OpenStudy (snitchseeker1496):

A proton enters a magnetic field of lux density 1.5 W b/m^(2) with a velocity if 2 x 10 ^(7) m/s at an angle of 30 degrees Celsius with the field. Compute the force on the proton.

OpenStudy (irishboy123):

you know this $${\mathbf {F}}=q{\mathbf {v}}\times {\mathbf {B}} \, \hat n$$ ??? for the magnitude $$|\mathbf {F}| =q{ |\mathbf {v}} | \, |{\mathbf {B}}| \sin \theta$$ $$q = 1.60217.....×10^{−19} C$$ make sure you get $$\theta$$ right way round. use the right hand rule if there is a drawing...