Movement of a Charge in a Magnetic Field

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Key facts

The force acting on a charge $q$ traveling at a speed $v$ in a magnetic field $B$ is given by:

$$F = q v B$$
(1)

If a charged particle of mass $m$ travels with a speed $v$ at right angles relative to a magnetic field $B$, then the charge will move in a circle of radius $r$ given by:

$$r = \frac{mv}{qB}$$
(2)

An analysis of the movement of an electric charge in a magnetic field, also discussing the particular case when the charge travels at right angles relative to the field

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Physics Magnetism

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In order to define the force acting on an electric charge placed in a magnetic field, consider a charge $q$ traveling at a speed $v$ in a magnetic field $B$. Also, assume that in a time $dt$ the charge traveled the distance $dl$, as highlighted in Figure 1.

Figure 1
Figure 1

We can thus write that:

$$v = \frac{dl}{dt}$$
(3)

from which:

$$dt = \frac{dl}{v}$$
(4)

Also, we know that the electric current $i$ can be calculated as:

$$i = \frac{q}{t}$$
(5)

where $q$ is the electric charge transferred during a time $t$. Using the expression of time from (4), equation (5) becomes:

$$i = \frac{qv}{dl}$$
(6)

The force which acts on an electric charge $q$ responsible for generating a current $i$, and which is placed in a magnetic field $B$, can be written as:

$$F = i \cdot dl \cdot B$$
(7)

By using the expression form of the electric current from (6) in (7), we get that:

$$F = \frac{qv}{dl} \cdot dl \cdot B$$
(8)

which leads to:

$$F = q v B$$
(9)

This force acts in the same direction as that of the lines of force of the magnetic field $B$. If the charge travels at right angles relative to the magnetic field, as for example in Figure 1, then the force will act perpendicular to the direction of the charge's movement, and thus the charge will move in a circle (see Figure 2).

Figure 2
Figure 2

As in this case the force $F=qvB$ is oriented inwards, it is equivalent to the centripetal force $F_c$ given by:

$$F_c = \frac{mv^2}{r}$$
(10)

where $m$ is the mass of the charged particle, and $r$ is the radius of the circle. Thus, we get that:

\calc{(m*v)/(q*B)} "Instant calculator eq(10)"

$$\frac{mv^2}{r} = qvB$$
(11)

from which we obtain the radius of the circle on which the charge moves:

$$r = \frac{mv}{qB}$$
(12)

Reference