Question

A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (Fig. E27.24). The beam travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field?

Short answer

The magnitude of the magnetic field is$\mathbf{B}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{T}$

Step-by-step solution

Radius of curvature for a particle moving in a magnetic field

The radius of curvature for a particle moving in a magnetic field is given by

$\mathbf{R}\mathbf{=}\frac{\mathbf{mv}}{\mathbf{qB}}$

Determine the magnitude of the magnetic field

The beam travels a distance of 1.8 cm

The particle covered the angle quarter of the circle

Therefore,$\mathbf{\theta }\mathbf{=}\mathbf{\pi }\mathbf{/}\mathbf{2}$

Therefore, the radius of the path will be

$$\begin{gathered} \mathbf{R}\mathbf{=}\frac{\mathbf{s}}{\mathbf{\theta }} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{18}}{\mathbf{\pi }\mathbf{/}\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{751} \end{gathered}$$

Therefore, the radius of curvature for a particle moving in a magnetic field is

$\mathbf{R}\mathbf{=}\frac{\mathbf{mv}}{\mathbf{qB}}$

Or

$\mathbf{B}\mathbf{=}\frac{\mathbf{mv}}{\mathbf{qR}}$

Substitute the given values we get

$$\begin{gathered} \mathbf{B}\mathbf{=}\frac{\left(1.67\times {10}^{-27}\mathrm{kg}\right)\left(1200m/s\right)}{\left(1.60\times {10}^{-19}C\right)\left(0.00751m\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{T} \end{gathered}$$

Therefore, the magnitude of the magnitude of the magnetic field is$\mathbf{B}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{T}$