Moving Charges and Magnetism

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XII, Chapter 4.

A wire twitches a compass needle. Why would a current care about a magnet?

In 1820 Hans Christian Oersted was setting up a lecture demonstration when he noticed a compass needle swing as he switched on a current in a nearby wire. Until that moment, electricity and magnetism were treated as two unrelated subjects. The needle did not point towards the wire or away from it. It lined up along circles around the wire. The conclusion was that moving charges produce magnetic fields, and, by symmetry, that moving charges feel forces in magnetic fields.

This chapter works both directions of that idea. First, what a magnetic field does to moving charges and to current-carrying wires. Second, what field a current creates. The two fit together so well that the force between two wires ends up defining the ampere.

The Lorentz force: the magnetic force on a moving charge

A charge q moving with velocity v through an electric field E and a magnetic field B feels the Lorentz force

$$\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$$

The magnetic part is $\vec{F} = q(\vec{v} \times \vec{B})$, with magnitude $F = qvB\sin\theta$, where θ is the angle between v and B. The SI unit of B is the tesla ($1\ \text{T} = 1\ \text{N s C}^{-1}\ \text{m}^{-1} = 1\ \text{N A}^{-1}\ \text{m}^{-1}$). A smaller unit is the gauss: $1\ \text{G} = 10^{-4}\ \text{T}$. Earth's field at the surface is about 3 × 10⁻⁵ T, or 0.3 G.

To find the direction for a positive charge, point your fingers along v, curl them towards B, and your thumb gives F. For a negative charge, reverse the result.

Three facts follow from the cross product, and each one is a favourite MCQ.

The force is always perpendicular to both v and B. That means it can turn the particle but can never speed it up. The magnetic force does no work, so the kinetic energy and speed of a particle in a purely magnetic field never change. Only the direction of motion does.

Pause and predict: an electron is fired along the axis of a long solenoid carrying a steady current. How does it move?

Check your answer: it moves in a straight line at constant speed. Inside a long solenoid the field is along the axis, so v is parallel to B and $v \times B = 0$. No magnetic force acts, and the particle carries on as if the solenoid were not there.

Combining the two fields: the velocity selector

If E and B are perpendicular to each other and to the velocity, the electric force $qE$ and the magnetic force $qvB$ can point in opposite directions. They cancel when $qE = qvB$, that is when

$$v = \dfrac{E}{B}$$

Only particles with this speed go straight through. Notice that the condition has no q and no m in it, so it does not matter what the particle is. Slower or faster particles are deflected away. A velocity selector is used before a mass spectrometer to give a beam of one speed.

Motion in a uniform magnetic field: circles and helices

When v is perpendicular to B, the magnetic force is always at right angles to v, so it acts as the centripetal force:

$qvB = mv^2/r$, giving $r = mv/(qB)$

Because the speed is constant, the particle moves at a constant rate on a circle. The angular speed and the period do not contain v or r at all:

$\omega = qB/m$, $T = 2\pi m/(qB)$, and the cyclotron frequency is $\nu = qB/(2\pi m)$

This is the surprising result of the chapter. A faster particle takes a bigger circle, but the extra distance exactly compensates the extra speed, so it takes the same time. If you write the radius in terms of momentum, $r = p/(qB)$, you also see that a heavier or faster particle has a larger radius for the same q and B.

It is often easier to start from the energy. If a particle of charge q is accelerated from rest through a potential difference V, then $1/2\,mv^2 = qV$, so $v = \sqrt{2qV/m}$, and

$$r = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$$

so for a given B, the radius scales as $\sqrt{V}$ and as $\sqrt{m/q}$.

Worked example: a proton in a magnetic field. A proton is accelerated from rest through 800 V and then enters a uniform field of 0.10 T perpendicular to its velocity. Find its speed, the radius of its path, and the work done by the magnetic field in one revolution. ($m_p = 1.67 \times 10^{-27}\ \text{kg}$, $e = 1.6 \times 10^{-19}\ \text{C}$)

Common wrong step: thinking the magnetic field adds energy to the proton and using its energy gain in the second stage. The field only bends the path. All the kinetic energy came from the 800 V.

Pause and predict: a proton and an alpha particle enter the same magnetic field perpendicular to it, both with the same speed. Which moves in the larger circle, and by what factor?

Check your answer: the alpha particle, by a factor of 2. Since $r = mv/(qB)$, $r \propto m/q$. The alpha particle has four times the mass and twice the charge of a proton, so m/q is two times larger. If instead both had the same kinetic energy, $r \propto \sqrt{m}/q$ would be equal for the two.

If v has a component along B as well, that component feels no force and stays constant. The particle then spirals along the field line in a helix. For a velocity v at angle θ to B:

radius $r = mv\sin\theta/(qB)$, pitch $p = v\cos\theta \times T = 2\pi m v\cos\theta/(qB)$

Worked example: a helical path. A proton moves at 4 × 10⁵ m s⁻¹ at 30° to a uniform field of 0.10 T. Find the radius and the pitch of its helix.

Common wrong step: using the full speed 4 × 10⁵ m s⁻¹ in $r = mv/(qB)$, which gives twice the correct radius.

