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Electromagnetism overview

Topic 4 of 7

Measure magnetic flux density

A current balance turns a magnetic force into a measurable change in a balance reading. The balance can weigh the magnet assembly while the force of interest acts on a separately supported wire.

Place a straight active wire length l perpendicular to the approximately uniform field between magnet poles. A low-voltage circuit supplies current through the wire, with an ammeter in series and a suitable current control. The magnet assembly rests on the balance; the wire and its support do not touch or rest on that assembly.

In the shown front view, current is right and B is into the page, from the front north pole towards the rear south pole. The wire feels an upward magnetic force. The magnet feels an equal downward reaction, increasing the balance reading. These forces act on different bodies.

The wire has its own supports

The wire has its own supportsA labelled front section omits the front north pole; the rear south pole is behind the active wire, so the gap field goes into the page. The wire crosses the gap without touching either pole. Separate insulated supports at its left and right stand on the bench outside the balance. A complete low-voltage loop contains a series control, a closed path through the ammeter, the active wire from left to right, and a return lead outside the field. Only the magnet and yoke rest on the balance pan. At 1.20 ampere the supplied model display is 150.60 grams, versus the separate zero-current baseline 150.00 grams. The active perpendicular length is 0.0500 metre. The white strip around the wire separates its foreground gap position from the projected rear pole; it is not a hole in or contact with that pole.Front N pole omittedRear S pole is behind the wire+-A1.20 ARear S150.60 gl = 0.0500 mOwn standOwn standCrosses: B into the pageLow-voltage d.c. circuit

This is a front section: the front N pole is omitted and the rear S pole lies behind the wire. Only the magnet assembly is weighed. The wire, its two insulated stands and the return leads do not touch that assembly.

Equal and opposite forces act on different bodies

Equal and opposite forces act on different bodiesThe top selected body is the supported active wire; its magnetic force is upward. The separate lower selected body is the magnet assembly; its magnetic reaction is downward. Both arrows have the same sixty-five-unit length to show equal magnitudes, 0.005886 newton at 1.20 ampere. Weight and support forces are omitted from these magnetic-interaction views, so they are not complete free-body diagrams. The downward reaction increases the balance display without changing the magnet mass.WireUp0.005886 NMagnet assemblyDown0.005886 N

These are the magnetic interaction forces only. Do not put both arrows on one body or infer that the magnet gains mass. Its additional downward force raises the balance reading.

Infer the field from the force-current gradient

Infer the field from the force-current gradientThe graph uses current in amperes horizontally and upward wire force in millinewtons vertically. Horizontal coordinate is sixty plus one hundred and fifty times current; vertical coordinate is three hundred and thirty minus thirty times force in millinewtons. The five supplied model points lie on the line F equals 4.905 I in millinewtons. Their coordinates include zero,zero; 0.4,1.962; 0.8,3.924; 1.2,5.886; and 1.6,7.848. The gradient is 4.905 millinewtons per ampere, or 0.004905 newtons per ampere.0246800.40.81.21.6Wire force upward / mNCurrent I / A

The gradient is 4.905 mN/A = 0.004905 N/A. Divide by the active length 0.0500 m to obtain B = 0.0981 T. A real record needs its own intercept and uncertainty assessment.

The labelled apparatus section omits the front north pole so the independently supported wire is visible; the rear south pole is behind it. The reaction panel separates the wire and magnet forces. The graph uses the supplied force change against current, with its millinewton scale stated.

Subtract the current-off baseline

The display is calibrated as a mass-equivalent value. A reading increase does not mean the magnet has gained physical mass. If Δm is the increase converted to kilograms, the additional downward force on the magnet is gΔm, equal in magnitude to the upward force on the wire.

A change around 0.5 g suggests a force around 0.005 N. With current around 1 A and active length around 0.05 m, this suggests B of order 0.1 T. Check that the actual balance resolution can resolve such a difference.

For the following supplied ideal model, use l = 0.0500 m and g = 9.81 N/kg. Positive current is right and positive wire force is up:

Generated current-balance values with a 150.00 g current-off baseline
I / ABalance display / gChange / g
0.00150.000.00
0.40150.200.20
0.80150.400.40
1.20150.600.60
1.60150.800.80
At 1.20 A: Δm = 0.60 g = 0.00060 kg
Fwire, up = gΔm = (9.81)(0.00060)
= 0.005886 N
B = F/(Il) = 0.005886/[(1.20)(0.0500)]
= 0.0981 T before rounding, about 0.098 T

The total 150.60 g display is not the magnetic change. Using it without subtracting the baseline would include the ordinary load supported by the balance. Reversing current to -1.20 A in this model gives 149.40 g: the wire force is now down and the magnet reaction up.

Use a gradient and investigate the intercept

At fixed field, active length and perpendicular angle, F = BlI. Plot signed wire force F = gΔm vertically against signed current I. The expected gradient is Bl, so:

Gradient = 0.004905 N/A = 4.905 mN/A
B = gradient/l = 0.004905/0.0500 = 0.0981 T

Convert a graph gradient in mN/A to N/A before using it. If plotting display change directly, convert its gradient to kg/A and multiply by g. The zero intercept follows from this ideal model; an actual nonzero intercept is something to investigate, not a reason to force the fit through the origin.

A complete measurement method

  1. Choose a low-voltage supply, current control, ammeter and wire within their ratings. Switch off before changing the wiring or active length. Measure the straight length that lies in the approximately uniform field.
  2. Support the wire independently, set it perpendicular to B, and keep its return conductors outside the active field region. Ensure no wire, lead or support touches the weighed magnet assembly.
  3. Check balance and ammeter zero, range and resolution. Record a current-off baseline and wait for a settled reading. The balance range must accommodate the assembly as well as the small magnetic change.
  4. Record several actual current settings and matching balance readings, with repeats. Keep active length, angle, wire position and magnet arrangement fixed. Monitor heating and baseline drift.
  5. Repeat with reversed current using the stated sign convention. Convert display differences to force, plot F against I, and infer B from the gradient divided by l.

Lead forces or contact with the magnet can change the balance reading without measuring only the intended active segment. Field nonuniformity and fringing make the effective field along that segment less simple; using an overstated active length would underestimate B for the same force/current reading. Current reversal helps distinguish a current-dependent force from a fixed offset, but does not remove every systematic error.

Subtraction also affects precision. If rounding alone places each of the two display readings within 0.005 g, their difference can be within 0.010 g. Relative to the 0.60 g change, this alone is about 1.7%, before current, length, alignment and other uncertainties. These are stated rounding bounds, not universal balance specifications.

Keep the original readings, including suspected anomalies. Investigate their cause and explain any selected fitting subset. The measurement-record method shows how to preserve actual readings separately from generated examples.

Optional check An independently supported wire carries 1.20 A through an active perpendicular length of 0.0500 m. The magnet assembly below it changes the balance display from 150.00 g to 150.60 g. Use g = 9.81 N/kg. What does this imply?
An independently supported wire carries 1.20 A through an active perpendicular length of 0.0500 m. The magnet assembly below it changes the balance display from 150.00 g to 150.60 g. Use g = 9.81 N/kg. What does this imply?