9478 / 2027
Electromagnetic Forces overview

Full chapter

Electromagnetic Forces

All 9 topics and the revision summary on one page.

01

Motion in a uniform electric field

An electric field exerts a force according to charge sign. In a uniform field, that force has a fixed direction, so resolve the motion along and perpendicular to the field.

Electric field direction is the force direction on a positive test charge. For a particle with signed charge Q, the vector relation is F = QE. Its force magnitude is |Q|E: positive charge is forced along the field and negative charge against it. Field strength E has unit N/C, equivalent to V/m.

With unchanged charge and mass and no other significant force, Newton's second law gives constant acceleration a = QE/m in a chosen direction. The electric force exists even if the particle is initially stationary. Its direction does not depend on the particle's initial velocity.

Parallel and perpendicular entry

A positive particle already moving along E speeds up. Moving against E, it slows and can reverse. A negative particle moving along E instead slows and can reverse, because its force is opposite E; moving against E, it speeds up. State both the charge sign and the velocity direction before deciding.

With initial velocity perpendicular to E, the component perpendicular to the field remains constant while the component along the field changes. Constant motion in one direction combined with constant acceleration in the other produces a parabola. Use the constant-acceleration equations separately for the two components.

Worked electron beam

Find the complete exit state

An electron enters horizontally right at 2.00 × 107 m/s into a uniform field 4000 N/C downward. The field region is 0.0600 m long and the vertical plate separation is 0.0300 m. Entry is halfway between the plates. The upper plate is positive relative to the lower.

Use electron charge Q = -e, with e = 1.60 × 10-19 C, and electron mass me = 9.11 × 10-31 kg. Neglect gravity, collisions, fringing and radiation, and use the classical motion model. The field strength is supplied directly.

A small deflection at equal position scales

A small deflection at equal position scalesThe plate edges run from x sixty to three hundred drawing units, with upper edge at one hundred and sixty and lower at two hundred and eighty. Four thousand drawing units represent one metre on both physical axes. The electron enters at sixty, two hundred and twenty and leaves at three hundred, 207.354555, so its upward displacement is 3.161361 millimetres. Filled points at one, two and three nanoseconds lie on the exact parabola. The electric field points down and the electron force up. Outside the right edge, the trajectory continues on its true exit tangent. Force and field arrows have separate schematic lengths, not position or common magnitude scales.Given E = 4000 N/C down+ + + + +- - - - -Ev inFField length: 0.0600 mPlate gap: 0.0300 m

Positions use the same scale horizontally and vertically. The electron remains inside the 0.0300 m gap. Its horizontal speed stays constant while its total speed increases; the straight continuation is the exit tangent.

The electron is forced upward in the downward electric field. Equal horizontal and vertical distance scales preserve the small deflection; the path becomes a straight exit tangent outside the ideal field region. Field and force arrows represent different quantities.

Take right and up as positive. The field component is Ey = -4000 N/C, so the negative charge gives an upward force:

Fy = QEy
= (-1.60 × 10-19)(-4000)
= +6.40 × 10-16 N
ay = Fy/me
≈ +7.025 × 1014 m/s2

An acceleration of order 1015 m/s2 acting for a few nanoseconds suggests a displacement on a millimetre scale. The short transit time matters even though the acceleration is very large.

There is no horizontal force, so vx remains 2.00 × 107 m/s. Initially vy = 0, but the total initial velocity is not zero:

t = 0.0600/(2.00 × 107)
= 3.00 × 10-9 s = 3.00 ns
y = ½ayt2 = +0.003161 m
vy = ayt = +2.108 × 106 m/s
Calculated states of the supplied electric-beam model; vx is constant
t / nsy / mmvy / (106 m/s)
000
1.000.35130.7025
2.001.4051.405
3.003.1612.108

The exit displacement is 3.16 mm upward, less than the 15.0 mm half-gap, so the electron reaches the far edge without striking the upper plate. Its total exit speed is:

v = √(vx2 + vy2)
≈ 2.011 × 107 m/s
Exit angle = tan-1(vy/vx) ≈ 6.02° above right

Keeping extra digits makes the speed increase visible; to three significant figures it is 2.01 × 107 m/s. The field has done positive work on the electron. Constant horizontal velocity does not imply constant total speed.

Outside the ideal field, with no further force, the electron keeps its exit velocity and follows a straight tangent. Reversing the charge sign reverses the force for the same E; a quantitative path comparison also needs the particle's mass and entry velocity.

The electric-field deflection page connects this motion to the wider electric-potential and energy account.

Optional check An electron enters a uniform downward electric field moving horizontally right. Neglect other forces. While it is inside the field, which description is correct?
An electron enters a uniform downward electric field moving horizontally right. Neglect other forces. While it is inside the field, which description is correct?

02

Magnetic sources and field patterns

Permanent magnets and electric currents produce magnetic fields. A field pattern describes direction at each point; it is separate from the force or path of a particular object placed there.

