Magnetic Flux Density

SPEC 3.7.5.1 & 3.7.5.2 • MAGNETIC FIELDS

Forces on current-carrying conductors and moving charges in magnetic fields

What is a Magnetic Field?

Definition

A magnetic field is a region of space where a magnetic material or a moving electric charge experiences a force.

Magnetic fields are created by:

Permanent magnets (aligned magnetic domains)

Electric currents (moving charges)

Key Concept

Unlike electric fields which act on all charges, magnetic fields only exert forces on moving charges. A stationary charge in a magnetic field experiences no magnetic force.

Representing Magnetic Fields

We represent magnetic fields using field lines:

Direction

Field lines point from North to South outside the magnet

Density

Closer lines = stronger field; further apart = weaker field

Continuous Loops

Field lines never start or end — they form closed loops (no magnetic monopoles exist in nature)

Never Cross

At any point, the field has only one direction. If lines crossed, there would be two directions — impossible.

3D Field Notation

When drawing magnetic fields in 2D, we use special symbols for fields pointing into or out of the page:

Memory Aid

Think of an arrow: a dot shows the point coming towards you; a cross shows the feathers going away.

Magnetic Field Around a Wire

When current passes through a wire, a magnetic field is induced around it. The field lines form concentric circles around the wire:

Key Features

Field lines are concentric circles centred on the wire

Field is stronger closer to the wire (lines more closely spaced)

Direction found using right-hand grip rule: thumb points in current direction, fingers curl in field direction

Magnetic Flux Density

Definition

Magnetic flux density (B) is the force per unit length on a wire carrying unit current perpendicular to the magnetic field.

B = \frac{F}{I l}

The SI unit of magnetic flux density is the tesla (T):

The Tesla

1 T = 1 N A⁻¹ m⁻¹

A field has flux density 1 T if it exerts a force of 1 N on a 1 m wire carrying 1 A perpendicular to the field.

SourceTypical Flux Density

Earth's magnetic field50 μT (5 × 10⁻⁵ T)

Bar magnet (at surface)0.01 – 0.1 T

MRI scanner1.5 – 3 T

Strong laboratory magnet10 – 20 T

Physical Meaning

B tells us "how strong" the field is — specifically, how much force a current or moving charge will experience. It's proportional to the density of field lines (lines per unit area). Higher B → more force → field lines packed more closely.

Force on a Current-Carrying Wire

When a wire carrying current is placed in a magnetic field, it experiences a force:

Formula

F = B I L

SymbolQuantityUnit

FForce on wireN

BMagnetic flux densityT

ICurrentA

LLength of wire in fieldm

Worked Example01

A wire of length 0.15 m carries a current of 3.5 A. It is placed perpendicular to a uniform magnetic field of flux density 0.80 T. Calculate the force on the wire.

1

Write down the known values: B = 0.80 T, I = 3.5 A, l = 0.15 m

2

Since the wire is perpendicular to the field, use F = BIL

3

Substitute: F = 0.80 × 3.5 × 0.15

4

Calculate: F = 0.42 N

Answer: F = 0.42 N

Force at an Angle

When the wire makes an angle θ with the magnetic field:

General Formula

F = B I L sin θ

where θ is the angle between the current direction and the magnetic field.

Exam Tip

Only the component of current perpendicular to the field produces a force. When θ = 0° (parallel), there's no perpendicular component, so no force. When θ = 90°, sin θ = 1 and F = BIL (maximum).

Why sin θ?

The magnetic force arises from charges cutting across field lines. The parallel component (along B) moves with the field lines — no cutting, no force. Only the perpendicular component (across B) cuts field lines and produces force.

Worked Example02

A 0.25 m wire carrying 4.0 A is placed at 30° to a uniform magnetic field. The force on the wire is 0.15 N. Calculate the magnetic flux density.

1

Write down known values: l = 0.25 m, I = 4.0 A, θ = 30°, F = 0.15 N

2

Use the formula F = BIL sin θ and rearrange: B = F / (IL sin θ)

3

Substitute: B = 0.15 / (4.0 × 0.25 × sin 30°)

4

Calculate: B = 0.15 / (4.0 × 0.25 × 0.5) = 0.15 / 0.5 = 0.30 T

Answer: B = 0.30 T

Fleming's Left-Hand Rule

This rule determines the direction of force on a current-carrying conductor in a magnetic field. Hold your left hand with thumb, first finger and second finger all mutually perpendicular (at right angles to each other):

How to Apply

1. Point your First finger in the direction of the magnetic Field (N → S)

2. Point your seCond finger in the direction of conventional Current (+ → −)

3. Your thuMb then shows the direction of Motion/Force on the conductorMemory AidFirst = Field

seCond = Current

thuMb = Motion

Important

This rule uses conventional current (+ to −). For electrons (which flow − to +), the force is in the opposite direction to what the thumb indicates.

