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 pependicurlar to its original direction (Fig. E27.24). The beam travels a distance of 1.10 cm while in the field. What is the magnitude of the magnetic field?

Short answer

$B=1.79\times {10}^{-3}\mathrm{T}$

Step-by-step solution

27.24

Recall that the force on a particle in a magnetic field is given by

$$\begin{gathered} \stackrel{\rightharpoonup }{F}=q\stackrel{\rightharpoonup }{v}\times \stackrel{\rightharpoonup }{B} \\ F=\left|q\right|vB\sin \left(\varphi \right) \end{gathered}$$

Where $q$is the charge of the particle, $v$is the particles velocity, $B$is the magnetic field, and $\varphi$is the angle between the velocity vector and the magnetic field.

In the magnetic field, our particles experience uniform circular motion, so

$F=m{a}_c$

Recall that the angular acceleration is given by

$a_c=\frac{v^2}{r}$

Thus we have that

role="math" localid="1650556775424" $$\begin{gathered} F=\left|q\right|vB\sin \left(\varphi \right) \\ F=qvB \\ m{a}_c=qvB \\ \frac{m{v}^2}{r}=qvB \\ B=\frac{mv}{rq} \end{gathered}$$

Now we just need to find the radius. The distance, $d$, is a fourth of the circumference, $C$, of a circle, so

$$\begin{gathered} d=\frac{1}{4}C \\ d=\frac{1}{4}2\pi r \\ d=\frac{\pi r}{2} \\ r=\frac{2d}{\pi } \end{gathered}$$

Now We plug this into our above equation and solve.

role="math" localid="1650557707264" $$\begin{gathered} B=\frac{mv}{rq} \\ B=\frac{mv}{\left({\displaystyle \frac{2d}{\pi }}\right)q} \\ B=\frac{\pi mv}{2qd} \\ B=\frac{\pi ·(1.67\times {10}^{-27}\mathrm{kg})·(1.20\times {10}^3\mathrm{m}/\mathrm{s})}{2·(1.60\times {10}^{-19}\mathrm{C})(1.10\times {10}^{-2}\mathrm{m})} \\ B=1.79\times {10}^{-3}\mathrm{T} \end{gathered}$$