A compass's north-seeking end indicates the local magnetic field direction. Outside a bar magnet, field lines run from north to south, returning inside it to form closed paths. Lines do not cross. Closer spacing can represent greater field strength within a consistently drawn pattern, but the lines are not material tracks.

The presence of a field does not mean every object experiences a nonzero force. For example, magnetic force on a charge depends on its velocity as well as the field. Keep the source of the field distinct from the body whose force is being considered.

Read the viewpoint and current direction

A dot means a named direction out of the page towards you, like an approaching arrowhead. A cross means into the page away from you, like the tail of an arrow. Check whether the symbol labels current, field or force. For an ordinary page view, use x right, y up and z out of the page; a separate end view must say where the observer is looking from.

Straight wire: look along its axis

Straight wire: look along its axisA brown central dot denotes conventional current out of the page. Three concentric teal circles have anticlockwise field arrows. At the point immediately to the right of the wire the field is upward, tangent to a circle. These are magnetic field lines, not particle paths.I outAt the right: B points up.

Brown dot: current towards you. Teal circular arrows: magnetic field anticlockwise. Reversing the current reverses the field.

Flat coil: view from the right along its axis

Flat coil: view from the right along its axisThe observer is on the right of the axial-section view, looking left towards the flat coil. Conventional current is anticlockwise in this face view, shown by brown arrows around the winding. The teal central dot denotes magnetic field towards this observer, which is rightward in the axial section. The circular brown path is the winding, not a circular magnetic-field line around the coil centre.B towards youCurrent anticlockwiseView from the right

Here the central dot means field towards you, not wire current. The observer stands on the right of the next axial-section view.

Flat coil: a section through its axis

Flat coil: a section through its axisThis axial section cuts the winding at top and bottom. Brown dot at the top means current out of the page; brown cross at the bottom means current in. Two closed noncrossing qualitative field paths point right through the interior and return left outside the coil. The central field arrow points right along the horizontal coil axis. The coil face in the preceding view was observed from the right.I outI inInterior B is rightward.Observer for face view is on the right.

The current symbols are wire cross-sections. Field lines pass through the interior along the axis and return outside; their shapes here are qualitative.

Long solenoid: an axial section

Long solenoid: an axial sectionRepresentative upper wire cuts carry current out of the page, and lower cuts carry current in. They belong to turns whose unseen parts connect outside the section; the drawing does not count the actual turns. Two noncrossing closed field paths run right through the interior, return left outside and bend at the ends. The left end is south and right end north. Approximately parallel central lines describe the interior away from the ends, not uniform field everywhere.SNRepresentative wire cutsNot a count of the actual turns

Top current out, bottom current in gives B right inside: S at the left, N at the right. The central region is approximately uniform; the ends and outside are not.

The straight-wire view and the flat-coil face view use their own stated viewing directions. In the matching coil and solenoid axial sections, top-wire current is out of the page and bottom-wire current into it, giving an interior field to the right. Return lines close the pattern outside.

Long straight wire

Field lines are concentric circles in planes perpendicular to the wire. Point your right thumb along conventional current; your curled fingers give the field direction. Current out of the page produces anticlockwise field lines in that face-on view. Immediately to the wire's right, the tangent field direction is upward.

Reversing current reverses every field direction. The lines are circular, not radial outward from the wire. A compass lies tangent to a local circle, not along the wire's current.

Flat circular coil

The field passes through the coil's centre along its axis and returns outside the coil. Viewed at a face, anticlockwise current gives a central field towards the observer; clockwise current gives it away. Curl the right-hand fingers with current around the coil and use the thumb for the central axial field.

The face and axial-section drawings describe the same arrangement from different directions. Do not draw the whole coil's field as concentric circles in the coil's plane: that is not the field pattern through its centre.

Long solenoid

A solenoid is a coil with many turns along a length. Well inside a long solenoid and away from its ends, the field is approximately uniform, with parallel, similarly spaced lines. Outside it, weaker field lines return from the north end towards the south end. The end region spreads out, so interior uniformity does not extend unchanged through the ends.

Curl the right fingers with current around the turns. The thumb gives the interior field and points towards the solenoid's north end. In the shown axial section, the interior field is right and the right end is north. Reversing current swaps the effective poles.

Map a field and distinguish its strength from flux

Move a small compass to several positions, record its north-end direction and connect smooth tangents. Keep current, source geometry and nearby magnetic objects fixed. Repeat with reversed current to compare the source contribution with Earth's or other background fields. Iron filings can suggest the pattern but do not by themselves identify its direction.

Magnetic flux density B describes the local field and is measured in teslas, T. Magnetic flux Φ describes the field through an area and is measured in webers, Wb. For this supplied uniform-field model with the area perpendicular to B, use Φ = BA.

