GoalCalculate the flux density B near a long straight wire with B = μ0I ÷ 2πd and give its direction.
New words
flux density (B) · the strength of a magnetic field, measured in tesla (T)tesla (T) · the SI unit of flux density: 1 T = 1 Wb/m² = 1 N/(A·m)permeability (μ0) · the number that says how a material carries a magnetic field; for air μ0 = 4π × 10⁻⁷ T·m/A
Explanation
The strength of a magnetic field is measured by its flux density B in tesla (T). At a distance d from a long straight wire carrying a current I in air, B = μ0I ÷ (2πd). The number μ0 is the permeability of air (free space), μ0 = 4π × 10⁻⁷ T·m/A; the book writes the same unit as Wb/(A·m). Because 4π ÷ 2π = 2, the formula simplifies to B = 2 × 10⁻⁷ × I ÷ d, with I in amperes and d in metres, so convert centimetres first. Read the formula as a story: B is proportional to I (double the current, double the field) and inversely proportional to d (double the distance, half the field). In this course π = 3.14, so μ0 = 1.256 × 10⁻⁶ T·m/A whenever it is needed on its own. The direction comes from the right-hand rule of Lesson 1.
ƒField of a long straight wire (air)
B = μ0 I ÷ (2π d) = 2 × 10⁻⁷ × I ÷ d
B in tesla, I in amperes, d in metres
Worked examples
d (m)
0.025
0.05
0.10
0.20
B (10⁻⁵ T)
16
8
4
2
I = 20 A and d = 0.05 m. B = 2 × 10⁻⁷ × 20 ÷ 0.05 = 8 × 10⁻⁵ T. At d = 0.10 m the distance is doubled, so B = 4 × 10⁻⁵ T.
I = 15 A and d = 6 cm = 0.06 m. B = 2 × 10⁻⁷ × 15 ÷ 0.06 = 5 × 10⁻⁵ T, about the size of the Earth’s own field at the surface.
Find the current that gives B = 1.2 × 10⁻⁵ T at d = 0.03 m. Rearrange: I = B·d ÷ (2 × 10⁻⁷) = (1.2 × 10⁻⁵ × 0.03) ÷ (2 × 10⁻⁷) = 1.8 A.
Class activity
Table task. For a wire carrying 10 A in air, calculate B at d = 1 cm, 2 cm, 4 cm and 8 cm (convert to metres first). Write what happens to B each time d is doubled. If you like, plot B against d on graph paper and notice that the curve falls quickly near the wire and slowly far away.
Practice
1
A long straight wire carries 10 A. Find B at 0.04 m from it in air.
B = 2 × 10⁻⁷ × 10 ÷ 0.04 = 5 × 10⁻⁵ T.
2
A long straight wire carries 50 A. Find B at 25 cm from it in air.
d = 0.25 m. B = 2 × 10⁻⁷ × 50 ÷ 0.25 = 4 × 10⁻⁵ T.
3
B is 8 × 10⁻⁶ T at 0.1 m from a long straight wire. Find the current.
I = B·d ÷ (2 × 10⁻⁷) = (8 × 10⁻⁶ × 0.1) ÷ (2 × 10⁻⁷) = 4 A.
4
The current in a vertical wire flows downward. Seen from above, in which direction does the field point at a place north of the wire?
The field is clockwise seen from above, so north of the wire it points east.
