A proton in a parallel accelerator has a speed of 5.0 x 10^6 m/s going downward.The proton encouters a magnetic field whose magnitude is 0.40 T and whose direction makes angle of 30.0° with respect to the protons velocity. See details for question?

Find the magnitude and direction of the magnetic force on the proton?

1 Answer
Jun 1, 2018

Given velocity of proton vecv=-5.0 xx 10^6 hatj\ ms^-1
Magnetic field whose direction makes angle of 30.0^@ with respect to the proton's velocity.
Magnetic field has components along the aexs as vecB=(0.40\ sin 30^@hati-0.40\ cos30^@hatj)\ T

From Lorentz force equation for a particle having charge q, we have

vecF=q(vecE+vecvxxvecB)

Since there is no electric field, only force on proton having charge 1.60×10^-19\ C is due to magnetic field which is given as

vecF_B=1.60×10^-19[(-5.0 xx 10^6 hatj)xx(0.40\ sin 30^@hati-0.40\ cos30^@hatj)]
=>vecF_B=1.60×10^-19[(-5.0 xx 10^6 hatj)xx(0.40\ sin 30^@hati)]

(because hatjxxhatj=0and hatjxxhati=-hatk)
=>vecF_B=(1.60×10^-19xx5.0 xx 10^6 xx0.40\ sin 30^@)hatk
=>vecF_B=1.60×10^-13hatk\ N