A rough field of 0.1 T through an area of order 10-3 m2 suggests flux of order 10-4 Wb. More precisely, a 0.0800 T perpendicular field through a 2.00 cm by 3.00 cm rectangle gives:

A = (0.0200)(0.0300) = 6.00 × 10-4 m2
Φ = BA = (0.0800)(6.00 × 10-4)
= 4.80 × 10-5 Wb = 48.0 µWb

Doubling this area at unchanged B doubles the flux, while local B stays the same. One microweber is 10-6 Wb. Using the force definition of the tesla, and J = N m and C = A s:

1 Wb = 1 T m2 = 1 N m/A
= 1 J/A = 1 V s
= 1 kg m2 s-2 A-1

Keep the quantity and unit together: B in T and Φ in Wb are different, even when one uniform field is used to calculate both.

Optional check A long straight wire carries conventional current out of the page towards you. What is the direction of its magnetic field immediately to the right of the wire?
A long straight wire carries conventional current out of the page towards you. What is the direction of its magnetic field immediately to the right of the wire?

03

Calculate a source magnetic field

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.

Optional check 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?
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?

04

Force on a current-carrying conductor

A magnetic field can exert a force on a current-carrying wire. Find its magnitude from the active length and angle, and its direction from the field and conventional current.

For a straight conductor with length l inside a uniform external field B, carrying current magnitude I:

F = BIl sin θ

The angle θ is between current and B. The active length l is the part in the field, not automatically the entire wire. Use the field produced by the magnet or another source; do not treat the wire's own field as though it pushes the wire sideways by itself.

Perpendicular current gives the maximum BIl. A conductor parallel or antiparallel to B gives zero magnetic force because sin θ = 0. A current-carrying conductor can therefore be in a nonzero field without experiencing a magnetic force.

Use Fleming's left-hand rule

Hold the left thumb, first finger and second finger mutually perpendicular. The first finger is field, the second is conventional current and the thumb is force on the conductor. If current is not perpendicular to B, use its perpendicular component for the direction rule; the sine factor supplies its magnitude.

For current right and B into the page, force is up. Reversing either current or field reverses the force; reversing both restores it. For current up and B right, the force is into the page. Name the selected conductor: the magnet's opposite reaction acts on a different body.

Perpendicular current: the wire force is upward

Perpendicular current: the wire force is upwardThe selected body is a straight conductor of active length 0.120 metre. Conventional current is 2.50 amperes rightward; crosses indicate an external field of 0.0800 tesla into the page. The purple force arrow is upward. Current and field are perpendicular, giving magnetic-force magnitude 0.0240 newton. The diagram shows the magnetic force only, not weight or support forces.IFB = 0.0800 T into pagel = 0.120 m; I = 2.50 AF = 0.0240 N upward

Brown: conventional current. Teal crosses: external B into the page. Purple: magnetic force on the chosen wire. The field, current and force use separate schematic arrow scales.

Parallel current and field: no magnetic force

Parallel current and field: no magnetic forceThe external magnetic field and conventional current both point right along parallel directions. The selected conductor still carries current, but the angle between current and field is zero and the magnetic force is zero. There is deliberately no force arrow. Reversing the current alone makes the angle one hundred and eighty degrees and still gives zero force.External BCurrent IAngle = 0°; F = 0

A current can be present while magnetic force is zero. Parallel and antiparallel orientations both give sin(theta) = 0.

A real 30-degree angle in the page

A real 30-degree angle in the pageIn this separate orientation, external B points vertically upward. The current direction has rightward component one half and upward component square root three divided by two. Its one-hundred-unit guide goes from 135,280 to 185,193.39746, making thirty degrees to B on equal horizontal and vertical direction scales. The purple dot at the conductor means force out of the page, towards the viewer. With the same current, active length and field magnitude, force is 0.0120 newton. The angle is not an in-page angle to an into-page field.BI30°Force outF = 0.0120 NSame B, I and active length l

This is a different physical orientation from the first panel. The angle is between I and B. Their cross-product direction gives force out of the page, and sin(30 degrees) halves the force magnitude.

The three panels show the perpendicular, parallel and separate 30-degree arrangements. The angle is genuinely between current and field: in the angled view B is upward and current tilts to the right, giving force out of the page. These are magnetic-force directions, not complete force balances.

Calculate and check the scale

A rough B of 0.1 T, I of 2 A and active length of 0.1 m suggest a perpendicular force around 0.02 N. For the supplied precise model, B = 0.0800 T into the page, I = 2.50 A right and l = 0.120 m:

F = (0.0800)(2.50)(0.120) = 0.0240 N upward

Now consider the separate arrangement with the same magnitudes and an actual 30.0° angle between current and B:

F = (0.0800)(2.50)(0.120) sin 30.0°
= 0.0120 N

In its shown view, B is up and current is 30.0° to the right of up, so the force is out of the page. An in-page arrow cannot make a 30° angle to a field directed straight into the page; those directions would be perpendicular.

