**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 q moving at constant velocity v in a uniform magnetic field B. 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:

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

F, with, vector, on top, equals, q, E, with, vector, on top, plus, q, v, with, vector, on top, times, B, with, vector, on top

\vec{F} = q\vec{v} \times \vec{B}

F =q v × B

F, with, vector, on top, equals, q, v, with, vector, on top, times, B, with, vector, on top

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 \thetaθtheta ( < 180^\circ<180∘is less than, 180, degree) between the velocity vector and the magnetic field vector:

\boxed{F = qvB\sin{\theta}}

F=qvBsinθ

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.

Sometimes we want to find the force on a wire carrying a current I 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 L we can write

qv = \frac{qL}{t}qv= t qL

q, v, equals, start fraction, q, L, divided by, t, end fraction

and since current is the amount of charge flowing per second,

qv = ILqv=ILq, v, equals, I, L

and therefore

\boxed{F = BIL\sin{\theta}}

F=BILsinθ

Force on a wire

Exercise 1a:

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~\mathrm{A}5 A to flow through the wire in the direction shown. If the magnetic field between the poles is known to be 0.2~\mathrm{T}0.2 T, what is the magnitude and direction of the force on the 10~\mathrm{mm}10 mm section of wire between the poles?

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.

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.

Exercise 2b:

If the electrons are known to be ejected from the electron gun horizontally at a speed vvv of 2\cdot 10^7~\mathrm{m/s}2⋅10

7

m/s, what is the strength of the magnetic field? Assume that the radius of the circular path can be approximated by L^2/2dL

2

/2dL, start superscript, 2, end superscript, slash, 2, d where LLL is the length of the tube and ddd is the horizontal deflection.

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