Question

Learning Goal: To apply Problem-SolvingStrategy20.1 Magnetic forces.

At a given instant, a particle with mass 5.00 x10^-3kg andcharge 3.5x10^-8 C has velocity ofmagnitude of2.0x10⁵m/s in the +ydirection. It is moving ina uniform magnetic field ofmagnitude 0.8T directed 40 degreesclockwise from the–x direction. What are the magnitude anddirectionof the magnetic force on the particle?

I found the angle between the velocity and magnitude ofthefield to be 50degrees, which is correct.

I calculated it as F=qvBsinθ:

3.5x10^-8*2.0x10⁵m/s*0.8T*sin50=4.29x10^-3

Now I need help with this part. I couldn'tfind any equationthat break B down into itscomponents.

Part B

To check your answer, you should calculate the magnitude oftheforce using the second formula: F=qvB_{perpendicular}.Whatis the magnitude of B_{perpendicular}?

IsB_perp=B_x?