Define magnetic flux density

Magnetic flux density is the force per unit current per unit length on a straight conductor placed perpendicular to the field:

B = F/(Il)   when θ = 90°
1 T = 1 N/(A m) = 1 kg s-2 A-1

One tesla gives 1 N on a 1 m perpendicular active length carrying 1 A. Without the perpendicular condition, F/(Il) would give B sin θ instead of B. A field value in teslas is not a force until the current, active length and orientation are known.

Optional check A straight active length 0.120 m carries 2.50 A in a uniform external field of 0.0800 T. The angle between current and field is 30.0 degrees. What is the magnetic force magnitude?
A straight active length 0.120 m carries 2.50 A in a uniform external field of 0.0800 T. The angle between current and field is 30.0 degrees. What is the magnetic force magnitude?

05

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.

Software exercise with the generated balance values

Convert, plot and infer the field

In a blank sheet, enter the five current/display pairs above in A2:B6, with A1 = Current / A and B1 = Balance display / g. Use numeric cells and retain the supplied precision. The motion-data instructions explain entry, formula copying and numeric XY plotting.

Put g / (N/kg) in cell H1 and 9.81 in cell H2. Put Active length / m in I1 and 0.0500 in I2. These are cell addresses. Use C1 = Display change / kg and D1 = Wire force upward / N:

Enter =(B2-$B$2)/1000 in C2 and =$H$2*C2 in D2.

Fill C2:D2 through row 6 only. The fixed B2 reference is the current-off baseline; division by 1000 converts grams to kilograms. D contains signed upward wire force, not force on the magnet.

Make an XY plot with A2:A6 horizontally and D2:D6 vertically, labelled current / A and wire force upward / N. Fit a line with a free intercept and display its equation. Use these result cells, with descriptive headers above them:

Current-balance fit and field-strength calculation
Cell and quantityFormula
F2: gradient / (N/A)=SLOPE(D2:D6,A2:A6)
G2: intercept / N=INTERCEPT(D2:D6,A2:A6)
J2: B / T=F2/$I$2
Compare the generated-model result

The gradient is 0.004905 N/A and the intercept is zero apart from possible floating-point roundoff. Cell J2 gives B = 0.0981 T. The force column is 0, 0.001962, 0.003924, 0.005886 and 0.007848 N.

This agreement checks the entered model and units; it is not experimental verification. An actual record needs the instrument information, uncertainty and baseline-drift assessment described above.

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?

06

Forces between parallel currents

Each current-carrying wire produces a magnetic field at the other wire. Use that other-wire field to find each force; the source field and the responding conductor must be named separately.

Place two long parallel wires vertically in the page, wire 1 on the left and wire 2 on the right. Let both conventional currents initially point up. At wire 2, wire 1 produces a field into the page. An upward current in that field experiences a force left, towards wire 1.

At wire 1, wire 2's field is out of the page, so its upward current experiences a force right. Both forces point inward: parallel currents attract. These are paired forces on different wires, not two forces to add on one selected wire.

Parallel currents: the wires attract

Parallel currents: the wires attractTwo portions of long vertical wires have current one of four amperes upward at the left and current two of three amperes upward at the right. At wire two, the field from wire one remains into the page and has magnitude forty microtesla. At wire one, the field from wire two is out of the page and has magnitude thirty microtesla. The left wire force is right and the right wire force left, so they attract. The equal-length purple arrows show equal force magnitudes, sixty micro-newtons, on the two considered half-metre portions. Each force uses the other wire's field.Wire 1: 4.00 AWire 2: 3.00 AFrom wire 2From wire 130 µT out40 µT ind = 0.0200 m

Teal dot/cross symbols show the field made by the other wire at that location. Brown arrows show current. The two forces are equal and opposite despite the unequal currents and local fields; the considered length is 0.500 m.

Reverse one current: the wires repel

Reverse one current: the wires repelTwo portions of long vertical wires have current one of four amperes upward at the left and current two of three amperes downward at the right. At wire two, the field from wire one remains into the page and has magnitude forty microtesla. At wire one, the field from wire two is into the page and has magnitude thirty microtesla. The left wire force is left and the right wire force right, so they repel. The equal-length purple arrows show equal force magnitudes, sixty micro-newtons, on the two considered half-metre portions. Each force uses the other wire's field.Wire 1: 4.00 AWire 2: 3.00 AFrom wire 2From wire 130 µT in40 µT ind = 0.0200 m

Teal dot/cross symbols show the field made by the other wire at that location. Brown arrows show current. The two forces are equal and opposite despite the unequal currents and local fields; the considered length is 0.500 m.

Each local field label names the other wire as its source. Reversing only the right-hand current reverses that wire's response and the field it produces at the left-hand wire; both forces become outward.

