Finding Velocity and angle for magnetic field

[kaikaii]

Homework Statement

A proton moves in a horizontal plane with a velocity v. It is found that when there is a magnetic field of 10(power - 2)T (magnitude) pointing from west to east, the magnetic force on the proton has a magnitude of 3.2 x 10(power of -16)N and points downwards. It is also found of that when a magnetic field of 10(power - 2)T pointing from south to north, the magnetic force on proton has a magnitude of 3.2 x 10(power - 16)N and points downward direction. Determine the magnitude and direction of the velocity.

I guess I'm required to find velocity and also the angle. Any ideas?

Homework Equations

F = qvB sin (theta)

The Attempt at a Solution

3.2 x 10(power - 16) = 1.6 x 10 (power - 19) * v * 10(power - 2) sin (theta)

I'm stuck here. i don't know how to continue on to find the velocity and also the theta.

Is there any other ways to find it?

 

[tiny-tim]

welcome to pf!

hi kaikaii! welcome to pf!

show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[kaikaii]

Hi! I have editted the attempt part. Do help solve the problem. Been pondering over it for sometime.

 

[tiny-tim]

hi kaikaii!

(have a theta: θ and try using the X² tag just above the Reply box )

3.2 x 10(power - 16) = 1.6 x 10 (power - 19) * v * 10(power - 2) sin (theta)

i really don't understand what you're doing

(and why aren't there two equations for the two different fields?)

please write it out properly​

 

[kaikaii]

hi kaikaii!

(have a theta: θ and try using the X² tag just above the Reply box )


i really don't understand what you're doing

(and why aren't there two equations for the two different fields?)

please write it out properly​

I subsitute the values into F = qvB sin (theta)...

I didnt sub two since those values are the same isn't it?