# Question #23e80

Sep 9, 2017

$\text{(i)}$ $6.24 \times {10}^{- 18}$ $\text{N}$

$\text{(ii)}$ $9.46 \times {10}^{8}$ ${\text{m s}}^{- 2}$

$\text{(iii)}$ The speed increases

#### Explanation:

$\text{1}$

$\text{(i)}$ The force due to a magnetic field can be calculated using the formula $F = q v B \sin \left(\theta\right)$; where $F$ is the force, $q$ is the charge of the particle, $v$ is the particle's velocity, $B$ is the magnetic field, and $\theta$ is the angle between the velocity vector and magnetic field vector:

$R i g h t a r r o w F = 3.2 \times {10}^{- 19} \times 550 \times 0.045 \times \sin \left({52}^{\circ}\right)$

$R i g h t a r r o w F = 3.2 \times {10}^{- 19} \times 24.75 \times 0.7880107536$

$R i g h t a r r o w F = 3.2 \times {10}^{- 19} \times 19.503266152$

$R i g h t a r r o w F = 6.241045169 \times {10}^{- 18}$

$\therefore F = 6.24 \times {10}^{- 18}$

Therefore, the force acting on the particle due to the field is around $6.24 \times {10}^{- 18}$ $\text{N}$.

$\text{(ii)}$ Now, we must find the acceleration of the particle due to the magnetic force acting on it.

Let's use the formula $F = m a$:

$R i g h t a r r o w = 6.24 \times {10}^{- 18} = 6.6 \times {10}^{- 27} \times a$

$R i g h t a r r o w 945 , 454 , 545.45 = a$

$\therefore a = 9.46 \times {10}^{8}$

Therefore, the acceleration on the particle due to $\vec{{F}_{B}}$ is around $9.46 \times {10}^{8}$ ${\text{m s}}^{- 2}$.

$\text{(iii)}$ As the particle accelerates, it increases in speed.