After reversing wire 2's current, wire 1's field at wire 2 remains into the page, but the reversed current now feels force right. Wire 2's field at wire 1 changes to into the page, giving the unchanged upward current a force left. Thus antiparallel currents repel. This follows from the field and force rules, not from treating currents as north or south poles.

Different currents, equal force magnitudes

Use I1 = 4.00 A, I2 = 3.00 A, separation d = 0.0200 m and a considered common length l = 0.500 m, away from the wire ends. With μ0 = 4π × 10-7 H/m:

B1 at 2 = μ0I1/(2πd)
= 4.00 × 10-5 T
Fon 2 = B1 at 2I2l
= (4.00 × 10-5)(3.00)(0.500)
= 6.00 × 10-5 N

For parallel upward currents this force is left. Calculate the partner using the other source:

B2 at 1 = μ0I2/(2πd)
= 3.00 × 10-5 T
Fon 1 = B2 at 1I1l
= (3.00 × 10-5)(4.00)(0.500)
= 6.00 × 10-5 N

The local source fields are unequal, yet the forces are equal and opposite. Combining the two relationships gives a useful result for these long parallel wires:

F/l = μ0I1I2/(2πd)

The symmetric product I1I2 explains the equal magnitudes. Reversing one current changes attraction to repulsion while leaving the magnitudes unchanged if both current magnitudes and the geometry stay fixed.

Optional check Two long parallel wires carry unequal upward currents and attract. Only the right-hand current is reversed, with both current magnitudes unchanged. What happens to the interaction forces?
Two long parallel wires carry unequal upward currents and attract. Only the right-hand current is reversed, with both current magnitudes unchanged. What happens to the interaction forces?

07

Magnetic force on a charged particle

Magnetic force depends on a particle's velocity as well as its charge. Determine the force for positive-charge motion first, then reverse that force for a negative particle moving the same way.

For charge Q moving at speed v through magnetic flux density B, with angle θ between velocity and field, the force magnitude is:

F = B|Q|v sin θ

The form F = BQv sin θ uses Q as the charge magnitude when calculating a magnitude. If Q is signed, handle the direction separately rather than reporting a negative magnitude. At perpendicular entry, F = B|Q|v. A stationary charge, a neutral particle, or motion exactly parallel or antiparallel to B has zero magnetic force.

Keep velocity and charge sign separate

Use x right, y up and z out of the page. For a positive charge, its motion has the conventional-current direction, so Fleming's left-hand rule gives the force. For an electron with the same stated velocity, reverse that force once. Do not reverse the velocity as well and then reverse the force a second time.

Same velocity and field; opposite charge signs

Same velocity and field; opposite charge signsTwo distinct selected-particle cases have velocity right and the same magnetic field into the page. The positive charge on the left has upward force; the electron on the right has downward force. Blue arrows show their rightward velocities and purple arrows show force. The teal circle-and-cross above them is the common magnetic-field direction key, not a current. The arrow lengths are schematic, not a claim of equal charge magnitudes or accelerations.B into the page+q-eFFvvPositive chargeElectron

Apply the positive-charge direction first, then reverse it for an electron. The actual electron velocity remains rightward; do not reverse both velocity and charge.

Velocity up and field right

Velocity up and field rightThe two selected particles both move upward in a magnetic field directed right. For the positive charge, v cross B points into the page, shown by a purple circle-and-cross at the left particle location. For the electron, force is out of the page, shown by a purple circle-and-dot at the right particle location. The charge labels below the symbols identify the bodies; the purple symbols represent force, not magnetic field or current.B rightvv+q-eForce into pageForce out of page

Here purple cross means force into the page and purple dot means force out. A stationary charge, or one travelling along B, would instead have zero magnetic force.

Each comparison holds the stated velocity and field directions fixed while changing charge sign. The first view gives upward or downward forces; the second gives forces into or out of the page. Dot and cross labels identify which quantity points through the page.
  • Velocity right, B into the page: positive-charge force is up; electron force is down.
  • Velocity up, B right: positive-charge force is into the page; electron force is out of the page, towards the observer.

Reversing only velocity, only B or only charge sign reverses the force for a nonzero-force case. Reversing both velocity and B restores the original force for the same charge. The force is perpendicular to both velocity and B; it does not point along either.

Calculate a magnitude

An electron with |Q| = 1.60 × 10-19 C moves right at 2.00 × 107 m/s through a 1.00 mT field into the page. Convert 1.00 mT to 1.00 × 10-3 T:

F = B|Q|v
= (1.00 × 10-3)(1.60 × 10-19)(2.00 × 107)
= 3.20 × 10-15 N downward

At a different angle, retain sin θ. The angle is between the actual velocity vector and B, not between force and either of them.

Why the magnetic force does no work

The magnetic force is perpendicular to the particle's instantaneous velocity and therefore to its small displacement at that instant. It does no work on the particle. With magnetic force alone, speed and kinetic energy remain constant while velocity direction can change.

