An alpha particle travels at a velocity$\overrightarrow{\mathbf{v}}$of magnitude$\mathbf{\text{550}}$m/s through a uniform magnetic field of magnitude$\overrightarrow{\mathbf{B}}$=0.045T. (An alpha particle has a charge of${\mathbf{\text{+3.2×10}}}^{\mathbf{\text{-19}}}$C and a mass ofkg.) The angle between$\overrightarrow{v}$and$\overrightarrow{\mathit{B}}$is 52°. (a)What is the magnitude of the force$\overrightarrow{{\mathbf{F}}_B}$acting on the particle due to the field? (b)What is the acceleration of the particle due to$\overrightarrow{{\mathbf{F}}_B}$?

(c)Does the speed of the particle increase, decrease, or remain the same?

The magnitude of alpha particle’s speed,$v=\text{550m/s}$

The angle between velocity and magnetic field,$f=52°$

The magnetic field on the particle,$B=\text{0.045T}$

Charge of alpha particle,$q={\text{+3.2×10}}^{\text{-19}}\text{C}$

Mass of alpha particle,$m{\text{=6.6×10}}^{\text{-27}}\text{\hspace{0.17em}kg}$

Findthe force$\overrightarrow{{\mathbf{F}}_B}$acting on the particle due to the fieldusing the formula for magnetic force. Then, equating this force with force according to Newton’s second law, get theacceleration of the particle due to. $\overrightarrow{\mathbf{}{\mathbf{F}}_B}$Lastly, the speed of the particle can be predicted from work done by$\overrightarrow{\mathbf{}{\mathbf{F}}_B}$.

Newton's second law states that the rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.

Formulae are as follows:

$$\begin{gathered} F_B=qvBsinf \\ {F}_{net}=ma \end{gathered}$$

Where Fis force, v is velocity, m is mass, a is acceleration, B is a magnetic field, and q is the charge of the particle.

The force experienced by proton due to the magnetic field is-

$F_B=qvBsinf$

For the given values, the equation becomes-

$F_B=(+3.2\times {10}^{-19}\text{\hspace{0.17em}C})(550\text{\hspace{0.17em}m}/\text{s})\left(0.045\text{\hspace{0.17em}T}\right)\text{sin}(52°)$

$F_B={\text{6.2×10}}^{\text{-18}}\text{N}$

Therefore, the force $\overrightarrow{F_B}$acting on the particle because of the field is${\text{6.2×10}}^{\text{-18}}\text{N.}$

According to Newton’s second law,

$F_{net}=ma$

In this case,

$F_{net}=F_B$

Hence,

$$\begin{gathered} a=\frac{62.41\times {10}^{-19}\text{\hspace{0.17em}N}}{6.6\times {10}^{-27}\text{\hspace{0.17em}kg}} \\ a={\text{9.5×10}}^{\text{8}}{\text{m/s}}^{\text{2}} \end{gathered}$$

Therefore, the acceleration of the particle due to.$\overrightarrow{F_B}{\text{is9.5×10}}^{\text{8}}{\text{m/s}}^{\text{2}}$

We know that speed is perpendicular to$\overrightarrow{F_B}$. Hence, the work done by$F_B$is zero.

Hence, the speed of the particle remains the same.