Find the charge of this particle moving in a magnetic field

[fight_club_alum]

Homework Statement

A particle (mass 6.0 mg) moves with a speed of 4.0 km/s and a direction that makes an
angle of 37° above the positive x axis in the xy plane. A magnetic field of (5.0i) mT
produced an acceleration of (8.0k) m/s2. What is the charge of the particle?

Relevant Equations

F = ma
F = QBxV = Q BV sin(theta)

F = ma
F = (6x10^-6) * 8
F = 4.8 * 10^-5
F = QBVsin(theta)
F/(BVsin(theta) = Q
(4.8 x 10^-5) / (5 x 10^-3) (4000) (sin(37)) = 3.98 x 10^-6 ~ 4 uc <---- THE RIGHT ANSWER IS -4 uc

 

[kuruman]

What does the right hand rule say about the direction of the force (and hence the acceleration) when the velocity and field are in the given directions? Specifically, in what direction is $\vec v \times \vec B~$? By the the way, $\vec F=q\vec v \times \vec B~$, not what you have.

 

[fight_club_alum]

I think I understand, now.
Thank you so much, but why didn't the question say an acceleration of -8k m/s^2 or acceleration of magnitude 8

 

[kuruman]

... but why didn't the question say an acceleration of -8k m/s^2 or acceleration of magnitude 8

If you ask this question, your understanding needs to become clearer. The acceleration is given as $\vec a= 8.0~ \mbox{(m/s)}\hat k$. The velocity is given as $\vec v=4.0 ~\mbox{(km/s)}[\cos(37^o)~\hat i+\sin(37^o)~\hat j]$ and the magnetic field is given as $\vec B =5.0 \mbox{(mT)}~\hat i$.
This is what you need to do
1. Express the vector $q\vec v \times \vec B$ in unit vector notation.
2. Ask yourself, for what sign of $q$ will the direction of this vector be in the same direction as the given acceleration?