A uniform B does not mean a fixed force direction: as the velocity turns, the perpendicular force turns too. For perpendicular entry this leads to the circular path. Any velocity component parallel to B remains unaffected by the magnetic force, so not every entry direction gives a circle in the page.

Optional check An electron moves upward in the page through a uniform magnetic field directed right. What is the direction of its magnetic force?
An electron moves upward in the page through a uniform magnetic field directed right. What is the direction of its magnetic force?

08

Circular motion in a magnetic field

With perpendicular entry into a uniform magnetic field, a charged particle keeps its speed while its velocity turns. The magnetic force supplies the inward resultant required for circular motion.

For constant speed v on a circle of radius r, inward acceleration has magnitude v2/r. With magnetic force as the only significant force, its perpendicular-entry magnitude B|Q|v therefore gives:

B|Q|v = mv2/r
r = mv/(|Q|B)

The magnetic force is the actual inward force; do not add a second force called centripetal force. At each point, velocity is tangent to the circle and magnetic force points towards its centre. The force does no work, so this model has constant speed, not constant horizontal velocity.

Worked magnetic beam

Use the radius to find the exit state

An electron enters at (0, 0), moving right at 2.00 × 107 m/s. A uniform 1.00 mT field into the page occupies x > 0, with an ideal sharp boundary at x = 0. Outside that region B is zero. Take right and up as positive.

Use me = 9.11 × 10-31 kg and |Q| = 1.60 × 10-19 C. Neglect electric fields, gravity, collisions, fringing and radiation, and use the classical model. Roughly, m of order 10-30 kg, v of order 107 m/s, |Q| of order 10-19 C and B of order 10-3 T suggest r of order 0.1 m.

F = B|Q|v = 3.20 × 10-15 N
r = [(9.11 × 10-31)(2.00 × 107)]
/[(1.60 × 10-19)(1.00 × 10-3)]
= 0.113875 m ≈ 0.114 m

The electron follows a true semicircle

The electron follows a true semicircleA uniform one-millitesla magnetic field points into the page only to the right of the ideal sharp boundary at x eighty. Position uses one thousand drawing units per metre in both directions. The electron enters at eighty,eighty-five travelling right and follows a right-hand semicircle of radius 113.875 drawing units about eighty,198.875. At the rightmost point 193.875,198.875, velocity is down and force left. At the exit eighty,312.75, velocity is left; its force immediately inside the field is up. Outside the field it continues left along the straight tangent with unchanged speed. Equal blue velocity arrows and equal purple force arrows are separate quantity scales. The radius is 0.113875 metre and the entry-to-exit separation is the diameter, 0.227750 metre.B = 1.00 mT into the pagevvvFFFrr = 0.113875 mEntry to exit: 0.227750 mDashed line: ideal field boundary

The circular path uses equal position scales. Blue velocities are tangent; purple forces are inward. Force arrows at entry and exit refer to the field side of the boundary. Beyond it, the ideal magnetic force is zero and the path is straight.

Equal horizontal and vertical scales show the electron's true right-hand semicircle. Velocity is tangent and force inward at the marked positions. The field stops at the labelled ideal boundary, so the exit continues along a straight leftward tangent.

The electron's initial force is down, so the circle centre is (0, -r). At the rightmost point (r, -r), velocity is down and force is left. The electron next reaches the boundary at (0, -2r):

Entry-to-exit vertical separation = 2r
= 0.227750 m ≈ 0.228 m

It leaves below entry, travelling left, still at 2.00 × 107 m/s. Immediately before exit, magnetic force is upward towards the centre. Outside the ideal field, with no further force, it travels straight along that leftward tangent.

Compare the electric and magnetic paths

In the uniform electric-field example, force has a fixed upward direction, horizontal velocity stays constant and the growing vertical component increases total speed. The path is a parabola within that field.

In the magnetic-only example, force continually turns with velocity and stays perpendicular to it. The path is circular at constant speed, while both horizontal and vertical velocity components change. Use the field type and initial direction to choose the motion model; a charged particle does not always follow one standard curve.

Optional check An electron enters at (0,0) moving right into a uniform into-page magnetic field occupying x > 0. Its circular radius is 0.113875 m, and other forces are negligible. What happens when it next leaves this field?
An electron enters at (0,0) moving right into a uniform into-page magnetic field occupying x > 0. Its circular radius is 0.113875 m, and other forces are negligible. What happens when it next leaves this field?

09

Select a beam speed

Electric and magnetic forces can oppose one another for particles travelling in a chosen direction. Only the matching incident speed makes their magnitudes equal and leaves the charged beam undeviated.

Let the incident velocity be right, E downward and B into the page. These directions are mutually perpendicular. A positive charge feels electric force down and magnetic force up. An electron has both force directions reversed: electric force up and magnetic force down.

