What Angle Gives a Magnetic Force of 1.8 Times the Original Force?

[kpangrace]

When a charged particle moves at an angle of 19° with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90°) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 1.8F?



can someone please tell me why doing a simple 19/f=x/1.8f cross-multiply and divide won't work?


it's my last problem on this dang homework!

 

[Tide]

Why should it work? You have no basis for writing that proportion.

You need use the Lorentz force equation for charged particles moving in a magnetic field. Does this look familiar: $\vec F = q \vec v \times \vec B$? Given $\vec v$ and $\vec B$ could you determine the magnitude of $\vec F$? How does it relate to the angle between the vectors?