Question

(a) Show that the period of the circular orbit of a charged particle moving perpendicularly to a uniform magnetic field is \({\rm{T = 2\pi m/(qB)}}\) (b) What is the frequency \({\rm{f}}\)? (c) What is the angular velocity\({\rm{\omega }}\) ? Note that these results are independent of the velocity and radius of the orbit and, hence, of the energy of the particle. \(\left( {{\rm{Figure 22}}{\rm{.64}}{\rm{.}}} \right)\)

Short answer

1. velocity = \(\frac{{{\rm{2\pi m}}}}{{{\rm{qB}}}}\).
2. Frequency =\(\frac{{{\rm{qB}}}}{{{\rm{2\pi m}}}}\).
3. Relationship is \(\frac{{{\rm{qB}}}}{{\rm{m}}}\).

Step-by-step solution

Define magnetic field

The magnetic impact of electric charges in relative motion and magnetized objects is described by a magnetic field, which is a vector field.

Evaluating the velocity

(a)

Given that the charged particle will move in a circular motion, the following equation may be used to calculate its velocity in terms of radius\({\rm{r}}\)and period\({\rm{T}}\):\({\rm{v = }}\frac{{{\rm{2\pi r}}}}{{\rm{T}}}\).In addition, the radius of a circular route is linked to the magnetic field by the formula:\({\rm{r = }}\frac{{{\rm{mv}}}}{{{\rm{qB}}}}\).To get the period\({\rm{T}}\), use these two equations.

\({\rm{T = }}\frac{{{\rm{2\pi }}}}{{\rm{v}}}{\rm{r}}\)

\({\rm{ = }}\frac{{{\rm{2\pi }}}}{{\rm{v}}}{\rm{ \times }}\frac{{{\rm{mv}}}}{{{\rm{qB}}}}\)

\({\rm{ = }}\frac{{{\rm{2\pi m}}}}{{{\rm{qB}}}}\)

Therefore, velocity = \(\frac{{{\rm{2\pi m}}}}{{{\rm{qB}}}}\).

Evaluating the frequency of periodic motion

(b)

The number of complete revolutions a particle does per unit time is the frequency\({\rm{ f}}\)of a periodic motion. The frequency is proportional to the period\({\rm{T}}\)according to the following formula:\({\rm{f = }}\frac{{\rm{1}}}{{\rm{T}}}\)

\({\rm{f = }}\frac{{\rm{1}}}{{\rm{T}}}\)

\({\rm{ = }}\frac{{{\rm{qB}}}}{{{\rm{2\pi m}}}}\)

Therefore, frequency = \(\frac{{{\rm{qB}}}}{{{\rm{2\pi m}}}}\).

Relation between the angular frequency and frequency

(c)

The relationship between the angular frequency\({\rm{\omega }}\)and the frequency\({\rm{f}}\)is:\({\rm{\omega = 2\pi f}}\).

\({\rm{\omega = 2\pi f}}\)

\({\rm{ = }}\frac{{{\rm{qB}}}}{{\rm{m}}}\)

Therefore, the relationship is \(\frac{{{\rm{qB}}}}{{\rm{m}}}\).