Current & Magnetic Fields

Explore how electric current generates magnetic fields, and master the laws that govern their relationship: from the right-hand rule to Ampere's law.

Physics

What Is a Magnetic Field?

A magnetic field is an invisible region of influence surrounding magnets and moving electric charges. It exerts a force on other moving charges and magnetic materials. We represent it using the symbol \( \vec{B} \), measured in Tesla (T).

Electric Current Creates Magnetic Fields

In 1820, Hans Christian Oersted discovered that a wire carrying electric current deflects a compass needle. This was the first evidence that electricity and magnetism are related: a moving charge produces a magnetic field.

Why Does This Matter?

The relationship between current and magnetic fields is the foundation of electromagnetism: the technology behind electric motors, generators, MRI machines, and wireless communication.

Electric Motors

Current in a coil creates a magnetic field that interacts with permanent magnets to produce rotation.

Generators

The reverse: a changing magnetic field induces current in a wire, converting mechanical to electrical energy.

MRI Scanners

Powerful electromagnets create fields thousands of times Earth's to image the human body.

The Right-Hand Rule

To determine the direction of the magnetic field around a current-carrying wire, use the right-hand rule: point your thumb in the direction of current, and your fingers curl in the direction of \( \vec{B} \).

Magnetic Field of a Straight Wire

For an infinitely long straight wire, the magnitude of the magnetic field at a perpendicular distance \( r \) is given by:

\( B \)

Magnetic field strength in Tesla (T)

\( \mu_0 \)

Permeability of free space: \( 4\pi \times 10^{-7} \) T·m/A

\( I \)

Current through the wire in Amperes (A)

\( r \)

Perpendicular distance from the wire in metres (m)

Ampere's Law

Ampere's Law generalises the relationship between current and magnetic fields for any closed loop:

The line integral of \( \vec{B} \) around a closed Amperian loop equals \( \mu_0 \) times the total enclosed current \( I_{\text{enc}} \). This is especially powerful for high-symmetry configurations.

Solenoids vs Straight Wires

	Straight Wire	Solenoid
Geometry	Single conductor	Coil of many loops
Field shape	Concentric circles	Uniform inside, like a bar magnet outside
Formula	\( B = \frac{\mu_0 I}{2\pi r} \)	\( B = \mu_0 n I \)
Depends on	Distance \( r \)	Turn density \( n = N/L \)
Field inside	Decreases with \( r \)	Nearly constant

Inside a Solenoid: \( B = \mu_0 n I \)

How to Solve Problems

1

Identify the geometry. Is it a straight wire, loop, solenoid, or toroid?

2

Choose the right formula. Straight wire: \( B = \frac{\mu_0 I}{2\pi r} \). Solenoid: \( B = \mu_0 n I \).

3

Determine direction. Use the right-hand rule: thumb along current, fingers curl with \( \vec{B} \).

4

Substitute values. Plug in \( \mu_0 = 4\pi \times 10^{-7} \) T·m/A and given quantities.

5

Check units and direction. Verify your answer is in Tesla and the field direction is physically reasonable.

Example 1: Straight Wire

A long straight wire carries a current of \( I = 5 \) A. Find the magnetic field at a distance \( r = 0.02 \) m from the wire.

Example 2: Solenoid

A solenoid has \( N = 400 \) turns and length \( L = 0.20 \) m. It carries a current of \( I = 3 \) A. Find the magnetic field inside.

Example 3: Force on a Current in a Field

A wire of length \( L = 0.50 \) m carries \( I = 4 \) A through an external field of \( B = 0.30 \) T perpendicular to the wire. Find the force.

Field at the Centre of a Current Loop

A circular loop of radius \( R \) carrying current \( I \) produces a magnetic field at its centre:

Common Mistakes to Avoid

✕ Wrong Hand

Using the left hand instead of the right hand reverses the field direction entirely.

✕ Forgetting \( 2\pi \)

Writing \( B = \frac{\mu_0 I}{r} \) instead of \( B = \frac{\mu_0 I}{2\pi r} \). The \( 2\pi \) comes from the circular symmetry.

✕ Mixing Up \( n \) and \( N \)

For solenoids, \( n = N/L \) is turns per metre, not the total number of turns \( N \).

Summary of Key Formulas

Configuration	Formula	Key Variable
Straight Wire	\( B = \frac{\mu_0 I}{2\pi r} \)	\( r \): distance from wire
Solenoid	\( B = \mu_0 n I \)	\( n = N/L \): turns per metre
Circular Loop (centre)	\( B = \frac{\mu_0 I}{2R} \)	\( R \): loop radius
Force on Wire	\( F = BIL\sin\theta \)	\( \theta \): angle between \( \vec{I} \) and \( \vec{B} \)
Ampere's Law	\( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \)	\( I_{\text{enc}} \): enclosed current

Question 1 of 5

A long straight wire carries a current of 10 A. What is the magnetic field at a distance of 0.05 m from the wire?

\( B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi \times 10^{-7})(10)}{2\pi(0.05)} = \frac{4\pi \times 10^{-6}}{0.1\pi} = 4 \times 10^{-5} \) T.

Question 2 of 5

According to the right-hand rule, if current flows upward in a vertical wire, the magnetic field directly to the right of the wire points:

Point your right thumb upward (direction of \( I \)). Your fingers curl counterclockwise when viewed from above. At a point directly to the right of the wire, the field direction is into the page.

Question 3 of 5

A solenoid has 800 turns and is 0.40 m long. If it carries 2 A, what is the magnetic field inside?

\( n = 800/0.40 = 2000 \) turns/m. \( B = \mu_0 n I = (4\pi \times 10^{-7})(2000)(2) = 16000\pi \times 10^{-7} \approx 5.03 \times 10^{-3} \) T.

Question 4 of 5

A wire carrying 6 A is placed in a uniform magnetic field of 0.50 T. The wire is 0.30 m long and at \( 30^\circ \) to the field. What is the force on the wire?

\( F = BIL\sin\theta = (0.50)(6)(0.30)\sin 30^\circ = (0.90)(0.50) = 0.45 \) N. Remember \( \sin 30^\circ = 0.5 \).

Question 5 of 5

If you double the current in a long straight wire, and also double the distance from the wire, how does the magnetic field change?

\( B = \frac{\mu_0 I}{2\pi r} \). If \( I \to 2I \) and \( r \to 2r \), then \( B \to \frac{\mu_0(2I)}{2\pi(2r)} = \frac{\mu_0 I}{2\pi r} \). The field is unchanged.