Question

A proton travels at a speed of 2.0 × 106 meters/second. Its velocity is at right angles with a magnetic field of strength 5.5 × 10-3 tesla. What is the magnitude of the magnetic force on the proton?

Answer 1

So here we are given that the the velocity of the proton ( V ) is 2.0 × $10^6$ meters / second, with a magnetic field of strength 5.5 × $10^{-3}$  tesla. If they each form a right angle, they are hence perpendicular to one another, such that ....

F = q( V × B ),

F = q v B( sin ∅ ),

F = q v B( sin( 90 ) )

.... they form the following formula. Let's go through each of the variables in our formula here -

{ F = Magnetic Force ( which has to be calculated ), q = charge of proton (has charge of 1.602 × $10^{19}$ coulombs ), B = magnetic field }

All we have to do now is plug and chug,

F = ( 1.602 × $10^{19}$ )( 2.0 × $10^6$ )( 5.5 × $10^{-3}$ ) = ( About ) 1.8 × $10^{-15}$ Newtons