Question

In an experiment with comic rays, vertical beam of particles that have charges of magnitude 3e and mass 12 times the proton mass enters a uniform horizontal magnetic field of and is bent in a semicircle of diameter 95.0 cm , as in figure (a) find the speed of the particles and the sign of their charge. (b) Is it reasonable to ignore the gravity force on the particles? (c) How does the speed of particles as they enter the field compare to their speed as they exit the field.

Short answer

(a) The speed of the particles is$2.84\times {10}^6\mathrm{m}/\mathrm{s}$ and the sign of their charge is negative.

(b) Yes, it is reasonable, on the particles gravitational force can be ignore.

(c) The speed of particles is constant.

Step-by-step solution

Definition of the magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

Determine the speed of the particles and the sign of their charge

force acting on it is

${\overrightarrow{\mathrm{F}}}_{\mathrm{B}}=\mathrm{q}\left(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}}\right)$

The magnetic field is perpendicular to the velocity so

${\overrightarrow{\mathrm{F}}}_{\mathrm{B}}=\left|\mathrm{q}\right|\mathrm{vB}$ …….(1)

according to Second law of motion the magnitude of the force is product of mass and acceleration

${\mathrm{F}}_{\mathrm{B}}=\mathrm{ma}$

And radial acceleration is $\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{r}}\mathrm{so}{\mathrm{F}}_{\mathrm{B}}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}$ ……

Now form (1) equation${\overrightarrow{\mathrm{F}}}_{\mathrm{B}}=\mathrm{evBsin}\left(\mathrm{\theta }\right)\mathrm{where}\mathrm{\theta }=90°\mathrm{So}{\mathrm{F}}_{\mathrm{B}}=\mathrm{evB}$

From (1) and (2)

$$\begin{gathered} \left|\mathrm{q}\right|\mathrm{vB}=\frac{{\mathrm{mv}}^2}{\mathrm{r}} \\ \mathrm{v}=\frac{\left|\mathrm{q}\right|\mathrm{vBR}}{\mathrm{m}} \\ \mathrm{v}=\frac{3\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(0.250\mathrm{T}\right)\left(0.475\mathrm{m}\right)}{12\left(1.67\times {10}^{-27}\mathrm{kg}\right)} \\ \mathrm{v}=2.84\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

According to the right-hand rule Multiply by negative sign to the direction of in order to the direction of force. The charge is negative.

Determine the Gravitational force on the particle can be ignore or not

First calculate magnetic force acting on the particle

$$\begin{gathered} {\mathrm{F}}_{\mathrm{B}}=\left|\mathrm{q}\right|\mathrm{vBsin}\left(\mathrm{\theta }\right) \\ {\mathrm{F}}_{\mathrm{B}}=3\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(2.84\times {10}^6\mathrm{m}/\mathrm{s}\right)\left(0.250\mathrm{T}\right)\sin \left(90°\right) \\ {\mathrm{F}}_{\mathrm{B}}=3.41\times {10}^{-13}\mathrm{N} \end{gathered}$$

And the gravitational force acting on the particle can be calculated as

$$\begin{gathered} {\mathrm{F}}_{\mathrm{g}}=\mathrm{mg} \\ {\mathrm{F}}_{\mathrm{g}}=12\left(1.67\times {10}^{-27}\mathrm{kg}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right) \\ {\mathrm{F}}_{\mathrm{g}}=1.96\times {10}^{-25}\mathrm{N} \end{gathered}$$

Conclude that${\mathrm{F}}_{\mathrm{B}}\gg {\mathrm{F}}_{\mathrm{g}}$

Hence, yes, it is reasonable, on the particles gravitational force can be ignore.

Determine the speed of particles

The angle between magnetic field and the path is $90°$.so it do-not changed magnitude but change velocity therefore speed of the particles is constant.

Hence the speed of particles is constant.