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Electromagnetism

Magnetic flux density, the force on a current and on a moving charge

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01

Magnetic flux density

Definition

Magnetic flux density B — The force per unit current per unit length on a conductor placed at right angles to the field. Measured in tesla, where 1 T = 1 N A⁻¹ m⁻¹.

At earlier levels a magnetic field is drawn with lines and left there. To calculate anything you need a number for how strong the field is, and that number is the magnetic flux density, B.

It is defined through the force it produces. Put a wire of length L carrying current I at right angles to a field, and the force on it is F = BIL. Rearranged, B = F/(IL) — which is what the unit N A⁻¹ m⁻¹ is saying.

The tesla is a large unit. The Earth's field is about 50 microtesla, a fridge magnet around 5 millitesla, and a hospital MRI scanner between 1.5 and 3 T. If a calculation gives you a field of hundreds of tesla, something has gone wrong.

When the wire is not perpendicular to the field, only the perpendicular component counts, so the force becomes F = BIL sin θ. A wire lying along the field feels no force at all — a fact worth checking your working against.

F = B I L sin θB = F / (I L)θ is the angle between the wire and the field; the force is zero when they are parallel
F
forceN
B
flux densityT
I
currentA
L
length in the fieldm

Flux density is really a statement about how crowded these lines are. Bring the poles closer and the lines bunch — the field between them strengthens, and so would the force on any current placed there.

02

The direction of the force

The force is perpendicular to both the current and the field, which is why it needs a three-dimensional rule rather than a diagram in the plane of the page.

Fleming's left-hand rule gives it. Hold the thumb and first two fingers of the left hand mutually at right angles: First finger along the Field (north to south), seCond finger along the Current (conventional), and the thuMb then points along the Motion.

Reversing either the current or the field reverses the force. Reversing both leaves it unchanged — a common exam question, and one you can answer by applying the rule twice rather than memorising the result.

Left hand for the force on a current; right hand for the current induced by motion. Using the wrong hand reverses every answer in the question, so it is worth saying "left for motor" under your breath each time.

Left hand, right hand

Left hand — the motor effect: a current in a field produces motion. Right hand — the generator effect: motion in a field produces a current. Fleming named them so the pair could be told apart; using one for both is the single most common error in this half of the syllabus.

03

The force on a moving charge

A current is charge in motion, so a single moving charge in a magnetic field feels a force too. Following the definition through gives F = qvB sin θ.

This force has an unusual property: it is always perpendicular to the velocity. A force at right angles to the motion changes direction but never speed — so a magnetic field can bend a charged particle's path but can never do work on it or speed it up.

A charge entering a uniform field at right angles therefore travels in a circle, with the magnetic force acting as the centripetal force. Setting qvB = mv²/r and rearranging gives the radius, and that relation is what mass spectrometers and particle accelerators are built on.

It also explains the aurora. Charged particles from the Sun spiral along the Earth's field lines and are funnelled towards the poles, which is why the lights appear there and not over the equator.

F = q v B sin θq v B = m v² / rr = m v / (q B)the radius grows with momentum and shrinks with field strength
q
chargeC
v
speedm s⁻¹
r
radius of the circular pathm
m
masskg
Worked example 16 marks

An electron of mass 9.11 × 10⁻³¹ kg and charge 1.60 × 10⁻¹⁹ C enters a uniform field of 0.012 T at right angles, moving at 3.0 × 10⁶ m s⁻¹. Find the force on it and the radius of its path.

  1. The velocity is perpendicular to the field, so sin θ = 1.Always check the angle before dropping the sine.
  2. F = qvB = 1.60 × 10⁻¹⁹ × 3.0 × 10⁶ × 0.012.
  3. F = 5.8 × 10⁻¹⁵ N.Tiny in newtons, but the electron is tiny too.
  4. This force is centripetal, so qvB = mv²/r.The magnetic force does no work — it only turns the particle.
  5. r = mv/(qB) = (9.11 × 10⁻³¹ × 3.0 × 10⁶) / (1.60 × 10⁻¹⁹ × 0.012).
  6. r = 1.4 × 10⁻³ m, about 1.4 mm.A tight circle — which is why bubble-chamber tracks curl so sharply.

F = 5.8 × 10⁻¹⁵ N, r = 1.4 mm

04

Fields made by currents

Every current makes a magnetic field of its own, and three arrangements matter.

Around a long straight wire the field lines are concentric circles, weakening with distance. The right-hand grip rule gives the direction: thumb along the conventional current, curled fingers along the field.

