IB DP Physics · Theme D · Fields

Motion in electromagnetic fields.

D.3 · Charged particles in electric and magnetic fields, and the forces on currents
Standard level and higher level

SL + HL · 6 hours

The journey

Three parts.

Part 01

Charges in electric fields

Constant force: speeding up, and parabolic paths

Part 02

Charges in magnetic fields

Circular paths, charge-to-mass ratio, crossed fields

Part 03

Forces on currents

A wire in a field, and two parallel wires

D.3 has no additional higher level content: every slide is for SL and HL.

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Part 01 · SL + HL

Charges in electric fields.

A uniform field gives a constant force, so a constant acceleration

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Along the field

A charge in a uniform electric field.

AN ELECTRON BETWEEN TWO PLATESPRESS → TO RELEASE

Constant force \(F = qE\), so constant acceleration
\(a = \dfrac{qE}{m}\)
Work done by the field becomes kinetic energy

\(qV = \dfrac{1}{2} m v^{2}\)

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Two ways to find the final speed

The force is constant, so the equations of motion from A.1 work: \(v^{2} - u^{2} = 2as\) with \(a = \frac{qE}{m}\). But energy is quicker. The field does work \(qV\) on the charge whatever the distance between the plates, and all of it becomes kinetic energy. The speed depends only on the potential difference crossed, not on how far apart the plates are.

Why gravity is ignored

For an electron in a typical laboratory field, the electric force is more than \(10^{13}\) times its weight. The electron falls by far less than the width of an atom while it crosses the apparatus.

An electron accelerated from rest through \(V\) volts has a kinetic energy of \(V\) electronvolts. A “2 keV electron” has been through 2000 V.

Sending a negative charge along the field. An electron accelerates towards the positive plate, against the field lines.

Worked example

An electron accelerated through 2000 V.

From rest · \(V = 2000 \,\mathrm{V}\) \(e = 1.60 \times 10^{-19} \,\mathrm{C}\) · \(m = 9.11 \times 10^{-31} \,\mathrm{kg}\) \(eV = \dfrac{1}{2} m v^{2}\) so \(v = \sqrt{\dfrac{2eV}{m}}\) \(v = \sqrt{\dfrac{2 \times 1.60 \times 10^{-19} \times 2000}{9.11 \times 10^{-31}}}\) \(v = \class{res}{2.65 \times 10^{7} \,\mathrm{m} \,\mathrm{s}^{-1}}\)

Kinetic energy gained: \(2000 \,\mathrm{eV} = 3.2 \times 10^{-16} \,\mathrm{J}\)
About 9% of the speed of light
Four times the voltage gives twice the speed

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Why speed goes as the square root of voltage

The energy gained is proportional to \(V\), and kinetic energy is proportional to \(v^{2}\). So \(v \propto \sqrt{V}\): to double the speed you need four times the accelerating voltage.

A different particle

A proton through the same 2000 V gains the same energy, 2000 eV, because its charge has the same size. But its mass is 1836 times the electron’s mass, so its speed is about \(\frac{1}{43}\) of the electron’s speed (\(\sqrt{1836} = 43\)): \(6.2 \times 10^{5} \,\mathrm{m} \,\mathrm{s}^{-1}\).

Above about 10% of the speed of light this equation starts to overestimate the speed, and relativity (A.5, HL) is needed. Exam questions stay below that.

Across the field

Entering an electric field at right angles.

AN ELECTRON BETWEEN CHARGED PLATESPRESS → TO FIRE

Along the plates: no force, constant velocity
Across the plates: constant force, constant acceleration
The path is a parabola, like a projectile
Outside the field: a straight line

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Projectile motion with a different force

In A.1 a ball thrown horizontally keeps its horizontal velocity while gravity accelerates it downwards, and it follows a parabola. Here the constant force is \(qE\) and the acceleration is \(\frac{qE}{m}\) instead of \(g\). Everything else is the same: treat the two directions separately and link them through the time.