Force on Moving Charges

We can derive the force on a single moving charge from F = BIL:

Derivation

Current is charge per unit time:

I = \frac{Q}{t}

Length travelled in time t:

L = v t

Substituting into F = BIL:

F = B \times \frac{Q}{t} \times v t = BQ v

Formula

F = BQ v

Force on a charge Q moving at velocity v perpendicular to field B. General case:

F = BQ v sin θ

Direction of Force on Charges

Using Fleming's Left Hand Rule:

Important

For electrons, the force is in the opposite direction to what you'd get using Fleming's LHR with the electron's velocity as the current direction.

03Worked Example

An electron moves at 2.5 × 10⁷ m s⁻¹ perpendicular to a uniform magnetic field of flux density 0.40 T. Calculate the force on the electron.

Electron charge e = 1.60 × 10⁻¹⁹ C

1

Write down known values: v = 2.5 × 10⁷ m s⁻¹, B = 0.40 T, Q = 1.60 × 10⁻¹⁹ C

2

Use F = BQv (perpendicular, so sin θ = 1)

3

Substitute: F = 0.40 × 1.60 × 10⁻¹⁹ × 2.5 × 10⁷

4

Calculate: F = 1.6 × 10⁻¹² N

Answer: F = 1.6 × 10⁻¹² N

Magnetic Force Does No Work

Critical Concept

The magnetic force on a moving charge is always perpendicular to its velocity.

This has important consequences:

Work done = Force × distance moved in direction of force

Since F ⊥ v, the particle moves perpendicular to the force

Therefore, the magnetic force does zero work on the particle

This means kinetic energy is unchanged

Speed remains constant; only direction changes

Important

A magnetic field can change a particle's direction but cannot change its speed. The magnetic force cannot speed up or slow down a charged particle.

Circular Motion in Magnetic Fields

When a charged particle enters a uniform magnetic field perpendicular to its velocity:

v = velocity (tangent to circle)

F = force (towards centre)

Why Circular?

The magnetic force is always perpendicular to velocity, so it acts as a centripetal force, continuously changing direction without changing speed. The particle moves in a circle at constant speed.

Radius of Circular Path

For circular motion, the magnetic force provides the centripetal force:

Derivation

Magnetic force = Centripetal force

BQ v = \frac{m v ^{2}}{r}

Rearranging for r:

r = \frac{m v}{BQ}

Key Relationships

r ∝ m — heavier particles have larger radius

r ∝ v — faster particles have larger radius

r ∝ 1/B — stronger field means smaller radius

r ∝ 1/Q — greater charge means smaller radius

Note

The quantity mv is the momentum p, so we can also write:

r = \frac{p}{BQ}

Physical Insight

A faster particle has more momentum — it's harder to "bend" its path. So the same magnetic force produces a gentler curve (larger r). Similarly, a stronger field or greater charge means more force, bending the path more sharply (smaller r).

04Worked Example

A proton travelling at 3.0 × 10⁶ m s⁻¹ enters a uniform magnetic field of 0.50 T perpendicular to its velocity. Calculate the radius of its circular path.

Proton mass = 1.67 × 10⁻²⁷ kg, proton charge = 1.60 × 10⁻¹⁹ C

1

Write known values: m = 1.67 × 10⁻²⁷ kg, v = 3.0 × 10⁶ m s⁻¹, B = 0.50 T, Q = 1.60 × 10⁻¹⁹ C

2

Use the formula: r = mv / BQ

3

Substitute: r = (1.67 × 10⁻²⁷ × 3.0 × 10⁶) / (0.50 × 1.60 × 10⁻¹⁹)

4

Calculate: r = (5.01 × 10⁻²¹) / (8.0 × 10⁻²⁰) = 0.063 m = 6.3 cm

Answer: r = 0.063 m (6.3 cm)

Period and Frequency

We can find the time period for one complete orbit:

Derivation

Circumference = 2πr, and T = distance / speed

T = \frac{2 π r}{v} = \frac{2 π}{v} \times \frac{m v}{BQ}

T = \frac{2 πm}{BQ}

Crucial Result

The period T is independent of velocity! All particles of the same type (same m and Q) in the same field have the same orbital period, regardless of their speed.

The frequency is:

f = \frac{1}{T} = \frac{BQ}{2 πm}

Why Does v Cancel?

A faster particle travels a larger circle (r ∝ v), but it also moves faster around that circle. These two effects exactly balance — the longer path is covered at higher speed, so the time is the same! This is why cyclotrons work: the AC supply uses a fixed frequency.

Worked Example05

Show that the orbital frequency of a proton in a magnetic field is independent of its speed. Hence calculate the frequency for a proton in a 1.2 T field.

