Question

PLEASE HELP ME!!!! ANSWER BOTH PARTS PLS!!! THANK YOU

1. A charged particle moves with velocity 5 km/s making angle 50° with the external magnetic field of magnitude of 0.6 T (see figure below). The field exerts the 3.5 N force on the charge. The direction of the force is out of the page/screen. What is the magnitude and sign of the charge?

Х 50° B

The magnetic force would be zero if:

1. v was parallel to the field;
2. v was directed out of page;
3. v was directed into the page
4. v=0;
5. A and D;
6. B and C.

Answer 1

Fm = q (DXB) Tx B is into the page but Fm is ma quB sino out of page so charge is negative. q- Fm 3.5 1 VB sino 5000 80.6 x sinsoo = 0.0015229 c magnetic force Fm= 2VB sino Fm=0 if o=0 parallel Vand B so Answer is Ag and D.