The method

Find the time inside the field from the motion along the plates, \(t = \frac{L}{v}\). Use that time for the motion across: deflection \(\frac{1}{2} a t^{2}\) and sideways velocity \(at\). After the plates, both components of velocity are constant, so the path is straight, along the tangent to the parabola at the exit.

The kinetic energy of the particle increases in an electric field: the field does work on it. Compare this with a magnetic field, in Part 2.

Worked example

How far is the electron deflected?.

\(v = 2.0 \times 10^{7} \,\mathrm{m} \,\mathrm{s}^{-1}\) along plates of length \(L = 5.0 \,\mathrm{cm}\) Field \(E = 2.0 \times 10^{4} \,\mathrm{V} \,\mathrm{m}^{-1}\) \(t = \dfrac{L}{v} = \dfrac{0.050}{2.0 \times 10^{7}} = 2.5 \times 10^{-9} \,\mathrm{s}\) \(a = \dfrac{eE}{m} = \dfrac{1.60 \times 10^{-19} \times 2.0 \times 10^{4}}{9.11 \times 10^{-31}}\) \(a = 3.5 \times 10^{15} \,\mathrm{m} \,\mathrm{s}^{-2}\)

Deflection across the plates \(y = \dfrac{1}{2} a t^{2}\) \(y = \dfrac{1}{2} \times 3.5 \times 10^{15} \times (2.5 \times 10^{-9})^{2}\) \(y = \class{res}{1.1 \times 10^{-2} \,\mathrm{m}}\)

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

What sets the deflection

Putting the steps together, \(y = \frac{eEL^{2}}{2mv^{2}}\). A faster electron spends less time in the field and is deflected less, by the square of its speed. A stronger field or longer plates deflect it more.

The velocity on leaving

The sideways velocity is \(at = 3.5 \times 10^{15} \times 2.5 \times 10^{-9} = 8.8 \times 10^{6} \,\mathrm{m} \,\mathrm{s}^{-1}\). The electron leaves at an angle \(\theta\) to its original direction with \(\tan \theta = \frac{8.8 \times 10^{6}}{2.0 \times 10^{7}}\), so \(\theta = 24^{\circ}\). Its speed is now \(\sqrt{(2.0 \times 10^{7})^{2} + (8.8 \times 10^{6})^{2}} = 2.2 \times 10^{7} \,\mathrm{m} \,\mathrm{s}^{-1}\).

Check that the particle does not hit a plate: here the plates would need to be more than 2.2 cm apart for an electron entering midway between them.

Quick check

Charges in electric fields.

01What kind of acceleration does a charge have in a uniform electric field?

Constant

02A proton is accelerated from rest through 500 V. Kinetic energy gained?

500 eV = \(8.0 \times 10^{-17} \,\mathrm{J}\)

03What shape is the path when a charge enters at right angles to the field?

A parabola

04Which component of its velocity stays constant?

The one at right angles to the field

05The accelerating voltage is increased to nine times its initial value. The final speed?

Three times the initial speed

06Which way does an electron accelerate, relative to the field lines?

Against them

PRESS → TO REVEAL EACH ANSWER

Part 02 · SL + HL

Charges in magnetic fields.

A force at right angles to the motion: it changes the direction, never the speed

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

The magnetic force

The force on a moving charge.

\(F = qvB \sin \theta\)

\(B\): magnetic field strength, in tesla (T)
\(\theta\): the angle between the velocity and the field
Greatest at 90°. Zero when moving along the field, and zero at rest

A POSITIVE CHARGE, AT RIGHT ANGLES TO B

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

How this differs from the electric force

An electric field pushes any charge, moving or not, along the field lines. A magnetic field pushes only a moving charge, and the push is sideways: at right angles to the field and at right angles to the velocity. A charge at rest in a magnetic field feels nothing at all.

