Learn what magnetic force is and how to calculate it.

What is the magnetic force?

The magnetic force is a consequence of the electromagnetic force, one of the four fundamental forces of nature, and is caused by the motion of charges. Two objects containing charge with the same direction of motion have a magnetic attraction force between them. Similarly, objects with charge moving in opposite directions have a repulsive force between them.
In our article on magnetic fields we learned how moving charge surrounds itself with a magnetic field. In this context the magnetic force is a force that arises due to interacting magnetic fields.

How to find the magnetic force?

Consider two objects. The magnitude of the magnetic force between them depends on how much charge is in how much motion in each of the two objects and how far apart they are. The direction of the force depends on the relative directions of motion of the charge in each case.
The usual way to go about finding the magnetic force is framed in terms of a fixed amount of charge qq moving at constant velocity vv in a uniform magnetic field BB. If we don't know the magnitude of the magnetic field directly then we can still use this method because it is often possible to calculate the magnetic field based on the distance to a known current.
The magnetic force is described by the Lorentz Force law:
Actually, this is just the magnetic part of the Lorentz Force Law, in full it also describes the force due to the electric force. The electric force works between stationary charge.
F=qE+qv×B\vec{F} = q\vec{E} + q\vec{v} \times \vec{B}
F=qv×B\vec{F} = q\vec{v} \times \vec{B}
In this form it is written using the vector cross product. We can write the magnitude of the magnetic force by expanding the cross product. Written in terms of the angle θ\theta (<180 < 180^\circ) between the velocity vector and the magnetic field vector:
F=qvBsinθ\boxed{F = qvB\sin{\theta}}
The direction of the force can be found using the right-hand-slap rule. This rule describes the direction of the force as the direction of a 'slap' of an open hand. As with the right-hand-grip rule, the fingers point in the direction of the magnetic field. The thumb points in the direction that positive charge is moving. If the moving charge is negative (for example, electrons) then you need to reverse the direction of your thumb because the force will be in the opposite direction. Alternatively, you can use your left hand for moving negative charge.
There are a few alternative versions of left/right hand rule using different parts of the hand to represent different quantities. All are equivalent, though we prefer the hand-slap version because it keeps the same relationship between the fingers as the right-hand-grip rule used for magnetic fields and it is natural that the 'slap' direction is the force.
Note that the right-hand-grip rule is defined with the thumb pointing in the direction of conventional current flow which for historical reasons is opposite to the direction of the electron flow.
Using the right-hand-slap rule for the force due to a positive charge moving in a magnetic field.
Figure 1: Using the right-hand-slap rule for the force due to a positive charge moving in a magnetic field.
Sometimes we want to find the force on a wire carrying a current II in a magnetic field. This can be done by rearranging our previous expression. If we recall that velocity is a distance / time then if a wire has length LL we can write
qv=qLtqv = \frac{qL}{t}
and since current is the amount of charge flowing per second,
qv=ILqv = IL
and therefore
F=BILsinθ\boxed{F = BIL\sin{\theta}}

Force on a wire

Exercise 1a:
Figure 2: Magnetic force on a wire.
Figure 2: Magnetic force on a wire.
Figure 2 shows a wire running through the north and south poles of a horseshoe magnet. A battery is connected to the wire which causes a current of 5 A5~\mathrm{A} to flow through the wire in the direction shown. If the magnetic field between the poles is known to be 0.2 T0.2~\mathrm{T}, what is the magnitude and direction of the force on the 10 mm10~\mathrm{mm} section of wire between the poles?
We begin with the Lorentz force law applied to a current carrying wire:
F=BILsinθF = BIL\sin{\theta}
In this exercise, the magnetic field lines (which run from north to south) form a right angle (9090^\circ) with the direction of the current flowing in the wire. So sinθ=1\sin{\theta}=1 and this term can be removed. The force can then be evaluated:
F=(0.2 T)(5 A)(0.01 m)=0.01 N\begin{aligned} F &= (0.2~\mathrm{T})(5~\mathrm{A})(0.01~\mathrm{m}) \\ &= 0.01~\mathrm{N}\end{aligned}
We can now use the hand-slap rule to find the direction of the force. The electrons are moving up in the diagram (reverse of conventional current) and the magnetic field is directed to the right. This makes the force on the wire vertically out of the page.
Exercise 1b:
Suppose the magnet was shifted a little to the left so that the wire is now closer to the south pole of the magnet. Would you expect any change in the force on the wire?
No, to a first approximation the magnetic field is uniform between the poles. We would expect the same force.
Exercise 1c:
Suppose the strength of the magnet was not known. Can you suggest a way to modify this experiment to measure the strength of the magnetic field? Assume you have a ruler, string and some calibrated weights available.
If the wire is slightly flexible then it will be deflected by the magnetic force when the current is flowing. We could measure this deflection with a ruler. Later we could tie weights to the wire and find the mass required to produce the deflection measured earlier. This would tell us the force on the wire due to current flow. If the current and length of the wire under strain are known then it is possible to find the strength of the magnet by rearranging the Lorentz force law.

