1. Feb 7, 2012

### jhosamelly

1. The problem statement, all variables and given/known data

Find the magnitude and direction of the magnetic force on a charged particle with charge -4nC and velocity

$\vec{v}$ = $(2.5\times 10^{4}) \hat{i}$ + $(1.1 \times 10^{4}) \hat{j}$ (m/s)

if the magnetic field in the region is given by

$\vec{B}$ = $(1.2\times 10^{-3}) \hat{i}$ + $(5.6 \times 10^{-3}) \hat{j}$ - $(3.2 \times 10^{-3}) \hat{k}$ (T).

2. Relevant equations

$\vec{F}$ = q ($\vec{v}$ $\times$ $\vec{B}$)

3. The attempt at a solution

$\vec{F}$ = q ($\vec{v}$ $\times$ $\vec{B}$)

$\vec{F}$ = ($-4 \times 10^{-9} C$) $\left[\left((2.5\times 10^{4}) \hat{i} + (1.1 \times 10^{4}) \hat{j}\right) \times \left((1.2\times 10^{-3}) \hat{i} + (5.6 \times 10^{-3}) \hat{j} - (3.2 \times 10^{-3}) \hat{k} \right) \right]$

$\vec{F}$ = ($-4 \times 10^{-9} C$)$\left[\left[(1.1 \times 10^{4})(-3.2 \times 10^{-3})\right] \hat{i} - \left[(2.5 \times 10^{4})(-3.2 \times 10^{-3})\right] \hat{j}+ \left[(2.5 \times 10^{4})(5.6 \times 10^{-3}) - (1.1 \times 10^{4}) (1.2 \times 10^{-3})\right] \hat{k}\right]$

$\vec{F}$ = ($-4 \times 10^{-9} C$) ($-35.2 \hat{i} + 80 \hat{j} + 126.8 \hat{k}$)

$\vec{F}$ = $(1.4 \times 10^{-7}) \hat{i}$ - $(3.2 \times 10^{-7}) \hat{j} - (5.0 \times 10^{-7})\hat{k}$ (N)

Last edited: Feb 7, 2012
2. Feb 7, 2012

### tiny-tim

hi jhosamelly!

looks ok so far

now the question asks for the "magnitude and direction"

3. Feb 7, 2012

### jhosamelly

Yah.., that's what I was thinking ..

I know how to get the magnitude.. Its

$\left|\vec{F}\right|$ = $\sqrt{(A_x)^2 + (A_y)^2 + (A_z)^2 }$

but the question also asks for the direction.. which I think is the i-hat, j-hat, and k-hat..

so., I think my answer earlier is already the final answer.. If not. How can I find the direction?

4. Feb 7, 2012

### tiny-tim

hi jhosamelly!

yes, that's the magnitude

my guess is that they want the direction defined by the unit vector

5. Feb 7, 2012

### jhosamelly

hmmm.. meaning the question is asking if the force is pointing along the + or - , x- y- or z- axis.....?? so i have three directions?? + x-axis, - y-axis and - z-axis? This is what confused me... I know this is not possible.. so, what is the direction of the force?

6. Feb 7, 2012

### tiny-tim

nooo …

the is a unit vector in every direction

every vector r is|r| (the magnitude) times the unit vector r^

7. Feb 7, 2012

### jhosamelly

ow yah! Right! Thanks for reminding me

so I need to do

$\hat{F}$ = $\frac{\vec{F}}{\left|\vec{F}\right|}$

8. Feb 7, 2012

Yup!

9. Feb 7, 2012

### jhosamelly

$\left|\vec{F}\right|$ = $\sqrt{(A_x)^2 + (A_y)^2 + (A_z)^2 }$

$\left|\vec{F}\right|$ = $\sqrt{(1.4 \times 10^{-7})^2 + (-3.2 \times 10^{-7})^2 + (-5.0 \times 10^{-7})^2 }$
$\left|\vec{F}\right|$ = $6.18 \times 10^{-7}$

$\hat{F}$ = $\frac{\vec{F}}{\left|\vec{F}\right|}$

$\hat{F}$ = $\frac{(1.4 \times 10^{-7}) \hat{i} - (3.2 \times 10^{-7}) \hat{j} - (5.0 \times 10^{-7})\hat{k}}{6.18 \times 10^{-7}}$

$\hat{F}$ = $.23 \hat{i}$ - $.52 \hat{j} - .82 \hat{k}$

so, again I still have 3 components.. Hmmmm... What's the direction then?

10. Feb 7, 2012

### tiny-tim

d'oh!!

that unit vector is the direction!

11. Feb 7, 2012

### jhosamelly

Is that so?? hmmmm... I know that only happens if only one component is left?? O well, I was mistaken.. thanks for your help.

12. Feb 7, 2012