Angle between magnetic field B and velocity of charge

[ivanwho49]

Homework Statement

Find magnitude and direction of F of a charge moving through a magnetic field.

velocity: 5 m/s
charge: 2 C
B: 3 T (pointing out of the page when drawn)

Homework Equations

(vector)F=qv x B
(magnitude)F=qvBsinθ

The Attempt at a Solution

I know to use the equation for magnitude but I'm confused as to what the angle would be between v and B. Since the magnetic field is pointing out of the page, would the angle be 90°, making sin(90)=1?

Also, don't know what direction that would be. Since I don't know the direction of B, I'm not sure how to use the right hand rule to figure this out.

 

[rude man]

You do know the direction of B. It's out of the page.

Unfortunately, the problem does not define the direction of v so you can't solve it.

 

[ivanwho49]

Sorry, forgot to mention that: the velocity moves straight from left to right.

 

[rude man]

Sorry, forgot to mention that: the velocity moves straight from left to right.

OK, so place a sheet of paper on the table, draw an arrow from left to right = v, then hold a pencil at the back end of the arrow pointing straight up (B).
What is the angle between the arrow and the pencil?

 

[HalfLostAndSearching]

I would suggest using the right hand rule to determine the direction of the force. Place your right hand so that your fingers point in the direction of the velocity and your thumb points in the direction of the magnetic field. The direction your palm faces is the direction of the force. In this case, the force would be perpendicular to both the velocity and the magnetic field, so the angle between them would indeed be 90°. Therefore, the magnitude of the force would be qvBsin(90) = 2 C * 5 m/s * 3 T * 1 = 30 N. The direction of the force would depend on the direction of the magnetic field, which is not specified in the problem. If the magnetic field is pointing towards the left, for example, then the force would be pointing downwards according to the right hand rule.