---
title: "Thermal power in rod moving in field"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:rod-on-rails-thermal-energy-rate"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/rod-on-rails-thermal-energy-rate
site: BCT Engineering
---

# Thermal power in rod moving in field

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2067 Asar, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/rod-on-rails-thermal-energy-rate
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

[Figure: a conducting rod of length L is pulled along horizontal, frictionless conducting rails at a constant velocity v; the rails are connected at one end with a metal strip; a uniform magnetic field B directed out of the page fills the region in which the rod moves.] Derive an expression for the rate of thermal energy being generated in the rod.

## Short answer

Induced emf ε = BLv, current I = BLv/R, thermal power P = I^2R = B^2 L^2 v^2 / R, which equals the mechanical power supplied by the external agent.

## Answer

Let the closed circuit (rod + rails + strip) have total resistance R. The rod has length L and moves at constant speed v in field B (out of the page).

```text
   . . . . . . . . . .
   ======+----------+====
   strip |  I       |-> v
   ======+----------+====
   . . . . . . . . . .   B out of page
```

#### Derivation
1. Area enclosed grows as the rod moves: dA/dt = Lv, so the flux Φ = BLx changes at dΦ/dt = BLv.
2. Faraday's law: induced emf ε = BLv.
3. Induced current: I = ε/R = BLv/R. Lenz's law gives a clockwise current for B out of the page and the area increasing.
4. Rate of thermal energy (Joule heating):

   P = I²R = (BLv/R)² R

**P = B²L²v²/R**

#### Check by energy conservation
- Magnetic force on the rod: F = ILB = B²L²v/R, opposing the motion.
- The agent pulling at constant speed must supply F_ext = F, so its power is F v = B²L²v²/R.

This equals I²R, so all the mechanical work appears as heat in the resistance. If only the rod has resistance, then R is the rod's resistance.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The emf in the rod moving at constant v is:
   A. ε = BLv
   B. ε = BL/v
   C. ε = B v/L
   D. ε = BLv/R
   Answer: A. dΦ/dt = B L v. (Key point: Induced emf)

2. The induced current in a circuit of resistance R is:
   A. I = BLvR
   B. I = BLv/R
   C. I = R/(BLv)
   D. I = B²L²v/R
   Answer: B. Ohm's law. (Key point: Current)

3. The rate of thermal energy generation is:
   A. P = BLv/R
   B. P = B²L²v/R
   C. P = I²R = B²L²v²/R
   D. P = BLv²R
   Answer: C. (BLv/R)²R. (Key point: Thermal rate)

4. Check by energy conservation:
   A. Pulling power is less than I²R
   B. Pulling power is zero at constant speed
   C. Heat comes from the magnetic field's own energy
   D. The pulling agent's power Fv equals I²R, so all mechanical work becomes heat
   Answer: D. F = B²L²v/R opposes motion; Fv = B²L²v²/R. (Key point: Energy conservation)

## Related in Engineering Physics

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