The tesla

Rearranging, \(B = \frac{F}{qv}\) for motion at right angles to the field. A field of 1 T gives a force of 1 N on a charge of 1 C moving across it at 1 m s⁻¹. One tesla is a strong field: the Earth’s field is about \(5 \times 10^{-5} \,\mathrm{T}\), a small bar magnet gives about 0.01 T close to a pole, and a hospital MRI scanner uses 1.5 T or more.

A numerical case: an electron crossing a 2.0 mT field at \(2.0 \times 10^{7} \,\mathrm{m} \,\mathrm{s}^{-1}\) feels \(1.60 \times 10^{-19} \times 2.0 \times 10^{7} \times 2.0 \times 10^{-3} = 6.4 \times 10^{-15} \,\mathrm{N}\).

Which way?

The direction of the magnetic force.

At right angles to both the field and the velocity
Left-hand rule, “FBI”: thumb \(F\), index finger \(B\), middle finger \(I\)
\(I\), the current, is the way a positive charge moves

For an electron, point the middle finger opposite to its velocity.

THE LEFT HANDF · B · I

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Using the rule

Hold the thumb, index finger and middle finger of your left hand at right angles to one another. Point the index finger along the field \(B\) and the middle finger along the conventional current \(I\). The thumb then shows the force \(F\). Thumb, index, middle: F, B, I. Check it on the last slide: field into the page, positive charge moving right, force up the page.

Negative charges

An electron moving to the right is a conventional current to the left. Either point the middle finger opposite to the electron’s velocity, or use the rule as for a positive charge and reverse the answer.

Any correct rule earns the marks. Some books use a right-hand rule with the palm: fingers along the field, thumb along the current, and the palm pushes in the direction of the force. Choose one and always use it.

Mixing this up with the right-hand grip rule from D.2. The grip rule gives the direction of the field made by a current. This rule gives the force on a current in a field.

At right angles to the field

Circular motion in a magnetic field.

Magnetic force = centripetal force \(qvB = \dfrac{m v^{2}}{r}\) \(r = \dfrac{mv}{qB}\)

The force does no work: the speed and kinetic energy stay constant

A POSITIVE PARTICLE, FIELD INTO THE PAGE

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Why a circle

The force is always at right angles to the velocity, so it can never speed the particle up or slow it down. It only turns it. A force of constant size, always at right angles to the motion, is exactly the condition for uniform circular motion from A.2.

Why no work is done

Work is force times the distance moved in the direction of the force. The particle never moves in the direction of the magnetic force, not even slightly, so the work done is zero and the kinetic energy cannot change.

Reading \(r = \frac{mv}{qB}\)

The top line is the momentum. A particle with more momentum is harder to turn, so its circle is larger. A stronger field or a larger charge gives a tighter circle. A negative particle circles the opposite way round.

Writing that the magnetic field accelerates the particle “to a higher speed”. It does accelerate it, because the direction of the velocity changes, but the speed is unchanged.

Worked example

An electron in a 2.0 mT field.

\(v = 2.0 \times 10^{7} \,\mathrm{m} \,\mathrm{s}^{-1}\), at right angles to \(B\) \(r = \dfrac{mv}{eB}\) \(r = \dfrac{9.11 \times 10^{-31} \times 2.0 \times 10^{7}}{1.60 \times 10^{-19} \times 2.0 \times 10^{-3}}\) \(r = \class{res}{5.7 \times 10^{-2} \,\mathrm{m}}\)

Time for one circle \(T = \dfrac{2 \pi r}{v} = \dfrac{2 \pi m}{eB}\) \(T = \dfrac{2 \pi \times 9.11 \times 10^{-31}}{1.60 \times 10^{-19} \times 2.0 \times 10^{-3}}\) \(T = \class{res}{1.8 \times 10^{-8} \,\mathrm{s}}\)

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

The surprising thing about the period

Substitute \(r = \frac{mv}{eB}\) into \(\frac{2 \pi r}{v}\) and the speed cancels: \(T = \frac{2 \pi m}{eB}\). A faster particle moves round a bigger circle, and the two effects cancel exactly, so fast and slow particles of the same kind take the same time for one orbit. The period depends only on the field and on the charge-to-mass ratio of the particle.

