The Biot-Savart Law and the Field of a Current Element
For steady currents in vacuum, the Biot-Savart law adds vector field contributions from current elements. Each contribution depends on current, element length, angle and distance. Integration gives the fields of straight wires and circular loops.
What a learner can do afterwards
- Applies the Biot-Savart law to find the field of a straight wire and on the axis of a loop
- Uses the cross product to fix the direction of each contribution
- Finds the force between two parallel currents and says when it is attractive
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For steady current in a thin wire in vacuum, dB = (mu0 I/(4 pi r^2))(dl cross rhat), a vector equation. mu0 is vacuum permeability. dl follows conventional current; rhat is a unit vector from the element to the field point, a distance r away. Add contributions by vector integration. For a nonzero cross product, point your right index finger along dl and your middle finger along rhat; your thumb gives dB.
The contribution's magnitude is dB = mu0 I dl sin(theta)/(4 pi r^2). Here dl is element length and theta is the angle between dl and rhat. For a nonzero contribution with current, length and angle fixed, doubling r divides dB by four. Parallel or antiparallel vectors give zero contribution. Treating a finite short section as one element is an approximation: distance and angle must vary very little across it.
The bars compare a nonzero contribution with its value at twice the distance. Current, element length and angle stay fixed.
For an infinite straight wire, r is perpendicular distance to the wire and x measures position along it from the nearest point. Element distance is sqrt(x^2+r^2), and sin(theta) = r/sqrt(x^2+r^2). Here dx is a tiny wire length. An integral is the limit of sums as those lengths shrink to zero. At a fixed field point, all contributions share a direction, so B = (mu0 I/(4 pi)) integral from -infinity to infinity of r dx/(x^2+r^2)^(3/2).
For r/(x^2+r^2)^(3/2), an antiderivative is x/[r sqrt(x^2+r^2)]. Its derivative gives the integrand. At the infinite limits, subtract lower from upper: 1/r minus (-1/r) = 2/r. Thus B = mu0 I/(2 pi r). At fixed current, doubling perpendicular distance halves B. Field lines circle the wire in perpendicular planes. Point your right thumb along current; your curled fingers follow B.
The whole-wire result follows 1/r, unlike one fixed element's 1/r^2.
At an arc's centre, dl is perpendicular to rhat and every contribution has the same axial direction. For radius R, use dl = R dphi. Integrating gives B = (mu0 I/(4 pi R)) integral from 0 to theta of dphi = mu0 I theta/(4 pi R). Theta must be in radians. A full loop has theta = 2 pi, giving B = mu0 I/(2R). Other circuit parts may add field.
For a loop of radius R, x is axial distance from its centre. Split each field into components along the axis and perpendicular to it (transverse). Opposite elements cancel transverse components; axial components add. Each axial contribution is mu0 I R dl/[4 pi(R^2+x^2)^(3/2)]. Since integral dl = 2 pi R, B = mu0 I R^2/[2(R^2+x^2)^(3/2)]. At x = 0, B = mu0 I/(2R). Curl your right fingers with loop current; your thumb gives the axial field direction.
At the centre, the axial formula agrees with the full-circle arc integral.
Two long, thin, parallel wires in vacuum are separated by r. Wire 1 makes B1 = mu0 I1/(2 pi r) at wire 2, perpendicular to its current. The wire-force relation gives F = I2 L B1, so F/L = mu0 I1 I2/(2 pi r), using current magnitudes. Same-direction currents attract; opposite-direction currents repel. L is the length whose force is calculated.
For a force-per-length comparison, constants cancel. Multiply the factors for I1 and I2, then divide by the separation factor. For example, tripling I1 while leaving I2 fixed and halving separation multiplies F/L by 3 divided by 1/2, or 6. Decide attraction or repulsion separately from the current directions.
Add current-element fields as vectors. One fixed element follows 1/r^2; an infinite wire follows 1/r. Loop symmetry selects axial components. Parallel-current directions set attraction or repulsion.
Lesson, questions and flashcards written for Marble. Physics reference: University Physics Volume 2, sections 12.1 to 12.4.
Original readings and flashcards written for Marble, using University Physics Volume 2, sections 12.1 to 12.4 for physics fact study. The linked video is identified as Khan Academy material by the supplied source record; its license is unknown.
More to read
Build the field at one chosen point
Choose a field point before working along the wire. For each element, draw dl along conventional current and rhat from the element toward that point. Moving to another element usually changes both its distance and rhat.
For steady currents in vacuum, dB = (mu0 I/(4 pi r^2))(dl cross rhat). Its magnitude is mu0 I dl sin(theta)/(4 pi r^2). Reversing the order of the cross product reverses its direction. An element aimed directly toward or away from the field point contributes zero because its sine factor is zero.
Add the contributions as vectors. A scalar sum works only after you establish that their directions agree or select a common component. A short finite section can be approximated by one element when its distance and angle change very little across that section.
Keep two distances separate
In the infinite-wire formula B = mu0 I/(2 pi r), r is the shortest distance to the wire. It is not the distance to every current element.
Let x measure position along the wire from its nearest point. An element there is sqrt(x^2+r^2) from the field point. Its sine factor is r/sqrt(x^2+r^2), so the integrand contains r/(x^2+r^2)^(3/2). All elements contribute in the same direction at that fixed point.
The antiderivative x/[r sqrt(x^2+r^2)] changes from -1/r to +1/r across the infinite limits. This produces the integrated 1/r field. One fixed element still has inverse-square dependence when its current, length and angle are held fixed. The two distance comparisons describe different source arrangements. For a finite wire section, use its actual endpoints rather than infinite limits.
Pair opposite parts of a loop
A point on a circular loop's axis is the same distance from every wire element. Pair elements on opposite sides of the loop. Their transverse field components cancel, while their axial components reinforce each other.
With radius R and axial distance x from the centre, the result is B = mu0 I R^2/[2(R^2+x^2)^(3/2)] for steady current in vacuum. Setting x to zero gives mu0 I/(2R). Curl your right fingers along conventional current; your thumb gives the field direction on the axis.
The centre result can also be obtained by integrating an arc. At its centre, each element contributes along the same axis and dl = R dphi. An angle theta in radians gives B = mu0 I theta/(4 pi R). This calculates the arc's contribution. Return wires and other circuit parts must be considered separately when a total field is requested.
Use one wire's field at the other wire
For two long, thin, parallel wires in vacuum, first calculate the field of wire 1 at wire 2. If their separation is r, that magnitude is B1 = mu0 I1/(2 pi r).
The field crosses wire 2 at right angles. Substituting it into F = I2 L B1 gives F/L = mu0 I1 I2/(2 pi r). Use positive current magnitudes in this expression. Its unit is newtons per metre.
Current directions decide the force directions: matching directions attract and opposite directions repel. When comparing magnitudes, track three factors separately: the change in I1, the change in I2 and the change in separation. Multiply the current factors and divide by the separation factor. Changing direction can reverse attraction or repulsion without changing the magnitude calculated from the current magnitudes.
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