Suggested languages for you:

Americas

Europe

Q16 PE

Expert-verified
Found in: Page 811

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# What radius circular path does an electron travel if it moves at the same speed and in the same magnetic field as the proton in Exercise $${\rm{22}}{\rm{.13}}$$?

The radius of the circular path of the electron is obtained as: $${\rm{4}}{\rm{.36}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ m}}$$.

See the step by step solution

## Circular motion of moving charged particle in a magnetic field

In the presence of a magnetic field, a force acts on the moving charged particle. The direction of this force can be obtained using Fleming’s right-hand rule and can be numerically expressed as

$${\rm{F}} = {\rm{q}}\left( {\vec v \times \vec B} \right)$$

where$${\rm{q,}}\;{\rm{v,}}\;{\rm{and B}}$$ are given as the amount of charge, the velocity of the charged particle, and the strength of the magnetic field.

## Evaluating the radius of the circular path of the electron

To obtain the circular path of the electron.

The equation used is: $${\rm{r = }}\frac{{{\rm{mv}}}}{{{\rm{qB}}}}$$.

The value of $${\rm{m}}$$ is the mass of the electron.

The value of $${\rm{q}}$$ is the charge of the electron.

The value of $${\rm{e}}$$ and $${\rm{B}}$$ is the strength of the magnetic field obtained in the exercise $${\rm{22}}{\rm{.13}}$$, which is $${\rm{0}}{\rm{.98 T}}$$.

Now, solve the previous equation for the value of $${\rm{r}}$$ which will be the radius of the path.

\begin{aligned}{\rm{r}} &= \frac{{{\rm{mv}}}}{{{\rm{eB}}}}\\ &= \frac{{\left( {{\rm{9}}{\rm{.1 \times 1}}{{\rm{0}}^{{\rm{ - 31}}}}\;{\rm{kg}}} \right){\rm{ \times }}\left( {{\rm{7}}{\rm{.5 \times 1}}{{\rm{0}}^{\rm{7}}}{\rm{ m/s}}} \right)}}{{\left( {{\rm{1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}\;{\rm{C}}} \right) \times \left( {{\rm{0}}{\rm{.98 T}}} \right)}}\\ &= {\rm{4}}{\rm{.36}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ m}}\end{aligned}

Therefore, the radius of the circular path of the electron is: $${\rm{4}}{\rm{.36}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ m}}$$.