Beyond the textbook. For the exam: a charge in a field moves in a circle or helix, and the magnetic force does no work. In reality: Earth's magnetic field traps charged particles from the Sun in exactly this helical way. They spiral along the field lines towards the poles, and when they collide with atmospheric atoms the atoms glow. That glow is the aurora, which is why it is seen near the polar regions.

The cyclotron

A cyclotron accelerates charged particles using two hollow D shaped metal boxes in a strong magnetic field, with an alternating electric field in the gap between them. Each time the particle crosses the gap the electric field gives it a push, and inside a D the magnetic field bends it in a semicircle. Since the time for a half circle is independent of speed, a fixed frequency of the alternating field stays in step with the particle. The particle spirals outward as it gains energy, until it leaves at the rim.

The cycle in one line: gap pushes the particle (energy up) → semicircle in the D (same time every time) → gap pushes again → bigger radius → exit at the rim.

The resonance condition is that the oscillator frequency equals the cyclotron frequency $qB/(2\pi m)$.

Beyond the textbook. For the exam: the cyclotron frequency does not depend on speed, so a fixed frequency works. In reality: this holds only when the speed is much smaller than the speed of light. As a particle approaches that speed its effective mass grows, its frequency drops and it falls out of step with the oscillator. This limits the simple cyclotron, and machines that adjust the frequency or the field were built to go beyond it.

Force on a current carrying wire

Current is moving charge, so a wire in a magnetic field feels the sum of the Lorentz forces on its carriers. For a straight wire of length l:

$\vec{F} = I\vec{l} \times \vec{B}$, with magnitude $F = BIl\sin\theta$

The direction of l is along the current, and the same right-hand rule applies as for charges. The force is largest when the wire is perpendicular to B and zero when the wire is parallel to B.

For a curved wire in a uniform field, the net force depends only on the straight line joining its ends. A wire in the shape of a semicircle of radius R with its ends 2R apart, in a field perpendicular to its plane, feels a net force $F = BI(2R)$, the same as for the straight diameter. A closed loop of any shape in a uniform field feels zero net force.

The Biot–Savart law: the field of a current

Each small piece of current, or current element $I\,dl$, produces a field at a point P given by

$dB = (\mu_0/4\pi) \times I\,dl \times \hat{r}/r^2$, with magnitude $dB = (\mu_0/4\pi) \times I\,dl\sin\theta/r^2$

Here r is the distance from the element to P and θ is the angle between dl and r. The constant is $\mu_0/4\pi = 10^{-7}\ \text{T m A}^{-1}$, so $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$, the permeability of free space. The field is perpendicular to both dl and r, and it is zero along the line of the element itself ($\sin 0 = 0$). To find the field of a whole conductor, add up all the elements.

This law is the magnetic counterpart of Coulomb's law, but with a twist. Coulomb's field falls as $1/r^2$ along the radius; the Biot–Savart field circles around the current. The direction comes from the right-hand thumb rule: point the thumb along the current in a straight wire, and your fingers curl in the direction of B.

Field at the centre of a circular coil. Every element is at the same distance R and at 90° to the radius, so adding them gives

$B = \mu_0 N I/(2R)$ for N turns

On the axis at a distance x from the centre of a single loop of radius R, $B = \mu_0 I R^2/[2(R^2 + x^2)^{3/2}]$. Far away (x much greater than R), B falls as $1/x^3$ and the loop behaves like a bar magnet.

Field at the centre of an arc. The field at the centre is proportional to the fraction of a full circle that carries current. For an arc subtending angle φ (in radians) at the centre, $B = \mu_0 I\varphi/(4\pi R)$. For a semicircle this is $\mu_0 I/(4R)$, and for a quarter circle $\mu_0 I/(8R)$. Straight wires that lie along a radius through the centre give nothing, since dl is parallel to r and $\sin\theta = 0$.

Worked example: two concentric semicircles. A wire carrying 4 A forms two concentric semicircular arcs of radii 10 cm and 20 cm, joined by straight segments along a common diameter. Find the field at the common centre.

Common wrong step: adding the two arc fields, which gives 1.9 × 10⁻⁵ T. The senses of the two arcs are opposite, so the fields subtract and the smaller arc (inner, stronger) decides the direction.

Ampere's circuital law: the shortcut

Biot–Savart always works, but the sum can be painful. When the situation is symmetric, Ampere's circuital law is much faster:

$$\oint B \cdot dl = \mu_0 I$$

where I is the total current passing through the loop that you choose (the Amperian loop). The trick is to choose a loop on which B has a constant magnitude along the path.

Long straight wire. A circular loop of radius r around the wire gives $B \times 2\pi r = \mu_0 I$, so

$$B = \dfrac{\mu_0 I}{2\pi r}$$

Thick wire carrying uniform current (radius a). Outside, the field is the same as for a thin wire, $B = \mu_0 I/(2\pi r)$, falling as $1/r$. Inside, the enclosed current is $Ir^2/a^2$, so $B = \mu_0 I r/(2\pi a^2)$, which grows in a straight line from zero at the axis to its maximum at the surface.