A flat coil concentrates that field through its centre, and more turns give a stronger field.

A solenoid — a long coil — produces a field very like a bar magnet's: nearly uniform inside, spreading out at the ends. Which end is north again comes from a grip rule: curl the right fingers the way the current goes round, and the thumb points to north.

Two parallel wires therefore exert forces on each other, because each sits in the other's field. Currents in the same direction attract; opposite currents repel — the reverse of what most people guess.

Key points

  1. F = BIL sin θ — zero when the wire lies along the field.
  2. Left hand for the motor effect, right hand for induction.
  3. F = qvB is always perpendicular to v, so it turns a charge without speeding it up.
  4. r = mv/(qB) — faster or heavier means a wider circle.
  5. Parallel currents in the same direction attract.

Practice questions

6 questions · 24 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Define magnetic flux density and state its unit.
Model answer

The force per unit current per unit length on a conductor placed at right angles to the field, B = F/(IL). Its unit is the tesla (T), equal to 1 N A⁻¹ m⁻¹.

Examiner tip. The condition "at right angles" belongs in the definition — leave it out and the statement is wrong.

SQ2[2 marks]
Explain why a magnetic field cannot change the speed of a charged particle.
Model answer

The force qvB always acts at right angles to the velocity. A perpendicular force does no work, so the kinetic energy and therefore the speed are unchanged — only the direction alters.

Examiner tip. "Does no work" is the phrase the mark scheme wants.

SQ3[2 marks]
A wire carrying a current lies parallel to a magnetic field. State and explain the force on it.
Model answer

The force is zero, because F = BIL sin θ and sin 0° = 0. Only the component of the wire perpendicular to the field experiences a force.

Examiner tip. Quote the sine term. A bare "zero" earns one mark of two.

Solved numericals

1 · 5 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[5 marks]
A wire of length 0.25 m carries a current of 4.0 A at right angles to a field of 0.15 T. Find the force. Then find the force if the wire is turned to 30° to the field.
Full working
  1. Uses F = BIL with sin 90° = 1[1]
  2. F = 0.15 × 4.0 × 0.25[1]
  3. F = 0.15 N[1]
  4. At 30°: F = BIL sin 30° = 0.15 × 0.5[1]
  5. F = 0.075 Nhalf, because sin 30° = 0.5[1]

0.15 N, then 0.075 N

Examiner tip. The angle is measured between the wire and the field, not between the wire and the normal.

Long questions

1 · 8 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[8 marks]
A proton of mass 1.67 × 10⁻²⁷ kg and charge 1.60 × 10⁻¹⁹ C moves at 2.0 × 10⁶ m s⁻¹ perpendicular to a uniform field of 0.35 T.
  1. Calculate the force on the proton. [2]
  2. Explain why it moves in a circle. [3]
  3. Calculate the radius of that circle. [3]
Mark scheme
  1. Uses F = qvB[1]
  2. F = 1.60 × 10⁻¹⁹ × 2.0 × 10⁶ × 0.35 = 1.12 × 10⁻¹³ N[1]
  3. The force is always perpendicular to the velocity[1]
  4. So it changes the direction of motion but not the speed[1]
  5. A constant force at right angles to a constant speed is centripetal, giving circular motion[1]
  6. Sets qvB = mv²/r[1]
  7. r = mv/(qB) = (1.67 × 10⁻²⁷ × 2.0 × 10⁶) / (1.60 × 10⁻¹⁹ × 0.35)[1]
  8. r = 0.060 m, about 6 cm[1]

(a) 1.12 × 10⁻¹³ N (c) 0.060 m

Examiner tip. Part (b) is three marks of explanation for one line of physics — perpendicular force, no speed change, therefore circular. Write all three steps.

Exam questions

1 · 5 marks

Multi-part questions with a full mark scheme.

Q1[5 marks]
Two long parallel wires 5.0 cm apart carry currents in the same direction.
  1. State whether they attract or repel. [1]
  2. Explain your answer using the field of one wire and the force on the other. [3]
  3. State what happens if one current is reversed. [1]
Mark scheme
  1. They attractthe opposite of what most people expect[1]
  2. Each wire sits in the magnetic field produced by the other[1]
  3. The field of the first wire at the second is perpendicular to that wireright-hand grip rule[1]
  4. Fleming's left-hand rule then gives a force on the second wire directed towards the first[1]
  5. Reversing one current makes them repel[1]

attract; reversing one current makes them repel

Examiner tip. Work it through with the two rules rather than trying to remember the result — the grip rule then the left-hand rule.