🌀

Topic 9 of 15

Sources of Magnetic Field

Just as electric charges are the sources of electric fields, electric currents are the sources of magnetic fields. The Biot–Savart law gives the field of any current distribution by integration; Ampère's law gives it directly for highly symmetric geometries. These results underpin every electromagnet, transformer, and electric motor ever built.

Key Concepts

Biot–Savart Law

The magnetic field contribution from a small current element

I\,d\vec{l}

at distance

r

d\vec{B} = (\mu_0/4\pi)\,I\,d\vec{l}\times\hat{r}/r^2

. The full field is found by integrating over the entire current path.

\mu_0 = 4\pi\times10^{-7}

T·m/A is the permeability of free space.

Field of a Long Straight Wire

A long straight wire carrying current

I

produces a magnetic field that circles the wire:

B = \mu_0 I/(2\pi r)

at perpendicular distance

r

. Direction given by right-hand rule: thumb along current, fingers curl in the direction of

\vec{B}

Force Between Parallel Wires

Two parallel wires carrying currents

I_1

and

I_2

separated by distance

d

attract each other (same direction currents) or repel (opposite direction) with force per unit length

F/L = \mu_0 I_1 I_2/(2\pi d)

. This defines the ampere.

Ampère's Law

The line integral of

\vec{B}

around any closed path equals

\mu_0

times the net current enclosed:

\oint\vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}

. Analogous to Gauss's law — useful only when symmetry makes

B

constant along the path.

Solenoid and Toroid

A solenoid of

n

turns per unit length carrying current

I

has a uniform interior field

B = \mu_0 nI

and nearly zero field outside. A toroid (solenoid bent into a ring) has field

B = \mu_0 NI/(2\pi r)

inside and zero outside.

Key Equations

Field of a long straight wire

B = \frac{\mu_0 I}{2\pi r}

B at perpendicular distance r from a long wire carrying current I. μ₀ = 4π×10⁻⁷ T·m/A.

Force per unit length between wires

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

Force per unit length between two long parallel wires separated by d. Attractive for same-direction currents.

Ampère's Law

\oint\vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}

Line integral of B around a closed Ampèrian loop equals μ₀ times the enclosed current.

Field inside a solenoid

B = \mu_0 nI

Uniform field inside an ideal solenoid; n = N/L is turns per unit length. Field outside ≈ 0.

Worked Example

Magnetic Field of a Long Straight Wire and Force Between Wires

Problem

Two long parallel wires carry $I_1 = 5.0$ A and $I_2 = 3.0$ A in the same direction, separated by $d = 0.10$ m. Find (a) the field from wire 1 at wire 2's location, and (b) the force per unit length between them.

Solution

(a) Magnetic field of wire 1 at distance $d = 0.10$ m:

B_1 = \frac{\mu_0 I_1}{2\pi d} = \frac{(4\pi\times10^{-7})(5.0)}{2\pi(0.10)} = \frac{20\pi\times10^{-7}}{0.20\pi} = 1.0\times10^{-5}\text{ T} = 10\,\mu\text{T}

(b) Force per unit length on wire 2 in the field of wire 1:

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi\times10^{-7})(5.0)(3.0)}{2\pi(0.10)} = \frac{60\pi\times10^{-7}}{0.20\pi} = 3.0\times10^{-5}\text{ N/m} = 30\,\mu\text{N/m}

Same-direction currents attract, so the wires pull toward each other.

Answer B₁ at wire 2 = 10 μT; F/L = 30 μN/m (attractive).

Practice

Exercises

7 problems

1 of 7

A long straight wire carries $I = 10$ A. What is the magnetic field magnitude (in $\mu$ T) at a perpendicular distance $r = 0.050$ m? Use $\mu_0 = 4\pi\times10^{-7}$ T·m/A.

Your answer

μT

2 of 7

A solenoid has $n = 1000$ turns/m and carries $I = 2.0$ A. What is the magnetic field (in mT) inside it?

Your answer

3 of 7

Two long parallel wires carry $I_1 = 4.0$ A and $I_2 = 6.0$ A, separated by $d = 0.20$ m. What is the force per unit length (in $\mu$ N/m) between them?

Unlock Exercise 3