Solenoid. A solenoid is a long coil of closely wound turns. Inside, far from the ends, the field is uniform and along the axis; outside it is nearly zero. With n turns per unit length:

$$B = \mu_0 n I$$

Toroid. A solenoid bent into a ring. For N turns, the field inside the core at radius r is $B = \mu_0 N I/(2\pi r)$, and it is zero in the hollow centre and in the space outside, since the net current enclosed by any circle there is zero.

Worked example: a null point between two wires. Two long parallel wires 20 cm apart carry currents of 6 A and 3 A in the same direction. Find the point between them where the net field is zero, and the force per unit length between them.

Common wrong step: putting the null point at the midpoint, 10 cm from each. The stronger current produces the stronger field, so the null point lies closer to the weaker current.

Force between parallel currents and the definition of the ampere

The field of wire 1 at wire 2 is $\mu_0 I_1/(2\pi d)$, and it pushes on wire 2 with force per unit length

$$f = \dfrac{\mu_0 I_1 I_2}{2\pi d}$$

Parallel currents attract. Antiparallel currents repel. This is the opposite of the rule for electric charges, where like charges repel. If $I_1 = I_2 = 1$ A and $d = 1$ m, $f = 2 \times 10^{-7}$ N m⁻¹. The ampere is therefore the steady current which, flowing in each of two long parallel wires 1 m apart in vacuum, produces a force of 2 × 10⁻⁷ N per metre of length.

Beyond the textbook. For the exam: the ampere is defined by the force of 2 × 10⁻⁷ N per metre between two wires 1 m apart. In reality: since 2019 the SI defines the ampere by fixing the value of the elementary charge e. The two wire setup is no longer the official definition, but it still gives the same result to a very high accuracy, and it is how the ampere was defined for decades. As a result, μ₀ is now a measured quantity, equal to 4π × 10⁻⁷ N A⁻² to within about one part in 10⁹.

Torque on a current loop and the magnetic dipole

In a uniform field, the forces on opposite sides of a rectangular loop are equal and opposite, so the net force is zero. But the forces on the two sides perpendicular to the field act along different lines, forming a couple. The torque is

$\tau = NIAB\sin\theta$, or $\tau = m \times B$, where $m = NIA$

m is the magnetic moment of the loop, a vector along the area vector (given by the right-hand rule: curl your fingers along the current, and the thumb gives m), and θ is the angle between m and B. Here θ is the angle between the normal to the coil and B, not between the plane and B.

The potential energy is $U = -m \cdot B$. A current loop is a magnetic dipole, and the correspondence with an electric dipole is complete.

Worked example: torque on a coil. A square coil of side 5.0 cm and 50 turns carries 2.0 A in a uniform field of 0.40 T. Find the torque when the plane of the coil is parallel to the field, and when it is perpendicular to the field.

Common wrong step: taking the angle between the plane and the field as $\theta$. That swaps the two answers. Always use the angle between the normal (the direction of m) and B.

The electron's own magnetic moment. An electron of charge e circling an orbit with period T is a tiny current loop, $I = e/T$, so its orbital magnetic moment is $\mu = IA = evr/2$, which equals $(e/2m)L$, where L is its orbital angular momentum. With $L$ in units of $h/2\pi$, the smallest value is the Bohr magneton: $\mu_B = eh/(4\pi m) \approx 9.27 \times 10^{-24}\ \text{A m}^2$.

The moving coil galvanometer, ammeter and voltmeter

A moving coil galvanometer has a coil of N turns, area A, hung in a radial magnetic field produced by curved pole pieces around a soft iron core. The field is always in the plane of the coil, so the torque is $NIAB$ at every angle. A spring provides a restoring torque $k\varphi$. In balance, $NIAB = k\varphi$, so

$$\varphi = \dfrac{NAB}{k} I$$

The deflection is proportional to the current, so the scale is evenly spaced. The current sensitivity is $\varphi/I = NAB/k$, and the voltage sensitivity is $\varphi/V = NAB/(kR)$, where R is the resistance of the coil. If you double the number of turns, the current sensitivity doubles but the voltage sensitivity stays the same, because the coil's resistance also doubles.

A galvanometer is a very sensitive but delicate instrument that only handles small currents. To convert it:

Here G is the galvanometer resistance, Ig is the current for full scale deflection, I is the new full scale current and V is the new full scale voltage. An ammeter goes in series in a circuit, and has low resistance so it does not change the current. A voltmeter goes in parallel with the part being measured, and has high resistance so it does not draw current.

Worked example: building meters. A galvanometer has a coil resistance of 50 Ω and gives full scale deflection for 2 mA. How would you convert it into an ammeter reading up to 5 A, and a voltmeter reading up to 10 V?

Common wrong step: adding 5000 Ω in series for the voltmeter and forgetting that the coil already has 50 Ω. The total must be 5000 Ω, so only 4950 Ω is added.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

The source of every magnetic field here is the moving charge you met in Current Electricity. The next step is how materials respond to these fields, in Magnetism and Matter, and how a changing flux creates current, in Electromagnetic Induction. The dipole analogy repeats what you saw in Electric Charges and Fields, and the way a magnetic field relates to light is developed in Electromagnetic Waves.