Why that matters

A voltage that switches direction at this fixed rate can give a particle a push twice in every orbit, however fast the particle is already going. That is the idea behind the cyclotron, on the next slide.

A proton at the same speed in the same field has a radius 1836 times the electron’s radius: 105 m. Heavy particles need strong fields or large machines.

The story · Berkeley, 1930–1939

Background story · not assessed

Lawrence’s cyclotron.

Patent drawing of two hollow D-shaped electrodes seen from above, with a spiral path drawn from the center outwards across the gap between them, and a side view of the electrodes between the poles of a magnet — From Lawrence’s **patent**: a spiral path through two hollow “dees”

Photograph looking down into an open round metal vacuum chamber containing two D-shaped electrodes, with pipes and connections round its rim — The dees of the **27-inch cyclotron**, 1935

A magnetic field keeps the particles circling. The period does not depend on their speed
An alternating voltage across the gap speeds them up twice in every turn, so the circle grows
Ernest Lawrence’s 27-inch machine of 1932 reached 4.8 MeV. Nobel Prize, 1939

Patent drawing: Ernest O. Lawrence, 1934, public domain. Photograph: John B. Livingood, 1935, public domain.

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Identifying a particle from its path

The charge-to-mass ratio.

Accelerate through \(V\), then bend in \(B\) \(qV = \dfrac{1}{2} m v^{2}\) and \(qvB = \dfrac{m v^{2}}{r}\) \(v = \dfrac{qBr}{m}\) \(\dfrac{q}{m} = \dfrac{2V}{B^{2} r^{2}}\)

Measure \(V\), \(B\) and \(r\): the path gives \(\dfrac{q}{m}\).

A FINE BEAM TUBE

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

The algebra

From the circular motion, \(v = \frac{qBr}{m}\). Put this into the energy equation: \(qV = \frac{1}{2} m \left(\frac{qBr}{m}\right)^{2} = \frac{q^{2} B^{2} r^{2}}{2m}\). Divide both sides by \(q\) and rearrange to get \(\frac{q}{m} = \frac{2V}{B^{2} r^{2}}\). If the speed is known directly, the shorter form \(\frac{q}{m} = \frac{v}{Br}\) does the same job.

Why only the ratio

Charge and mass always appear together as \(\frac{q}{m}\) in the motion of a particle in electric and magnetic fields. No experiment of this kind can find either one separately. J. J. Thomson measured the ratio for cathode rays in 1897; it took Millikan’s oil drops to find \(e\) on its own, and then the mass of the electron followed.

How the tube works

The bulb holds gas at low pressure. Electrons that hit gas atoms make them glow, so the path shows as a thin bright ring whose diameter can be measured with a ruler. The uniform field comes from a pair of large coils, one on each side of the bulb.

A mass spectrometer uses the same physics to sort ions: for the same charge and speed, heavier ions follow wider circles.

Worked example

Measuring \(\frac{e}{m}\) for the electron.

\(V = 250 \,\mathrm{V}\) · \(B = 1.0 \,\mathrm{mT}\) Diameter of the ring 10.6 cm, so \(r = 5.3 \,\mathrm{cm}\) \(\dfrac{e}{m} = \dfrac{2V}{B^{2} r^{2}}\) \(\dfrac{e}{m} = \dfrac{2 \times 250}{(1.0 \times 10^{-3})^{2} \times (5.3 \times 10^{-2})^{2}}\) \(\dfrac{e}{m} = \class{res}{1.8 \times 10^{11} \,\mathrm{C} \,\mathrm{kg}^{-1}}\)

Accepted value: \(1.76 \times 10^{11} \,\mathrm{C} \,\mathrm{kg}^{-1}\)
Measure the diameter and halve it
\(r\) is squared: its uncertainty counts twice

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

How large this number is

One kilogram of electrons would carry a charge of \(1.76 \times 10^{11} \,\mathrm{C}\). For a hydrogen ion, a single proton, the ratio is \(9.6 \times 10^{7} \,\mathrm{C} \,\mathrm{kg}^{-1}\), \(\frac{1}{1836}\) of the electron’s ratio. When Thomson found such an enormous ratio for cathode rays, he concluded that the particles must be far lighter than any atom.

