Magnetic Forces on Charges and Currents, and Magnetic Flux – Magnetic Field and Magnetic Forces (Senior 5 Physics) | Ganzaa

Magnetic Field and Magnetic Forces

Lessons

Senior 5 · Physics

Magnetic Field and Magnetic Forces

4 lessons

Lesson 2 of 4

Magnetic Forces on Charges and Currents, and Magnetic Flux

25 min

Pushing a wire with a magnet

Hang a light aluminium rod between the poles of a strong horseshoe magnet and switch on a current: the rod jumps sideways, out of the gap. Reverse the current, and it jumps the other way. Reverse the magnet as well, and it jumps the first way again. This force on a current in a magnetic field turns every electric motor and loudspeaker, steers the electron beam in old television tubes and bends the paths of particles in accelerators. In this lesson you find it from its root: the force on a single moving charge.

The force on a moving charge

Experiments show that a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ feels a force that is

proportional to $q$ , to $v$ and to $B$ ;
proportional to $\sin\theta$ , where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$ : zero for motion along the field, greatest for motion across it;
always at right angles to both $\vec{v}$ and $\vec{B}$ .

In vector form, $\vec{F} = q\vec{v} \times \vec{B}$ , and its size is

F = qvB\sin\theta

This relation defines the tesla: a field of 1 T exerts a force of 1 N on a charge of 1 C moving at 1 m s⁻¹ at right angles to it, so $1\,\mathrm{T} = 1\,\mathrm{N\,s\,C^{-1}\,m^{-1}} = 1\,\mathrm{N\,A^{-1}\,m^{-1}}$ .

For a positive charge the direction is given by Fleming's left-hand rule: hold the thumb and first two fingers of your left hand at right angles; the First finger points along the Field, the seCond finger along the Current (the velocity of a positive charge), and the thuMb gives the Motion (force). For an electron, the force is in the opposite direction.

Because the force is always perpendicular to the velocity, a magnetic field does no work on a moving charge: it can change the direction of motion but never the speed or kinetic energy. If an electric field is also present, the total (Lorentz) force is $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$ .

Definition

The tesla (T) is the unit of magnetic flux density: a field of 1 T exerts a force of 1 N on a charge of 1 C moving at 1 m s⁻¹ perpendicular to the field. The magnetic force on a moving charge is $F = qvB\sin\theta$ , perpendicular to both $\vec{v}$ and $\vec{B}$ . Example: an electron moving at $1.0 \times 10^{7}$ m s⁻¹ across a 0.10 T field feels $(1.60 \times 10^{-19})(1.0 \times 10^{7})(0.10) = 1.6 \times 10^{-13}$ N.

Worked example 1: a proton crossing a field

A proton moves at $2.0 \times 10^{6}$ m s⁻¹ at 30° to a uniform 0.50 T field. Find the size of the force on it and the acceleration this gives ( $m_p = 1.67 \times 10^{-27}$ kg). What happens to its speed?

Solution

$F = qvB\sin\theta = (1.60 \times 10^{-19})(2.0 \times 10^{6})(0.50)\sin 30^\circ = 8.0 \times 10^{-14}$ N.

$a = \frac{F}{m} = \frac{8.0 \times 10^{-14}}{1.67 \times 10^{-27}} = 4.8 \times 10^{13}\,\mathrm{m\,s^{-2}}$ .

The speed does not change: the force is always perpendicular to the velocity, so it does no work. Only the direction changes.

The force on a current-carrying conductor

A current is a stream of moving charges, so a wire carrying a current in a field feels the sum of the forces on them. Take a length $L$ of wire of cross-section $A$ containing $n$ free charges $q$ per unit volume, drifting at speed $v_{d}$ . The number of charges is $n A L$ , each feeling $qv_dB\sin\theta$ , so

F = (nAL)(qv_dB\sin\theta) = (nAqv_d)LB\sin\theta

But $n A q v_{d}$ is the current $I$ . Therefore

F = BIL\sin\theta

where $\theta$ is the angle between the wire and the field. The force is perpendicular to both, in the direction given by Fleming's left-hand rule.

