Question

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?

Answer 1

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`