Posted by
**Sar** on
.

A uniform magnetic field points north; its magnitude is 3.5 T. A proton with kinetic energy 6.0 × 10-13 J is moving vertically downward in this field. What is the magnetic force acting on it?

This is what I've done. The answer is incorrect. Please help me find out why.

3.5T to the north

KE=1/2MV^2

2(6.0X10^-13J)/1.673X10^-27 = V^2

V^2 = 7.17x10^14

sqrt (7.17e14) = 2.6e7

Thennn

(1.60e-19C)(2.6e7)(3.5T)(sin90degrees)

= 1.45e-11 to the east.

To the east is correct, but my numerical answer is not. Help?