# What is the direction of the magnetic force on the electron? What is the magnitude of the magnetic force on the electron? What is the direction of the magnetic force on the proton? What is the magnitude of the magnetic force on the proton?

## A single electron is traveling south, with speed 575 km/s. The electron encounters a magnetic field of 190 mT going east, due to a large bar magnet. A single proton is traveling east, with speed 130 km/s. The proton encounters a magnetic field of 54.2 μT going north, due to a large bar magnet.

Apr 22, 2016

Faraday's left hand rule gives us the direction of the force on the particles, then $F = B q v$ gives the size of the force.

#### Explanation:

If we use Faraday's left hand rule on each particle, we can determine the effect of the magnetic field on the particle. The diagram shows the direction of the fields and the "current" so we can apply the rule.

If you are unsure about the rule, the link below will help
(https://socratic.org/questions/what-does-the-left-hand-rule-mean)

For the electron, the magnetic field will exert a force directed into the plane of the paper (or if travelling on the earth, this will be towards the earth).

For the proton, the magnetic field will exert a force out of the paper (or if travelling on the earth, this will be away from the earth).

The force on a particle of charge $q$ moving in a magnetic field, $B$, with a velocity $v$, is given by:

$F = B q v$
Both proton and electron have a charge of $1.6 \times {10}^{-} 19 C$

For the electron
$F = \left(190 \times {10}^{-} 3\right) \cdot \left(1.6 \times {10}^{-} 19\right) \cdot 575000 = 1.748 \times {10}^{-} 14 N$

For the proton
$F = \left(54.2 \times {10}^{-} 6\right) \cdot \left(1.6 \times {10}^{-} 19\right) \cdot 130000 = 1.127 \times {10}^{-} 18 N$