Aproton travels with a speed of 4.90 106 m/s at an angle of 61° with the direction of a magnetic field of magnitude 0.240 t in the positive x-direction. (a) what is the magnitude of the magnetic force on the proton? n (b) what is the proton's acceleration? m/s2

Our values are,

B = 0.240 T

θ= 61°

Where

V = Proton travels with a speed

B=Magnitude of magnetic field

θ=Angle

q=charge of proton

a)

Magnitude of the magnetic Force is given by,

b ) Protons Accelerationis is given from Second Newton's law,

a=d/t

a="triangle"v/t

explanation:

a = acceleration

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