Establish that the forces oppose before equating their magnitudes. For a nonzero charge travelling in the stated direction:

|Q|E = B|Q|v
v = E/B

The charge magnitude cancels, and mass does not enter this ideal force-balance condition. Neutral particles do not satisfy this nonzero-charge selection argument. Reversing the incident direction without changing either field makes the two forces reinforce rather than cancel.

Calculate the selected speed

A field E of a few thousand N/C divided by B of a few 10-4 T suggests a selected speed of order 107 m/s. Use the supplied E = 4000 N/C and B = 0.200 mT = 2.00 × 10-4 T:

v = 4000/(2.00 × 10-4)
= 2.00 × 107 m/s

This B is different from the 1.00 mT field in the magnetic semicircle example. For an electron with |Q| = 1.60 × 10-19 C at the selected speed, both opposing forces have magnitude 6.40 × 10-16 N.

The selected speed passes through both apertures

The selected speed passes through both aperturesAn electron beam enters right through two aligned apertures. The electric field is four thousand newtons per coulomb down and the magnetic field is 0.200 millitesla into the page. At twenty million metres per second, electric force is upward and magnetic force downward, both 6.40 times ten to the minus sixteen newton. Their sixty-four-unit arrows use the same force scale as the next two panels and cancel. The path is straight. Field and velocity arrows have separate schematic scales.E-eElectricMagneticCrosses: B into the pageAligned entrance and exit aperturesv = 2.00 × 10⁷ m/s

The opposing forces are each 6.40 x 10^-16 N, so this electron travels straight at 2.00 x 10^7 m/s. In all three panels, force-arrow length uses 10 drawing units per 10^-16 N; field and velocity arrows use separate scales.

Slower electron: initial force upward

Slower electron: initial force upwardOnly the initial forces are shown for an electron entering right at fifteen million metres per second. E is down and B into the page, unchanged from the selected-speed case. Electric force is 6.40 times ten to the minus sixteen newton upward, drawn sixty-four units long. Magnetic force is 4.80 times ten to the minus sixteen newton downward, drawn 48 units long. The initial resultant is upward with magnitude 1.60 times ten to the minus sixteen newton. There is no drawn trajectory or claim that the subsequent force direction remains fixed.Same E down and B into pageEB in-evElectricMagneticv = 1.50 × 10⁷ m/s

Initial electric force: 6.40 x 10^-16 N up. Initial magnetic force: 4.80 x 10^-16 N down. These arrows predict the initial deflection only; the selector does not set every incident speed to E/B.

Faster electron: initial force downward

Faster electron: initial force downwardOnly the initial forces are shown for an electron entering right at twenty-five million metres per second. E is down and B into the page, unchanged from the selected-speed case. Electric force is 6.40 times ten to the minus sixteen newton upward, drawn sixty-four units long. Magnetic force is 8.00 times ten to the minus sixteen newton downward, drawn 80 units long. The initial resultant is downward with magnitude 1.60 times ten to the minus sixteen newton. There is no drawn trajectory or claim that the subsequent force direction remains fixed.Same E down and B into pageEB in-evElectricMagneticv = 2.50 × 10⁷ m/s

Initial electric force: 6.40 x 10^-16 N up. Initial magnetic force: 8.00 x 10^-16 N down. These arrows predict the initial deflection only; the selector does not set every incident speed to E/B.

The selected electron beam has equal opposing forces and follows the straight aperture line. The slower and faster panels compare initial forces on the same local force scale; they are not calculated full paths through the combined fields.

The selected beam remains straight when gravity, collisions, fringing and interactions between beam particles are negligible. A selector does not automatically accelerate every incident particle to E/B; it identifies the speed for an undeviated path in this geometry.

Decide the initial deflection of other speeds

For the same electron charge and fields, electric force stays at 6.40 × 10-16 N upward. Magnetic force depends on incident speed:

  • At 1.50 × 107 m/s, magnetic force is 4.80 × 10-16 N down. The initial resultant is 1.60 × 10-16 N up.
  • At 2.50 × 107 m/s, magnetic force is 8.00 × 10-16 N down. The initial resultant is 1.60 × 10-16 N down.

Positive charges entering with these same velocities have the opposite initial force directions. These statements describe the initial resultants. Once a particle deflects, its velocity changes and so does its magnetic force; a complete off-speed trajectory is not generally a constant-acceleration parabola.

Optional check Electrons enter right through E = 4000 N/C downward and B = 0.200 mT into the page. The selected speed is 2.00 x 10^7 m/s. What happens initially to an electron entering at 1.50 x 10^7 m/s?
Electrons enter right through E = 4000 N/C downward and B = 0.200 mT into the page. The selected speed is 2.00 x 10^7 m/s. What happens initially to an electron entering at 1.50 x 10^7 m/s?

