Question

an alpha particle travels at a velocity of magnitude 700 m/s through a uniform magnetic field of magnitude 0.039 t. (an alpha particle has a charge of charge of +3.2×10^(-19) c and a mass 6.6×10^(-27) kg.) the angle between the particles direction of motion and the magnetic field is 70.0°. what is the magnitude of (a) the force acting on the particle due to the field, and (b) the acceleration of the particle due to this force? (c) does the speed of the particle increase, decrease, or remain the same?

Explanation:

Step1: Recall magnetic - force formula

The magnetic - force formula is $F = qvB\sin\theta$. Given $q = 3.2\times10^{- 19}\ C$, $v = 700\ m/s$, $B=0.039\ T$, and $\theta = 70^{\circ}$.

Step2: Calculate the force

$F=(3.2\times10^{-19}\ C)\times(700\ m/s)\times(0.039\ T)\times\sin(70^{\circ})$.
First, $\sin(70^{\circ})\approx0.9397$. Then $F=(3.2\times10^{-19})\times(700)\times(0.039)\times0.9397$.
$F=(3.2\times10^{-19})\times(700)\times0.0366483$.
$F=(3.2\times10^{-19})\times25.65381$.
$F = 8.2092192\times10^{-18}\ N$.

Step3: Recall the formula for acceleration

The force $F = ma$, so $a=\frac{F}{m}$. Given $m = 6.6\times10^{-27}\ kg$ and $F = 8.2092192\times10^{-18}\ N$.
$a=\frac{8.2092192\times10^{-18}\ N}{6.6\times10^{-27}\ kg}\approx1.243821\times10^{9}\ m/s^{2}$.

Step4: Analyze the effect on speed

The magnetic force is always perpendicular to the velocity of the charged particle. According to the work - energy theorem, $W=\Delta K$. Since the magnetic force does no work ($W = 0$ as $F\perp v$), the kinetic energy $K=\frac{1}{2}mv^{2}$ and the speed $v$ of the particle remain the same.

Answer:

(a) $8.21\times10^{-18}\ N$
(b) $1.24\times10^{9}\ m/s^{2}$
(c) Remain the same