Improving the experiment

A single reading gives one value. A better method is to change the voltage, measure the radius each time and plot \(r^{2}\) against \(V\). The graph is a straight line through the origin with gradient \(\frac{2m}{eB^{2}}\), and the gradient uses all the readings at once.

Leaving the field in millitesla or the radius in centimeters. Both are squared, so the error is a factor of a million or ten thousand.

Neither along nor across

Velocity at an angle to the field.

Component along \(B\): no force, carries on at constant velocity
Component across \(B\): circular motion
Together: a helix round the field lines

Only \(v \sin \theta\), across the field, appears in the force.

A HELICAL PATH

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

In the sky · seen from orbit

Background story · not assessed

The aurora.

Photograph from orbit of the curved edge of the Earth at night, with a glowing green band of aurora above the clouds and part of the space station in the foreground — The **aurora borealis** photographed from the International Space Station, 2014

The Sun sends out a stream of electrons and protons
They spiral along the Earth’s magnetic field lines
The lines dip into the atmosphere near the poles
There the particles hit oxygen and nitrogen, which glow

A helix round a field line, a hundred kilometers up.

Photograph: NASA / Samantha Cristoforetti, public domain.

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Both fields at once

Crossed electric and magnetic fields.

A VELOCITY SELECTORPRESS → TO FIRE THREE PARTICLES

The fields are at right angles, and the two forces oppose
Straight through when \(qE = qvB\)

\(v = \dfrac{E}{B}\)

The charge cancels: one speed passes, whatever the particle.

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Why only one speed gets through

The electric force \(qE\) is the same for every speed. The magnetic force \(qvB\) grows with speed. For a particle that is too fast, the magnetic force wins and it curves one way. For one that is too slow, the electric force wins and it curves the other way. Only at \(v = \frac{E}{B}\) do they balance, and the particle obeys Newton’s first law.

A worked case

With \(E = 2.0 \times 10^{4} \,\mathrm{V} \,\mathrm{m}^{-1}\) and \(B = 1.0 \,\mathrm{mT}\): \(v = \frac{2.0 \times 10^{4}}{1.0 \times 10^{-3}} = 2.0 \times 10^{7} \,\mathrm{m} \,\mathrm{s}^{-1}\). Changing either field selects a different speed.

Negative particles

For an electron both forces reverse, so they still oppose each other, and the same speed passes straight through.

Thomson used crossed fields to find the speed of his cathode rays, then switched one field off and measured the deflection to get \(\frac{e}{m}\).

Quick check

Charges in magnetic fields.

01What is the magnetic force on a charge moving along the field lines?

Zero

02What happens to the kinetic energy of a particle in a magnetic field?

It stays constant

03The speed of a particle is doubled. What happens to the radius of its circle?

It doubles

04And to the time for one circle?

It does not change

05Fields \(E = 600 \,\mathrm{V} \,\mathrm{m}^{-1}\) and \(B = 2.0 \,\mathrm{mT}\) are crossed. Which speed passes undeflected?

\(\frac{600}{2.0 \times 10^{-3}} = 3.0 \times 10^{5} \,\mathrm{m} \,\mathrm{s}^{-1}\)

06What path does a particle follow when its velocity is at 30° to the field?

A helix

PRESS → TO REVEAL EACH ANSWER

Part 03 · SL + HL

Forces on currents.

