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.

$F_B = qv \times B$

$F_B$ = 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':

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