Question

A proton travels at a velocity of 3.0x106 m/s in a direction perpendicular to a uniform magnetic field.

A proton entering a magnetic field

If the proton experiences a magnetic force of 1.15x10-13 N, what is the magnitude of the magnetic field?

Answer 1

Hi there!

Recall the equation for magnetic force on a moving particle.

`F_B = qv * B`

`F_B` = Magnetic Force (N)
q = Charge of particle (C)

v = Velocity of particle (m/s)
B = Magnetic Field Strength (T)

**This is a cross product, so the equation can be written as
`F_B` = qvBsinφ where φ is the angle between the velocity vector and magnetic field vector.

Since the proton is traveling perpendicular to the field, we can disregard the cross product. (sin90 = 1.)

Rearrange the equation to solve for 'B':

`B = (F_B)/(qv)\\\\B = (1.15 * 10^(-13))/((1.6 * 10^(-19))(3.0 * 10^6)) = \boxed{0.24 T}`