Worked example 2: a rod in a magnet's gap

A straight rod 20 cm long carries 5.0 A in a uniform field of 0.30 T. Find the force on it when it is (a) perpendicular to the field, (b) at 40° to the field, (c) parallel to the field. (d) The rod has mass 10 g and lies horizontally, perpendicular to a horizontal field. What field would make the magnetic force equal to its weight ( $g = 9.8\,\mathrm{m\,s^{-2}}$ )?

Solution

(a) $F = (0.30) (5.0) (0.20) = 0.30$ N.

(b) $F = 0.30 \times \sin 40^\circ = 0.19$ N.

(d) The force must be vertical, so the field is horizontal and across the rod: $B = \frac{mg}{IL} = \frac{(0.010)(9.8)}{(5.0)(0.20)} = 0.098$ T.

The force between two parallel currents

Each of two long parallel wires lies in the magnetic field of the other. Wire 1 makes a field $B_1 = \frac{\mu_0 I_1}{2\pi d}$ at wire 2, at right angles to it, so a length $L$ of wire 2 feels $F = B_{1} I_{2} L$ :

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

By Newton's third law wire 1 feels an equal and opposite force. Until 2019 this force defined the ampere: 1 A was the current that, in two long parallel wires 1 m apart in a vacuum, gave a force of $2 \times 10^{-7}$ N per metre. Since 2019 the SI defines the ampere by fixing the charge of the electron, $e = 1.602176634 \times 10^{-19}$ C, but the formula above is unchanged.

Worked example 3: busbars in a substation

Two long parallel conductors in an electrical substation are 0.50 m apart and each carries 1000 A in the same direction. Find the force per metre between them, and say whether they attract or repel.

Solution

$\frac{F}{L} = \frac{(4\pi \times 10^{-7})(1000)(1000)}{2\pi(0.50)} = 0.40\,\mathrm{N\,m^{-1}}$ , an attraction, since the currents are in the same direction.

During a short circuit the current can rise twenty-fold; the force rises with $I^{2}$ , four hundred-fold, to 160 N per metre, which is why busbars are firmly clamped.

Magnetic flux and Gauss's law for magnetism

The magnetic flux $\Phi$ through a flat area $A$ in a uniform field $B$ measures how much field passes through the area:

\Phi = BA\cos\theta

where $\theta$ is the angle between the field and the normal (perpendicular) to the area. Its unit is the weber, $1\,\mathrm{Wb} = 1\,\mathrm{T\,m^2}$ . This is why $B$ is called the flux density: $B = \frac{\Phi}{A}$ for an area at right angles to the field. Flux is central to electromagnetic induction (Senior 6).

Magnetic field lines never begin or end: they form closed loops, running outside a magnet from N to S and inside it from S back to N. So every line that leaves a closed surface must enter it somewhere else. This is Gauss's law for magnetism:

\oint \vec{B} \cdot d\vec{A} = 0

The net magnetic flux through any closed surface is zero. It expresses the fact that there are no isolated magnetic poles (monopoles): cut a magnet in half and you get two smaller magnets, each with an N and an S pole. Compare Gauss's law for electric fields, where the flux through a closed surface equals the enclosed charge divided by $\varepsilon_0$ : electric charges exist on their own, magnetic poles do not.

Definition

Magnetic flux $\Phi = BA\cos\theta$ is the product of the field and the area it crosses at right angles; unit the weber (Wb). Gauss's law for magnetism: the net magnetic flux out of any closed surface is zero, because magnetic field lines are closed loops and there are no magnetic monopoles. Example: a 0.50 T field at 60° to the normal of a 0.020 m² coil gives $\Phi = (0.50)(0.020)\cos 60^\circ = 5.0 \times 10^{-3}$ Wb.

