a)A proton enters a uniform field B in a perpendicular direction. The proton is

Question

a)A proton enters a uniform field B in a perpendicular direction. The proton is moving out of the page with a speed of 1.00 x 10^7 m/s when it enters the magnetic field. After entering the magnetic field the proton experiences an acceleration of 2.00 x 10^13 m/s^2 in +x direction. Determine the magnitude and direction of the magnetic field B

a)A proton enters a uniform field B in a perpendicular direction. The proton is moving out of the page with a speed of 1.00 x 10^7 m/s when it enters the magnetic field. After entering the magnetic field the proton experiences an acceleration of 2.00 x 10^13 m/s^2 in +x direction. Determine the magnitude and direction of the magnetic field B b) Find the direction of the magnetic field acting on an electron moving through two situations shown in the figure below. (Note: The directions of velocity and force are shown in each case.)

Explanation / Answer

a) F = q v B

F = m a

m a = q v B

1.67E-27*2.0E13 = 1.6E-19*1.0E7*B

B=0.0209 T

so to get direction v into B gives F

soout of the page and -y gives +x

so B is -y direction

b)

+x into -z gives + y

so direction is -z

ii)

-y

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