91Ó°ÊÓ

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

force acting on it is

${\stackrel{→}{\mathrm{F}}}_{\mathrm{B}}=\mathrm{q}\left(\stackrel{→}{\mathrm{v}}×\stackrel{→}{\mathrm{B}}\right)$

The magnetic field is perpendicular to the velocity so

${\stackrel{→}{\mathrm{F}}}_{\mathrm{B}}=\left|\mathrm{q}\right|\mathrm{vB}$ …â¶Ä¦.(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 $\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{r}}\mathrm{so}{\mathrm{F}}_{\mathrm{B}}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}$ …â¶Ä¦

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

From (1) and (2)

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

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.

First calculate magnetic force acting on the particle

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

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

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

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

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

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.