Warning

Left hand for force, right hand for field. Fleming's left-hand rule gives the force on a current; the right-hand grip rule gives the field round a current. And in $\Phi = BA\cos\theta$ , $\theta$ is measured from the normal to the area, not from the surface itself.

Tip

Like currents attract, the opposite of like charges and like poles. Picture two wires carrying current the same way: between them their fields cancel, so the stronger field outside pushes them together.

Force on current carrying wire in a magnetic field #1 (NCPQ), Dr Richard Walding. Watch the wire move as the current is switched on, and move the other way when the current is reversed.

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Practice

An alpha particle (charge $3.2 \times 10^{-19}$ C) moves at $1.5 \times 10^{7}$ m s⁻¹ at right angles to a 0.20 T field. Find the force on it.
Explain why a magnetic field cannot change the kinetic energy of a moving charge.
A wire 0.50 m long carrying 3.0 A experiences a force of 0.12 N when perpendicular to a uniform field. Find the field.
A wire carrying current due east lies in a field pointing vertically downwards. Use Fleming's left-hand rule to find the direction of the force.
Two parallel wires 2.0 cm apart carry 10 A and 15 A in opposite directions. Find the force per metre between them and state whether it is attractive or repulsive.
A square coil of side 10 cm is in a 0.40 T field. Find the flux through it when its normal makes (a) 0°, (b) 60°, (c) 90° with the field.
Use Gauss's law for magnetism to explain why breaking a bar magnet never produces an isolated north pole.
Mutesi says: "The force on a wire in a magnetic field is largest when the wire points along the field lines, because then it is following the field." Evaluate her statement.

Show answersHide answers

$F = (3.2 \times 10^{-19})(1.5 \times 10^{7})(0.20) = 9.6 \times 10^{-13}$ N.
The force is always perpendicular to the velocity, so no work is done ( $W = Fs\cos 90^\circ = 0$ ); with no work done, the kinetic energy and speed stay constant.
$B = \frac{F}{IL} = \frac{0.12}{(3.0)(0.50)} = 0.080$ T.
First finger down (field), second finger east (current): the thumb points north. The force is towards the north.
$\frac{F}{L} = \frac{(4\pi \times 10^{-7})(10)(15)}{2\pi(0.020)} = 1.5 \times 10^{-3}\,\mathrm{N\,m^{-1}}$ , repulsive.
$A = 0.010\,\mathrm{m^2}$ : (a) $4.0 \times 10^{-3}$ Wb; (b) $2.0 \times 10^{-3}$ Wb; (c) 0.
An isolated north pole would be a source of field lines with no return, giving a net outward flux through a surface round it. Gauss's law says the net flux through every closed surface is zero, so field lines always loop back and each piece of magnet has both poles.
Incorrect. $F = BIL\sin\theta$ is zero when the wire is parallel to the field ( $\theta = 0$ ) and greatest when the wire is perpendicular to it ( $\theta = 90^\circ$ ). The force acts across the field, not along it.

Key points

Moving charge: $F = qvB\sin\theta$ , perpendicular to $\vec{v}$ and $\vec{B}$ ; no work is done; defines the tesla.
Current-carrying conductor: $F = BIL\sin\theta$ ; direction by Fleming's left-hand rule.
Parallel wires: $\frac{F}{L} = \frac{\mu_0 I_1I_2}{2\pi d}$ ; like currents attract.
Flux $\Phi = BA\cos\theta$ (Wb); Gauss's law for magnetism: net flux through any closed surface is zero, so there are no monopoles.

Sources and credits

Video: Force on current carrying wire in a magnetic field #1 (NCPQ), Dr Richard Walding, embedded from YouTube.

Written by Ganzaa. The three diagrams, the worked examples and the practice exercises are by Ganzaa.

1Pushing a wire with a magnet

2The force on a moving charge

3The force on a current-carrying conductor

4The force between two parallel currents

5Magnetic flux and Gauss's law for magnetism

Pushing a wire with a magnet

The force on a moving charge

The force on a current-carrying conductor

The force between two parallel currents

Magnetic flux and Gauss's law for magnetism