1
$\begingroup$

I know that when a charge is moving through a magnetic field with all or some of its velocity being perpendicular to the field direction, it is influenced by the field. The Lorentz force - which is perpendicular to the field direction and the velocity of the charge - can be calculated with the following equation:

$$F=qBv \text{ (perpendicular force)}$$

with $v$ (perpendicular) being perpendicular to the field direction.

If the Lorentz force is a centripetal force, then it is caused by a homogeneous field. The force is constant because $q B$ and $v$ (perpendicular) are constant.

How can you prove that the Lorentz force is changing the direction of the velocity of the charge so that $v$ (perpendicular) stays the same?

$\endgroup$

2 Answers 2

1
$\begingroup$

Because the work done to the particle by the Lorentz force is zero.

The work is directly related to the change of kinetic energy and the speed: $$ W = \int \vec{F} \cdot d \vec{x} = m \int \vec{v}\cdot \frac{d\vec{x}}{dt} dt = \int \frac{1}{2} m v^2 dt. $$

Therefore, if the work is zero, then the speed will keep constant in time.

Examine the Lorentz force $\vec{F} = q \vec{v}\times\vec{B}$, work done is:

$$ W = q\int \vec{v}\times\vec{B} \cdot d \vec{x} = q\int \vec{v}\times\vec{B} \cdot \frac{d \vec{x}}{dt} dt = q\int \vec{v}\times\vec{B} \cdot \vec{v} dt . $$

Since the term $\vec{v}\times\vec{B}$ is perpendicuar to $\vec{v}$, the integrand is zero. This leads to a constant speed.

$\endgroup$
2
  • $\begingroup$ Ok, thanks. But apart from the work done, the position of the charge and the direction of its velocity changes, so how do we make sure that the share of the charges velocity that is vertical to the field direction stays the same? $\endgroup$
    – Patrick
    Jan 31, 2021 at 11:32
  • $\begingroup$ @Patrick By solving the equation of motion, $\vec{F}= m d\vec{v} / dt$. The solution renders a constant velocity in the $z$-direction (assume $\vec{B} = B \hat{z}$), and circular motion in $$xy$-plane. $\endgroup$
    – ytlu
    Jan 31, 2021 at 19:06
0
$\begingroup$

Since the magnetic field does not contribute any energy to the deflection of the charge, only the kinetic energy of the charge remains. And indeed, the velocity decreases, accompanied by EM radiation from the charge. This is not surprising because any charge radiates under acceleration (and a circular motion is an acceleration).

How can you prove that the Lorentz force is changing the direction of the velocity of the charge so that $v$ (perpendicular) stays the same?

You prove it by experiment and it turns out that the velocity changes from the initial value to zero, the circular path is a spiral path and the kinetic energy is used up by the emission of photons. The equation of the Lorentz force does not reflect this. Like any physical law, it only works to a certain approximation.

And by the way, the deflection of a charge in the magnetic field is stated but not explained. Sometimes it is sufficient to be satisfied with a certain level of abstraction. Anything beyond that is scientific curiosity, which seems to be marginal for this question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.