---
title: "Induced voltage in sliding bar"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:induced-voltage-sliding-bar"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/induced-voltage-sliding-bar
site: BCT Engineering
---

# Induced voltage in sliding bar

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2076 Chaitra, 3 marks
- Video: [Induced EMF In Moving Conductor, Sliding Bar Generator - Faraday's Law of Electromagnetic Induction](https://www.youtube.com/watch?v=ogPT-Py-ss0) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/induced-voltage-sliding-bar
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

A conducting bar can slide freely over two conducting rails placed at x = 0 and x = 10 cm. Calculate the induced voltage in the bar if the bar slides at a velocity V = 10a_y m/s and B = 3a_z mWb/m².

## Short answer

V_emf = ∮(v × B)·dl. v × B = 10 a_y × (3×10⁻³ a_z) = 0.03 a_x V/m. Integrated along 0.1 m gives V_emf = 3 mV. By Faraday's law, the rate of flux change is also 3 mWb/s, confirming 3 mV.

## Answer

The bar moves through a magnetic field, so a motional emf is induced:

```text
V_emf = ∮ (v × B)·dl
```
Here v = 10 a_y m/s, B = 3 a_z mWb/m² = 3×10⁻³ a_z T, and the bar lies along the x-direction between the rails at x = 0 and x = 10 cm = 0.1 m, so dl = dx a_x.

```text
v × B = (10 a_y) × (3×10⁻³ a_z) = 3×10⁻² (a_y × a_z) = 0.03 a_x  V/m

V_emf = ∫(x=0→0.1) 0.03 a_x · dx a_x = 0.03 × 0.1 = 3×10⁻³ V
```

Check with Faraday's law: the loop area increases at rate l v = 0.1 × 10 = 1 m²/s, so |emf| = B (dA/dt) = 3×10⁻³ × 1 = 3 mV.

**Answer: V_emf = 3 mV**, with the emf directed along +a_x in the bar.

## Check yourself

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

1. The motional emf is:
   A. ∮ B·dl
   B. ∮ (v × B)·dl
   C. ∮ v·dl
   D. B/v
   Answer: B. V_emf = ∮(v × B)·dl. (Key point: Motional emf formula)

2. With v = 10 a_y and B = 3×10^−3 a_z, v × B is:
   A. 0.03 a_x V/m
   B. 0.03 a_y V/m
   C. −0.03 a_x V/m
   D. 30 a_z V/m
   Answer: A. a_y × a_z = a_x, so (10)(3×10^−3) a_x = 0.03 a_x. (Key point: Cross product)

3. Faraday's check B·l·v gives:
   A. 3 mV
   B. 30 mV
   C. 0.3 mV
   D. 3 V
   Answer: A. 3×10^−3 × 0.1 × 10 = 3×10^−3 V. (Key point: Result and check)

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