Question

What is the direction of the magnetic force on the proton? What is the magnitude of the magnetic force on the proton? What is the direction of the magnetic force on the second proton? What is the magnitude of the magnetic force on the second proton?

A proton is traveling upward, with speed 9.0 x 10^5 m/s. The proton encounters a magnetic field of 0.3 T going west, due to a large bar magnet. A second proton is traveling north, with speed 8.0 x 10^5 m/s. The proton encounters a magnetic field of 0.3 T going east, due to a large bar magnet.

Answer 1

First proton: `=4.3254xx10^(−14)N` south.
Second proton: `=3.8448xx10^(−14)N` downwards.

Explanation:

Lorentz force equation is

`vecF=q[ vecE+vecvxxvecB]`

In the problem, `q=1.602xx10^(−19) C` charge of a proton.
`vecE=0, vecv=9.0 xx 10^5 m//s,` upwards; assumed coming out of the plane of the paper. `vecB=0.3T` going west

`vecF_"magnetic"=q[vecvxxvecB]`
`|vecF|_"magnetic"=1.602xx10^(−19)[9.0 xx 10^5 xx0.3]`
`=4.3254xx10^(−14)N`

Direction is given by cross product of the velocity and magnetic field vectors (right hand rule) and is due south.

Second proton is travelling North with speed of `8.0 xx 10^5 m//s.` And the magnetic field is `0.3T` going east.
In this case

`|vecF|_"magnetic"=1.602xx10^(−19)[8.0 xx 10^5 xx0.3]`
`=3.8448xx10^(−14)N` downwards, into the plane of paper.