A current is a stream of moving charges, so a wire in a field feels a force

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

A wire in a magnetic field

The force on a current-carrying conductor.

\(F = BIL \sin \theta\)

\(L\): the length of wire in the field
\(\theta\): the angle between the current and the field
Direction: the same left-hand rule

A WIRE BETWEEN THE POLES OF A MAGNET

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Where it comes from

The force on the wire is the total of the forces on all the moving charges inside it. In a time \(t\) a charge \(q = It\) passes along a length \(L = vt\) of wire. The force on that charge is \(qvB = (It) \left(\frac{L}{t}\right) B = BIL\). The two equations describe the same force, once for a single charge and once for a current.

Checking the direction in the diagram

Index finger along the field, from N to S: to the right. Middle finger along the current: out of the page, towards you. The thumb points up the page.

What it is used for

This force turns every electric motor and moves the cone of every loudspeaker. A coil carrying a current in a magnetic field feels forces on its two sides in opposite directions, and so it turns.

A wire lying along the field feels no force: \(\sin 0^{\circ} = 0\). The force is greatest when the wire is at right angles to the field.

Worked example

A wire in a 0.050 T field.

\(I = 3.0 \,\mathrm{A}\) · \(L = 0.20 \,\mathrm{m}\) of wire in the field \(F = BIL \sin \theta\) At right angles to the field: \(\theta = 90^{\circ}\) \(F = 0.050 \times 3.0 \times 0.20 \times \sin 90^{\circ} = \class{res}{0.030 \,\mathrm{N}}\)

The wire is turned to \(\theta = 30^{\circ}\) \(F = 0.050 \times 3.0 \times 0.20 \times \sin 30^{\circ}\) \(F = \class{res}{0.015 \,\mathrm{N}}\) Still at right angles to the wire and to the field

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Currents act on each other

The force between two parallel wires.

LONG STRAIGHT WIRESPRESS → FOR EACH STEP

\(\dfrac{F}{L} = \mu_{0} \dfrac{I_{1} I_{2}}{2 \pi r}\)

\(\dfrac{F}{L}\): the force on each meter of wire
\(r\): the separation of the wires
\(\mu_{0} = 4 \pi \times 10^{-7} \,\mathrm{T} \,\mathrm{m} \,\mathrm{A}^{-1}\)

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

The opposite of charges

Like charges repel, but like currents attract: two wires carrying current in the same direction pull together. Currents in opposite directions push apart.

Equal forces

The force on each meter of wire 1 is the same size as the force on each meter of wire 2, even if the currents are different. They are a Newton’s third law pair, and the equation is unchanged if \(I_{1}\) and \(I_{2}\) are swapped.

The constant

\(\mu_{0}\) is the permeability of free space. It does for magnetic fields what \(\varepsilon_{0}\) does for electric fields: it sets how strong a field a given current produces in a vacuum.

The two wires of a mains cable carry equal currents in opposite directions, so they repel. At ordinary currents the force is far too small to notice.

One wire in the field of the other

Why parallel currents attract.

Wire 1 makes a field at wire 2: \(B_{1} = \dfrac{\mu_{0} I_{1}}{2 \pi r}\)
Wire 2 carries a current in that field: \(F = B_{1} I_{2} L\)
So \(\dfrac{F}{L} = \mu_{0} \dfrac{I_{1} I_{2}}{2 \pi r}\)

Grip rule for the field, left-hand rule for the force.

SEEN END ONBOTH CURRENTS OUT OF THE PAGE

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Following the two rules

Wire 1 carries current out of the page, so by the right-hand grip rule its field circles counterclockwise. At wire 2, to the right of wire 1, that field points up the page. Now the left-hand rule at wire 2: index finger up the page, middle finger out of the page, and the thumb points to the left, towards wire 1.

