A uniform magnetic field points north...?

Question

A uniform magnetic filed pints north; its magnitude is 1.5 T. A proton with kinetic energy 8.0x10^-13 J is moving vertically downward in this field. What is the magnetic force acting on it?

Answer: 7.4x10^-12 N east

But that's not what I got. Here's what i did:

B = 1.5 T; K_E = 8.0x10^-13 J; mass of proton, m_p = 1.67x10^-27 kg

magnetic force, F_m = ?

(1/2) x m_p x v^2 = K_E

v^2 = (2K_E)/m_p

v = [(2K_E)/m]^1/2

= [2(8.0x10^-13)/(1.67x10^-27)]^1/2

= 3.095x10^7 m/s

F_m = qvB

F_m = (1.6x10^-19)(3.095x10^7)(1.5)

F_m = 2.3x10^-4 N

Where did I go wrong?

Answer 1

You did everything right.... maybe you just made a mistake with the calculator.  Check the numbers again.

UPDATE:  I checked with my calculator.... your mistake is just in the last step, in multiplying qvB ... plug it into your calculator again.