1

From r = mv/BQ, the radius increases with velocity

2

Period T = 2πr/v = 2π(mv/BQ)/v = 2πm/BQ

3

The v cancels out, so T depends only on m, B, and Q — not on v

4

Frequency f = BQ/2πm = (1.2 × 1.60 × 10⁻¹⁹)/(2π × 1.67 × 10⁻²⁷)

5

f = (1.92 × 10⁻¹⁹)/(1.05 × 10⁻²⁶) = 1.8 × 10⁷ Hz = 18 MHz

Answer: f = 1.8 × 10⁷ Hz (18 MHz)

The Cyclotron

How It Works

1. Ion source releases particles at centre

2. B field (⊗ into page) exerts F = BQv → particle curves in a circular path

3. E field in gap accelerates particle → gains KE and speed

4. Higher speed → larger radius (r = mv/BQ) → next circle is bigger

5. AC reverses each half-cycle → E field always accelerates

6. Particle spirals outward until it reaches the edge and exits

Why It Works

Period T = 2πm/BQ is independent of speed — particle always arrives at gap in sync with AC!

Applications

Medical isotopes (PET scans), cancer radiotherapy, particle physics research.

Cyclotron Frequency

The cyclotron works because of the constant orbital period:

Cyclotron Frequency

f_{c} = \frac{BQ}{2 πm}

This is the frequency of the AC supply needed to accelerate the particles.

As particles gain energy:

Their speed v increases

Their orbital radius r = mv/BQ increases

But their orbital period T = 2πm/BQ stays constant!

So they always arrive at the gap at the right time

Applications

Cyclotrons produce radioactive isotopes for medical imaging (PET scans), generate beams for cancer therapy, and conduct particle physics experiments.

Worked Example06

A cyclotron has a magnetic field of 0.80 T and accelerates protons. Calculate:

(a) the cyclotron frequency

(b) the radius when speed is 1.5 × 10⁷ m s⁻¹

(a)

Cyclotron frequency: f = BQ/(2πm)

•

f = (0.80 × 1.60 × 10⁻¹⁹)/(2π × 1.67 × 10⁻²⁷)

•

f = 1.28 × 10⁻¹⁹ / (1.05 × 10⁻²⁶) = 1.2 × 10⁷ Hz = 12 MHz

(b)

Radius: r = mv/BQ

•

r = (1.67 × 10⁻²⁷ × 1.5 × 10⁷)/(0.80 × 1.60 × 10⁻¹⁹)

•

r = (2.5 × 10⁻²⁰)/(1.28 × 10⁻¹⁹) = 0.20 m = 20 cm

Answers: (a) f = 1.2 × 10⁷ Hz; (b) r = 0.20 m

Practice

Question 1: A straight wire of length 0.20 m carries a current of 5.0 A. It is placed in a uniform magnetic field of flux density 0.60 T at an angle of 45° to the field direction. Calculate the force on the wire.

1

Known values: l = 0.20 m, I = 5.0 A, B = 0.60 T, θ = 45°

2

Use F = BIL sin θ

3

F = 0.60 × 5.0 × 0.20 × sin 45° = 0.60 × 5.0 × 0.20 × 0.707

4

F = 0.42 N

Answer: F = 0.42 N

Practice

Question 2: An alpha particle (charge +2e, mass 6.64 × 10⁻²⁷ kg) moves at 2.0 × 10⁶ m s⁻¹ perpendicular to a 0.25 T magnetic field. Calculate the radius of its circular path.

1

Known: Q = 2 × 1.60 × 10⁻¹⁹ = 3.20 × 10⁻¹⁹ C, m = 6.64 × 10⁻²⁷ kg, v = 2.0 × 10⁶ m s⁻¹, B = 0.25 T

2

Use r = mv/BQ

3

r = (6.64 × 10⁻²⁷ × 2.0 × 10⁶)/(0.25 × 3.20 × 10⁻¹⁹)

4

r = (1.33 × 10⁻²⁰)/(8.0 × 10⁻²⁰) = 0.17 m

Answer: r = 0.17 m (17 cm)

Practice

Question 3: Explain why a charged particle moving parallel to a magnetic field experiences no force, but one moving perpendicular to the field follows a circular path.

1

The magnetic force is given by F = BQv sin θ, where θ is the angle between velocity and field.

2

When moving parallel to field: θ = 0°, so sin θ = 0, giving F = 0 (no force).

3

When moving perpendicular: θ = 90°, so sin θ = 1, giving maximum force F = BQv.

4

This force is always perpendicular to velocity, so it acts as a centripetal force.

5

A constant centripetal force perpendicular to velocity causes circular motion at constant speed.

Summary

Key Formulas

FormulaDescription

F = BILForce on wire (perpendicular to field)

F = BIL sin θForce on wire at angle θ to field

F = BQvForce on moving charge (perpendicular)

r = mv/BQRadius of circular path

T = 2πm/BQPeriod (independent of speed)

v = E/BVelocity selector condition

Key Exam Skills

Use Fleming's Left-Hand Rule correctly for force direction

Remember: magnetic force does NO work (can't change speed)

Know that orbital period is independent of velocity

Understand velocity selector: v = E/B for undeflected path

Apply r = mv/BQ for circular motion problems