Computing for magnetic force. Please check my answer.

[jhosamelly]

Homework Statement

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).

Homework Equations

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

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)

Is this already the answer? Am I correct? Thanks in advance.

 

[tiny-tim]

hi jhosamelly!

looks ok so far

now the question asks for the "magnitude and direction" ​

 

[jhosamelly]

hi jhosamelly!

looks ok so far

now the question asks for the "magnitude and direction" ​

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?

 

[tiny-tim]

hi jhosamelly!

yes, that's the magnitude

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

 

[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?

 

[tiny-tim]

hmmm.. meaning the question is asking if the force is pointing along the + or - , x- y- or z- axis...??

nooo …

the is a unit vector in every direction

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

 

[jhosamelly]

nooo …

the is a unit vector in every direction

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

ow yah! Right! Thanks for reminding me

so I need to do

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

 

[tiny-tim]

Yup!

 

[jhosamelly]

Yup!

I already did it.

\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?

 

[tiny-tim]

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

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

d'oh!

that unit vector is the direction! ​

 

[jhosamelly]

d'oh!

that unit vector is the direction! ​

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

 

[tiny-tim]

perhaps i should add …

i j and k are the basis unit vectors​

 

[jhosamelly]

perhaps i should add …

i j and k are the basis unit vectors​

I see.. Big thanks So the final answer is that unit vector and the magnitude. I see. Thanks ))))))