Revision

Electromagnetic forces: revision summary

Name the source field, the selected body and the direction of conventional current or particle velocity. Calculate a force magnitude and a direction, then choose the motion or measurement model.

Uniform electric motion

Electric force magnitude = |Q|E
Signed component: Fy = QEy
ay = Fy/m

A positive charge is forced along E and a negative charge against E, including when initially stationary. With perpendicular entry and no other significant force, the field-perpendicular velocity component is constant while the field-parallel component changes. The path is parabolic and total speed can change. Outside the ideal field, continue along the exit tangent.

Field patterns and directions

  • Long straight wire: concentric field circles; right thumb follows current and curled fingers give B. Out-of-page current gives anticlockwise field in that view.
  • Flat circular coil: central axial field with external return paths. Anticlockwise current seen at a face gives central field towards that observer.
  • Long solenoid: approximately uniform interior away from ends, with weaker returning field outside. The grip-rule thumb gives the interior direction and points towards its north end.

Field lines are not particle tracks. A dot means the labelled quantity points out of the page and a cross means into it. Distinguish the views before comparing a face diagram with an axial section.

Source-field magnitudes

Long wire, perpendicular distance d:
B = μ0I/(2πd)

Thin flat coil, at its centre:
B = μ0NI/(2r)

Long solenoid, interior away from ends:
B = μ0nI   with n = N/L

Use the stated air/free-space assumptions and the supplied μ0. A suitable ferrous core becomes magnetised and can strengthen a solenoid's field, but there is no universal factor and μ0 alone does not calculate that core field.

Conductor force and the current balance

F = BIl sin θ
B = F/(Il) for a perpendicular active length

The angle is between conventional current and the external B. Fleming's left-hand rule uses first finger for field, second for conventional current and thumb for conductor force; use the current's perpendicular component for a non-right angle. Parallel or antiparallel current gives zero force. Reversing one of I or B reverses the force; reversing both restores it.

In the shown current balance, an upward wire force produces a downward reaction on the separately weighed magnet. Convert its baseline-subtracted display change from grams to kilograms and use F = gΔm for the stated upward wire-force reference. At fixed perpendicular length, the F-against-I gradient is Bl, so divide by l to obtain B.

Keep supports separate, return wires outside the active field and geometry fixed. Check current-off drift, ranges, resolution, heating, fringing and contact forces. A changed balance display is not a changed physical mass; actual nonzero intercepts and anomalies need investigation.

For two long parallel wires, use each wire's field at the other. Parallel currents attract and antiparallel currents repel. Their fields can differ when the currents differ, but the interaction force magnitudes are equal: F/l = μ0I1I2/(2πd).

Magnetic motion and velocity selection

Magnetic force magnitude = B|Q|v sin θ
Perpendicular uniform-field circle:
B|Q|v = mv2/r   so   r = mv/(|Q|B)

For positive-charge motion, use the conventional-current force direction; reverse it once for a negative charge with the same velocity. A stationary charge or velocity parallel to B gives zero magnetic force. Magnetic force does no work: it changes velocity direction while speed remains constant. It is the inward force, not an extra force in addition to a centripetal force.

For the electron semicircle in x > 0, the exit lies 2r below entry and its velocity is left. Radius and entry-to-exit separation are different. Outside the ideal field, the path is a straight tangent.

Opposing crossed-field forces:
|Q|E = B|Q|v   so   v = E/B

This requires the intended incident direction, mutually perpendicular velocity and fields, and nonzero charge. The selected speed is independent of mass and charge magnitude. Off-speed force comparisons give the initial deflection, not automatically a full path or a process that brings every particle to the selected speed.

Quantity and unit reference

Field, particle and measurement quantities used in this chapter
QuantitySymbolSI unit
Electric chargeQC = A s
Elementary-charge magnitude; electron charge is -eeC
Particle mass; electron massm; mekg
Electric field strengthEN/C = V/m
Magnetic flux densityBT = N/(A m)
Magnetic fluxΦWb = T m2 = V s
ForceFN
CurrentIA
Active conductor lengthlm
Speedvm/s
Path radiusrm
Balance display change, converted to kgΔmkg
Permeability of free spaceμ0H/m = N/A2
Total turns; turns per unit lengthN; nNo unit; m-1

The supplied electron constants are e = 1.60 × 10-19 C and me = 9.11 × 10-31 kg. Convert 1 mT = 10-3 T, 1 µT = 10-6 T, 1 µWb = 10-6 Wb, 1 ns = 10-9 s and 1 g = 10-3 kg before using SI equations.

For the supplied uniform perpendicular-area example, Φ = BA: local B and the flux through an area are distinct quantities. The units are 1 T = 1 kg s-2 A-1 and 1 Wb = 1 kg m2 s-2 A-1.

For μ0, H = Wb/A gives H/m = N/A2 = kg m s-2 A-2. Solenoid n is turns per metre, distinct from the carrier number density in the current model.