9478 / 2027
Electromagnetic Forces overview

Lesson 3 of 9

Calculate a source magnetic field

  • I can use the appropriate magnetic field expressions for a long straight wire, flat circular coil and long solenoid.Syllabus

    Syllabus 9478, 17(c). Use the appropriate magnetic field expressions for a long straight wire, flat circular coil and long solenoid.

  • I can explain how a ferrous core can influence the magnetic field of a solenoid.Syllabus

    Syllabus 9478, 17(d). Explain how a ferrous core can influence the magnetic field of a solenoid.

Make a guess. You are not marked.

Where is the magnetic field of a long solenoid most nearly uniform?

  1. Inside, well away from the ends, where the lines are parallel and evenly spaced.
  2. Just outside the solenoid, where the field lines spread out evenly around it.
  3. At the ends, where the field lines are closest together.
Show the answer

Inside, well away from the ends, where the lines are parallel and evenly spaced.

Inside a long solenoid, away from the ends, the field lines are straight, parallel and evenly spaced, so B = mu0 n I is the same everywhere there.

Choose a field formula by its source geometry and the location where the field is required. Each equation gives a magnitude; use current direction and the field pattern to complete the answer.

Use the supplied free-space constant μ0 = 4π × 10-7 H/m, called the permeability of free space. The following expressions use an air or free-space approximation. They give the contribution of the named current, not automatically the total field when other sources are present.

Wire field: measure from the wire axis

Wire field: measure from the wire axisConventional current three amperes is out of the page. Point P lies to the right, a perpendicular distance 0.0400 metre from the wire axis, so its magnetic field is upward. The distance dimension starts at the axis, not the wire surface. The long-wire free-space model gives fifteen microtesla at P, not the same value everywhere.I = 3.00 APBd = 0.0400 mB = 15.0 µT at P

For the long-wire model, d = 0.0400 m is the perpendicular axis-to-point distance. The field at P is 15.0 microtesla upward.

Coil field: use radius at the centre

Coil field: use radius at the centreA representative face outline marks the common radius of a forty-turn flat coil; it is not one actual turn or a count of all forty. Anticlockwise conventional current is 0.600 ampere. The radius from centre to winding is 0.0500 metre, not the diameter. The free-space field at the centre is 0.301593 millitesla towards the observer.N = 40; I = 0.600 Arr = 0.0500 mCentre B = 0.302 mT

Use N = 40 and r = 0.0500 m at the centre. The outline represents the common winding radius; it does not count the turns. The result is about 0.302 mT towards you.

Solenoid field: count turns per unit length

Solenoid field: count turns per unit lengthA schematic axial section of a long air-core solenoid has length 0.400 metre, eight hundred actual turns and current 0.300 ampere. Only representative top and bottom wire cuts are drawn. At a central point away from the ends, the magnetic field is rightward. Turns per metre is two thousand, and the model predicts 0.753982 millitesla. The drawn length-to-width ratio and number of symbols are schematic.BN = 800; I = 0.300 AL = 0.400 mInterior B = 0.754 mT

n = N/L = 2000 per metre. The air-core result is about 0.754 mT in the central region. A ferrous core needs a separate physical description; this calculation does not give its new field.

Each schematic identifies the distance or length used by its equation and the location of the calculated field. The labelled turn counts are supplied values; the drawings do not count out every physical turn.

Long straight wire: perpendicular distance from its axis

For current I in a long straight wire, at perpendicular distance d and away from its ends:

B = μ0I/(2πd)

A few amperes at a few centimetres suggest a field of order 10-5 T using the supplied μ0. For I = 3.00 A and d = 0.0400 m:

B = [(4π × 10-7)(3.00)]/[(2π)(0.0400)]
= 1.50 × 10-5 T = 15.0 µT

The distance d is measured from the wire axis, not along its length. At fixed current, halving d doubles B: this is an inverse-distance relationship, not an inverse square. The field remains tangent to the circular pattern.

Flat circular coil: the field at its centre

For N turns of approximately common radius r in a thin flat coil:

Bcentre = μ0NI/(2r)

With N = 40, I = 0.600 A and r = 0.0500 m:

Bcentre = [(4π × 10-7)(40)(0.600)]/[2(0.0500)]
= 3.016 × 10-4 T ≈ 0.302 mT

Use radius rather than diameter. This is the central value, not one value for every point in the surrounding pattern. The central direction is along the coil axis, given by the right-hand grip.

Long solenoid: turns per unit length

Well inside a long solenoid, away from its ends:

B = μ0nI    n = N/L

Here n is turns per metre, not the carrier number density used in the drift-current model. For N = 800 turns over L = 0.400 m and I = 0.300 A:

n = 800/0.400 = 2000 m-1
B = (4π × 10-7)(2000)(0.300)
= 7.540 × 10-4 T ≈ 0.754 mT

Increasing total turns increases this field only as described by N/L: doubling both N and L leaves n unchanged. The interior direction follows the winding and current, while the end regions need a different geometrical treatment.

Ferrous core and units

A ferrous core such as soft iron becomes magnetised and can add strongly to a solenoid's field. For a suitable core at unchanged current, the interior field commonly increases. There is no universal multiplier: the air-core formula with μ0 alone does not calculate the new core field.

Convert prefixes before substitution: 1 mT = 10-3 T and 1 µT = 10-6 T. For the units of μ0, the henry is H = Wb/A. Using Wb = T m2 and T = N/(A m):

1 H/m = 1 Wb/(A m) = 1 N/A2
= 1 kg m s-2 A-2

These equivalent units make the three equations dimensionally consistent. They do not replace the source, location and material assumptions.

Check your understanding

At perpendicular distance d from a long straight wire, its source field is B. Current stays fixed. What is the source field at distance d/2, away from the wire ends?

  1. B/2.
  2. 4B.
  3. 2B.
Show the answer

2B.

For this source and location, B = mu_0 I/(2 pi d). Halving d doubles its magnitude. Direction still follows the local circular tangent.