Reversing one current

If the current in wire 2 goes into the page, the middle finger reverses and so does the thumb: wire 2 is pushed away. You only need to know the field of a straight wire as a pattern; the expression for \(B_{1}\) is not in the data booklet, but the final equation is.

Apply the same steps the other way round, with wire 1 in the field of wire 2, and the force on wire 1 comes out equal and opposite.

Worked example

Two wires 2.0 cm apart.

\(I_{1} = 5.0 \,\mathrm{A}\) · \(I_{2} = 8.0 \,\mathrm{A}\), in the same direction \(\dfrac{F}{L} = \mu_{0} \dfrac{I_{1} I_{2}}{2 \pi r}\) \(\dfrac{F}{L} = \dfrac{4 \pi \times 10^{-7} \times 5.0 \times 8.0}{2 \pi \times 0.020}\) \(\dfrac{F}{L} = \class{res}{4.0 \times 10^{-4} \,\mathrm{N} \,\mathrm{m}^{-1}}\), attractive

The \(\pi\) cancels: \(\dfrac{4 \pi}{2 \pi} = 2\)
On 1.5 m of either wire: \(4.0 \times 10^{-4} \times 1.5 = 6.0 \times 10^{-4} \,\mathrm{N}\)
Halve the separation: double the force

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

The story · Paris, 1820

Background story · not assessed

Ampère and the ampere.

Engraved portrait of André-Marie Ampère, a man with curly hair in a high-collared coat — **André-Marie Ampère**, engraved in 1825

In September 1820 Ørsted’s discovery was shown in Paris
Ampère soon found that two parallel wires attract or repel, with no magnet involved
Until 2019 this force defined the ampere: \(2 \times 10^{-7} \,\mathrm{N}\) on each meter of two wires 1 m apart
The ampere is now defined from the charge of the electron

That old definition is where the value \(\mu_{0} = 4 \pi \times 10^{-7}\) comes from.

Portrait: Ambroise Tardieu, 1825, public domain.

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Quick check

Forces on currents.

01A wire lies along the field lines. Force on it?

Zero

020.10 m of wire carries 2.0 A at right angles to a 0.50 T field. Force?

\(0.50 \times 2.0 \times 0.10 = 0.10 \,\mathrm{N}\)

03Two parallel wires carry currents in the same direction. Attract or repel?

Attract

04Both currents are doubled. What happens to the force per meter?

It becomes four times its original value

05The separation of the wires is tripled. Force per meter?

One third

06Wire 1 carries 1 A and wire 2 carries 5 A. Which feels the larger force?

Neither: the forces are equal

PRESS → TO REVEAL EACH ANSWER

Summary

Motion in electromagnetic fields on one slide.

Electric field

Constant force

\(F = qE\), and \(qV = \frac{1}{2} m v^{2}\)
A parabola when entering at right angles

Magnetic field

A sideways force

\(F = qvB \sin \theta\)
No work done: speed constant

Circular paths

\(r = \frac{mv}{qB}\)

Period independent of speed
The path gives \(\frac{q}{m}\)

Crossed fields

\(v = \frac{E}{B}\)

Only one speed passes straight through

A wire in a field

\(F = BIL \sin \theta\)

Left-hand rule for the direction

Parallel wires

\(\frac{F}{L} = \mu_{0} \frac{I_{1} I_{2}}{2 \pi r}\)

Same direction: attract
Opposite directions: repel

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL

Summary

Key words.

Uniform fieldSame strength and direction everywhere

ElectronvoltEnergy of charge e through 1 V

Magnetic field strengthB, in tesla. Also called flux density

Tesla1 N on 1 C moving at 1 m s⁻¹

Left-hand ruleThumb F, index B, middle I

Charge-to-mass ratio\(\dfrac{q}{m}\), in C kg⁻¹

HelixCircling while moving along the field

Velocity selectorCrossed fields pass one speed

Permeability of free space\(\mu_{0}\), the magnetic constant

D.3 MOTION IN ELECTROMAGNETIC FIELDS · SL + HL