Question

A proton moves through a region containing a uniform electric field given by E with arrow = 54.0 ĵ V/m and a uniform magnetic field B with arrow = (0.200 î + 0.300 ĵ + 0.400 k) T. Determine the acceleration of the proton when it has a velocity v with arrow = 170 î m/s.

Answer 1

Answer:

acceleration is 9.58 × $10^{7}$ (- 14 ĵ + 51 k  ) m/s

Explanation:

given data

uniform electric field E  = 54.0 ĵ V/m

uniform magnetic field B = (0.200 î + 0.300 ĵ + 0.400 k) T

velocity v  = 170 î m/s.

to find out

acceleration

solution

we know magnetic force for proton is

i.e = e (velocity × uniform magnetic field)

magnetic force = e (170 î × (0.200 î + 0.300 ĵ + 0.400 k) )

magnetic force = e (- 68 ĵ + 51 k  )   ..................1

and now for electric force for proton i.e

= uniform electric field × e ĵ

electric force = e (54 ĵ )   ............2

so net force will be  add magnetic force + electric force

from equation 1 and 2

e (- 68 ĵ + 51 k  )  + e (54 ĵ )

e (- 14 ĵ + 51 k  )

so the acceleration (a)  for proton will be

net force = mass × acceleration

a = e (- 14 ĵ + 51 k  )  / 1.6 $10^{-19}$ / 1.67 × $10^{-27}$

acceleration = 9.58 × $10^{7}$ (- 14 ĵ + 51 k  ) m/s