The Biot–Savart law
The field of a current element
Every current-carrying wire, whatever its shape, can be cut into tiny straight pieces. Each piece, of length dl and carrying current I, is called a current element, written I dl.
Each element makes its own small magnetic field dB at a point P. Add up the contributions of all the elements, and you get the field of the whole wire.
Let r be the vector from the element to P, and θ the angle between dl and r. Biot and Savart found that dB is proportional to I, to dl, and to sin θ, and falls off as 1 over r².
In vector form: dB equals μ₀ over 4π, times I dl cross r, over r³. Its size is μ₀ over 4π, times I dl sin θ, over r².
μ₀ is the permeability of free space. μ₀ over 4π is exactly 10⁻⁷ tesla metre per ampere, so μ₀ is 4π × 10⁻⁷.
Which way does dB point? Perpendicular to the plane that holds dl and r. Turn a right-handed screw from dl towards r: the direction it advances is the direction of dB.
Walk round the element at a fixed distance. dB keeps turning with you, always at a tangent. So the field lines of an element are circles around its line.
Straight ahead of the element, along its own line, θ is 0 and sin θ is 0. There, dB is zero. It is largest at right angles to the element, where sin θ is 1.
Notice the likeness to Coulomb’s law: both fields fall as 1 over r², and both obey superposition. In fact μ₀ and ε₀ are linked to the speed of light: c equals 1 over √(μ₀ε₀).
Compared with Coulomb’s law
Put the two laws side by side. An electric field comes from a charge q, which is a scalar. A magnetic field comes from a current element I dl, which is a vector, with a direction of its own.
The electric field of a point charge points along r, straight away from the charge. The magnetic field of an element is perpendicular to r, and to dl as well.
And the magnetic field depends on the angle: it carries a sin θ. The electric field of a point charge is the same in every direction at the same distance.
Now an example. A small element of length 1 centimetre lies at the origin along the x-axis, and carries a large current of 10 amperes. Find the field at a point on the y-axis, 0.5 metre away.
Here θ is 90°, so sin θ is 1. dB equals 10⁻⁷ times 10 times 10⁻², over 0.5 squared.
That is 10⁻⁸ over 0.25, which is 4 × 10⁻⁸ tesla. A small field, because the element is so short.
Its direction is along dl cross r: x cross y, which is the z-axis. So the field points out of the page, along plus z.
Now move P to 0.5 metre along the x-axis, in line with the element. θ is 0, and the field is zero there.
Finally, keep the distance 0.5 metre but make θ equal to 30°. sin 30° is one half, so the field halves to 2 × 10⁻⁸ tesla.
To find the field of a whole wire, we add up such contributions from every element. In the next lesson we do this for a circular loop.
Try it: the field of an element
- The amber arrow dB is always perpendicular to the blue plane that holds dl and r. Drag the scene to check.
- Slide “Go round the element”: dB stays tangent to a circle around the element’s line.
- Set θ = 0° or 180°: P lies on the line of the element, and dB vanishes.
- At θ = 90°, double r from 0.5 m to 1 m. dB drops to a quarter: 1/r².
In line with the element
A current element I dl points along the x-axis at the origin. What is the magnetic field dB it produces at a point on the x-axis, 0.5 m away?
- Largest, because the point is straight ahead of the element
- Zero
- (μ₀/4π) I dl / r², along the x-axis
- Half its value on the y-axis at the same distance
Answer
B. dB = (μ₀/4π) I dl sin θ / r². On the line of the element, dl and r are parallel, so θ = 0 and sin θ = 0: the field is zero. Unlike the electric field of a charge, the magnetic field of an element is never along r.
Farther and at an angle
An element 1 cm long carrying 10 A makes dB = 4 × 10⁻⁸ T at a point 0.5 m away at right angles to it. What is dB at a point 1.0 m away along a line at 30° to the element?
- 5 × 10⁻⁹ T
- 8.7 × 10⁻⁹ T
- 1 × 10⁻⁸ T
- 2 × 10⁻⁸ T
Answer
A. Doubling r divides dB by 2² = 4, and sin 30° = ½ halves it again: 4 × 10⁻⁸ × ¼ × ½ = 5 × 10⁻⁹ T. Using cos 30° gives 8.7 × 10⁻⁹ T; using 1/r instead of 1/r² gives 1 × 10⁻⁸ T; forgetting the change in r gives 2 × 10⁻⁸ T.
Summary
- Biot–Savart law:
dB = (μ₀/4π) I dl × r / r³ - Size:
dB = (μ₀/4π) I dl sin θ / r² - Permeability of free space:
μ₀/4π = 10⁻⁷ T m A⁻¹, μ₀ = 4π × 10⁻⁷ T m A⁻¹ - Link with ε₀:
c = 1/√(μ₀ε₀)
- Each current element I dl makes a field dB perpendicular to the plane of dl and r (right-hand screw from dl to r).
- dB falls as 1/r² and is zero along the line of the element.
- Like Coulomb’s law it is long-range and obeys superposition; unlike it, the source I dl is a vector and the field depends on the angle θ.
- The field of any wire is the sum (integral) of the fields of all its elements.