotates with angular speed 100 rad/s in a magnetic field of 0.5 T. Find the induced EMF.

ε=½BωL²

ε = ½(0.5)(100)(0.4²)

ε=4 V

21. Important Formula Sheet

Quantity	Formula
Motional EMF	ε=BLv
Induced current	I=BLv/R
Magnetic force	F=BIL
Magnetic force in terms of v	F=B²L²v/R
Equation of motion	m dv/dt=mg−B²L²v/R
Terminal velocity	vt=mgR/(B²L²)
Terminal current	It=mg/(BL)
Heat dissipation	P=I²R=B²L²v²/R
Terminal heat power	Pt=m²g²R/(B²L²)
Time constant	τ=mR/(B²L²)
Rotating rod EMF	ε=½BωL²

Section A — 1 Mark MCQs
10 Questions

Q1. The motional EMF in a rod of length L moving with speed v perpendicular to B is: 1 Mark

A) B/Lv

B) BLv

C) Bv/L

D) Lv/B

Answer: B

Q2. The induced current in a circuit of resistance R is: 1 Mark

A) BLvR

B) BLv/R

C) R/BLv

D) B/RLv

Answer: B

Q3. The magnetic force on the sliding rod is: 1 Mark

A) BIL

B) BI/L

C) BL/I

D) B/I L

Answer: A

Q4. At terminal velocity, acceleration is: 1 Mark

A) g

B) 2g

C) Zero

D) Infinite

Answer: C

Q5. Terminal velocity of the rod is proportional to: 1 Mark

A) B²

B) 1/R

C) mR

D) 1/m

Answer: C

Q6. The magnetic force on a falling rod acts: 1 Mark

A) Downward

B) Upward

C) Horizontally only

D) Zero always

Answer: B

Q7. At terminal velocity, gravitational force is equal to: 1 Mark

A) Electric force

B) Magnetic force

C) Normal force only

D) Zero

Answer: B

Q8. The rotating rod of length L has motional EMF: 1 Mark

A) BωL

B) BωL²

C) ½BωL²

D) 2BωL²

Answer: C

Q9. Electrical power dissipated as heat is: 1 Mark

A) IR

B) I²R

C) I/R

D) R/I

Answer: B

Q10. At terminal velocity, gravitational power is converted into: 1 Mark

A) Only kinetic energy

B) Electrical heating

C) Potential energy

D) Nuclear energy

Answer: B

Section B — 2 Marks
10 Questions

Q11. Define motional EMF. 2 Marks

The EMF induced in a conductor due to its motion through a magnetic field is called motional EMF. For perpendicular motion:

ε=BLv

Q12. Write the expression for induced current in the sliding rod. 2 Marks

I=BLv/R

Q13. Why does the magnetic force oppose the motion of the rod? 2 Marks

According to Lenz's law, the induced current produces a magnetic effect that opposes the change responsible for producing it. Hence the magnetic force opposes the motion of the rod.

Q14. What is terminal velocity? 2 Marks

Terminal velocity is the constant velocity attained when the downward gravitational force becomes equal to the upward magnetic force.

Q15. Write the terminal velocity of the rod. 2 Marks

v_t=mgR/(B²L²)

Q16. Write the terminal current. 2 Marks

I_t=mg/(BL)

Q17. Write the expression for heat dissipation in the circuit. 2 Marks

P=I²R

Q18. How does terminal velocity change if magnetic field is doubled? 2 Marks

Since:

v_t∝1/B²

If B is doubled, terminal velocity becomes one-fourth.

Q19. How does terminal velocity change if the resistance is doubled? 2 Marks

Since:

v_t∝R

Doubling R doubles the terminal velocity.

Q20. Why does the rod not accelerate indefinitely? 2 Marks

As velocity increases, induced current and magnetic opposing force increase. Eventually magnetic force becomes equal to weight, making acceleration zero.

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the terminal velocity of a conducting rod sliding under gravity. 3 Marks

Motional EMF:

ε=BLv

Current:

I=BLv/R

Magnetic force:

F=BIL=B²L²v/R

At terminal velocity:

mg=B²L²v_t/R

Therefore:

v_t=mgR/(B²L²)

Q22. Derive the terminal current in the sliding rod. 3 Marks

We know:

I=BLv/R

At terminal velocity:

v_t=mgR/(B²L²)

Therefore:

I_t = BL/R × mgR/(B²L²)

Hence:

I_t=mg/(BL)

Q23. Derive the rate of heat dissipation in terms of velocity. 3 Marks

Electrical power:

P=I²R

Current:

I=BLv/R

Therefore:

P=(BLv/R)²R

Hence:

P=B²L²v²/R

Q24. Show that at terminal velocity, mechanical power supplied by gravity equals electrical heat power. 3 Marks

Mechanical power:

P_g=mgv_t

At terminal velocity:

mg=BIL

Multiplying both sides by v:

mgv=BILv

Since:

BLv=IR

Therefore:

mgv=I²R

Hence:

P_gravity=P_heat

Q25. A rod of mass 1 kg and length 0.5 m moves on rails in a magnetic field of 2 T. Total resistance is 4 Ω. Find terminal velocity. Take g=10 m/s². 3 Marks

v_t=mgR/(B²L²)

v_t = (1)(10)(4) / (2²)(0.5²)

v_t=20 m/s

Q26. A rod of mass 2 kg and length 0.5 m moves in a magnetic field of 1 T. Find terminal current. Take g=10 m/s². 3 Marks

I_t=mg/(BL)

I_t = (2)(10)/(1)(0.5)

I_t=40 A

Q27. A rod of length 0.5 m moves at 10 m/s in a 2 T magnetic field. If total resistance is 5 Ω, calculate induced EMF and current. 3 Marks

EMF:

ε=BLv

ε=(2)(0.5)(10)=10 V

Current:

I=ε/R=10/5

ε=10 V,   I=2 A

Q28. A rotating conducting rod of length 0.6 m rotates with angular speed 50 rad/s in a magnetic field of 0.8 T. Find the induced EMF. 3 Marks

ε=½BωL²

ε = ½(0.8)(50)(0.6²)

ε=7.2 V

Q29. A rod of mass 0.5 kg, length 1 m and resistance 2 Ω moves vertically in a magnetic field of 1 T. Find terminal velocity and terminal current. Take g=10 m/s². 3 Marks

Terminal velocity:

v_{t}=mgR/(B²L²)}

v_t = (0.5)(10)(2)/(1²)(1²) =10 m/s

Terminal current:

I_t=mg/(BL)

I_t=(0.5)(10)/(1)(1) =5 A

v_t=10 m/s,   I_t=5 A

Q30. Explain why increasing the magnetic field decreases terminal velocity but increases the magnetic opposing force. 3 Marks

Terminal velocity:

v_t∝1/B²

Therefore increasing B decreases terminal velocity. But magnetic force is:

F_B=B²L²v/R

Hence, for a given velocity, increasing B increases the magnetic force. At terminal velocity this increased magnetic force balances the same weight mg.

22. Competitive Exam Quick Tricks

1. Sliding rod:

ε=BLv

2. Current:

I=BLv/R

3. Magnetic force:

F=B²L²v/R

4. Terminal velocity:

v_t=mgR/(B²L²)

5. Terminal current:

I_t=mg/(BL)

6. Heat power:

P=I²R=B²L²v²/R

7. At terminal velocity:

mgv_t=I_t²R

8. Rotating rod:

ε=½BωL²

Motional EMF in Rotating and Sliding Systems under Gravity
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