Magnetic deflection of electrons in a cathode-ray tube

A cathode ray tube is an evacuated tube with an electron gun at one end and a phosphorescent screen at the other end. Electrons are ejected from the electron gun at high speed and impact the screen where a spot of light is produced on impact with the phosphor.
Because electrons have charge it is possible to deflect them in-flight with either the electric or magnetic force. Controlling the deflection allows the spot of light to be moved around the screen. Old style 'tube' televisions use this principle with magnetic deflection to form images by rapidly scanning the spot.
Exercise 2a:
Figure 3 shows a cathode ray tube experiment. A pair of coils are placed outside a cathode ray tube and produce a uniform magnetic field across the tube (not shown). In response to the field, the electrons are deflected and follow a path which is a segment of a circle as shown in the figure. What is the direction of the magnetic field?
Figure 3: Cathode ray tube experiment.
Figure 3: Cathode ray tube experiment.
Based on an understanding of circular motion, we know that there must be a centripetal force directed towards the center of the circular path that the electrons take. This force is provided by the magnetic force.
Applying the left-hand slap rule (electrons are negatively charged) with an upward direction of the force and thumb pointing along the direction of travel we see that the magnetic field must be directed out of the page.
Exercise 2b:
If the electrons are known to be ejected from the electron gun horizontally at a speed vv of 2107 m/s2\cdot 10^7~\mathrm{m/s}, what is the strength of the magnetic field? Assume that the radius of the circular path can be approximated by L2/2dL^2/2d where LL is the length of the tube and dd is the horizontal deflection.
From the left-hand slap rule, we know that the magnetic force is perpendicular to the velocity. This is also the condition that gives rise to circular motion. Equating the centripetal force on an electron of mass mem_e at velocity vv to the magnetic force (assuming the field is always at a right angle to the velocity),
qvB=mev2RqvB = \frac{m_e v^2}{R}
B=mevqRB = \frac{m_e v}{qR}
We can find the radius of the circular path using the given approximation,
R=L22d=(0.2 m)220.025 m=0.8 m\begin{aligned} R &= \frac{L^2}{2d} \\ &= \frac{(0.2~\mathrm{m})^2}{2\cdot 0.025~\mathrm{m}} \\ &= 0.8~\mathrm{m}\end{aligned}
and now substitute back into the original equation,
B=(9.11031 kg)(2107 m/s)(1.61019 C)(0.8 m)=1.42104 T\begin{aligned} B &= \frac{(9.1\cdot 10^{-31}~\mathrm{kg})\cdot( 2\cdot 10^7~\mathrm{m/s} ) }{(1.6\cdot 10^{-19}~\mathrm{C}) \cdot (0.8~\mathrm{m}) } \\ &= 1.42 \cdot 10^{-4}~\mathrm{T} \end{aligned}
Interestingly, the deflection is caused by quite a small magnetic field (less than ten times larger than the earth's field). For this reason we would expect the earth's field to have a significant effect on the deflection. Manufacturers of equipment employing cathode-ray tubes often attempt to mitigate this problem by including magnetic shielding and sometimes by providing controls to adjust the image if the external magnetic field changes after moving the equipment.