Unit 4 · Lesson 5b

5bElectromagnetism

Discover how electric currents create magnetic fields, how solenoids and electromagnets work, and the real-world applications that power modern civilization.

Electromagnetism is the foundation of modern technology — every electric motor, generator, transformer, and MRI machine depends on the relationship between electric currents and magnetic fields.

Lesson Overview

Electromagnetism reveals that electric currents create magnetic fields. In this lesson you will learn how a current-carrying wire produces a magnetic field, how solenoids and electromagnets amplify that field, and how Ampere's law relates the magnetic field to the current that creates it. You will also explore real-world applications including electric motors, MRI machines, and maglev trains.

Key Concepts

Field Around a Wire

B = μ₀I / (2πr); field forms concentric circles around the wire

Right-Hand Rule (Wire)

Wrap right hand around wire with thumb pointing in current direction — fingers curl in the direction of B

Solenoid

A coil of wire; field inside B = μ₀nI where n = turns per meter

Electromagnet

A solenoid with a ferromagnetic core; field is greatly amplified by the core

Ampere's Law

∮B·dl = μ₀I_enc; the line integral of B around a closed loop equals μ₀ times the enclosed current

Applications

Electric motors (force on current), MRI (strong uniform B), maglev trains (magnetic levitation)

Example 1

A long straight wire carries a current of 5.0 A. Calculate the magnetic field at a distance of 0.10 m from the wire. (μ₀ = 4π × 10⁻⁷ T·m/A)

Answer:B = μ₀I / (2πr) = (4π × 10⁻⁷ × 5.0) / (2π × 0.10) = (2.0 × 10⁻⁶) / (0.20) = 1.0 × 10⁻⁵ T.
Example 2

A solenoid has 500 turns wound over a length of 0.25 m and carries a current of 2.0 A. Find the magnetic field inside the solenoid.

Answer:n = 500/0.25 = 2000 turns/m. B = μ₀nI = (4π × 10⁻⁷)(2000)(2.0) = 5.0 × 10⁻³ T = 5.0 mT.
Example 3

Using the right-hand rule for a wire, determine the direction of the magnetic field directly above a wire carrying current to the right.

Answer:Point the right thumb to the right (current direction). The fingers above the wire curl out of the page. The magnetic field directly above the wire points out of the page.
Example 4

Explain how an electric motor uses electromagnetism to produce rotation.

Answer:A current-carrying coil is placed in a magnetic field. The magnetic force F = BIL sinθ acts on each side of the coil in opposite directions, creating a torque that rotates the coil. A commutator reverses the current direction every half-turn to maintain continuous rotation.
Example 5

A solenoid with an iron core has n = 1000 turns/m and carries 3.0 A. The relative permeability of iron is about 5000. Estimate the field inside.

Answer:B = μᵣμ₀nI = (5000)(4π × 10⁻⁷)(1000)(3.0) ≈ 18.8 T. The iron core amplifies the field by a factor of 5000 compared to air.
Guided Problem 1

A wire carries 8.0 A. At what distance from the wire is the magnetic field equal to 1.6 × 10⁻⁵ T?

Hint: Use B = μ₀I/(2πr) and solve for r.

Guided Problem 2

A solenoid of length 0.50 m has 1000 turns and carries 4.0 A. Calculate the magnetic field inside.

Hint: Find n = turns/length first, then use B = μ₀nI.

Guided Problem 3

Two parallel wires carry currents in the same direction. Do they attract or repel each other? Explain using the right-hand rule.

Hint: Find the direction of B from wire 1 at the location of wire 2, then find the force on wire 2 using F = BIL.

Guided Problem 4

Why does adding an iron core to a solenoid greatly increase the magnetic field?

Hint: Think about how the iron's magnetic domains respond to the external field.

Guided Problem 5

An MRI machine uses a superconducting solenoid with B = 1.5 T. If n = 10,000 turns/m, what current is required? (Ignore core effects.)

Hint: Rearrange B = μ₀nI to solve for I.

Key Vocabulary

Solenoid

A cylindrical coil of wire that produces a nearly uniform magnetic field inside when current flows through it.

Example: The solenoid in an MRI machine creates the strong, uniform field needed for imaging.

Electromagnet

A magnet created by passing electric current through a coil, often wound around a ferromagnetic core to amplify the field.

Example: Electromagnets in junkyards lift and release heavy metal objects by switching current on and off.

Permeability (μ₀)

A constant (4π × 10⁻⁷ T·m/A) that describes how easily a magnetic field forms in a vacuum.

Example: μ₀ appears in the formula for the magnetic field around a current-carrying wire: B = μ₀I/(2πr).

Ampere's Law

States that the line integral of B around any closed loop equals μ₀ times the net current passing through the loop: ∮B·dl = μ₀I_enc.

Example: Ampere's law is used to derive the magnetic field inside a solenoid: B = μ₀nI.

Interactive Practice — 5 Questions

1

What is the magnetic field at 0.20 m from a wire carrying 10 A? (μ₀ = 4π × 10⁻⁷ T·m/A)

2

The magnetic field inside a solenoid is given by:

3

Two parallel wires carrying current in the same direction:

4

Which device uses a strong, uniform magnetic field produced by a superconducting solenoid?

5

Ampere's Law states that ∮B·dl equals:

Independent Practice

1

Calculate the magnetic field 0.05 m from a wire carrying 15 A. (μ₀ = 4π × 10⁻⁷ T·m/A)

2

A solenoid is 0.40 m long, has 800 turns, and carries 5.0 A. Find the magnetic field inside.

3

Explain how a maglev train uses electromagnetism to levitate above the track.

4

Two parallel wires are 0.10 m apart and each carries 3.0 A in opposite directions. Do they attract or repel? Explain.

5

★ Derive the expression B = μ₀nI for the field inside an ideal solenoid using Ampere's Law. Draw the Amperian loop and show all steps.

Challenge
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Common Mistakes

Confusing the right-hand rule for a wire (field direction) with the right-hand rule for force on a charge.

For a wire: thumb along current, fingers curl in the direction of B. For force on a charge: fingers along v, curl toward B, thumb gives force direction.

Thinking the magnetic field inside a solenoid depends on the total number of turns N.

B = μ₀nI depends on n = N/L (turns per meter), not the total turns alone.

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Math Tips

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For B = μ₀I/(2πr): the field decreases as 1/r — doubling the distance halves the field.

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For a solenoid, B = μ₀nI is independent of the solenoid's radius — only n and I matter for an ideal solenoid.