What is the magnetic force (in newtons) on a particle traveling in a 1.5 t magnetic field if q = 7.5 microcoulombs and v = 1.75 × 106 m/s at a 45° angle to the magnetic field? answer with three signficant digits.

Given:

B(Magnetic field): 1.5 T                                                                                

q=  7.5 microcoulombs                                                                                

v= 1.75 x 10 ∧6 m/s                                                                                    

The angle ∅ between B and v is 45 °.                                                    

Now we know that F= qvB sin ∅

Substituting these values we get:

F= 7.5 x 10∧-6 x 1.75 x 10∧6 x 1.5 x sin 45                                                

 F= 16.752 N

answer: (431,2)

step-by-step explanation:

i already answered it by 34/s