Question

A proton, traveling with a velocity of 6.8 × 10⁶ m/s due east, experiences a maximum magnetic force of 8.0 × 10^-14 N. The direction of the force is straight down, toward the surface of the earth. What is the magnitude and direction of the magnetic field B, assumed perpendicular to the motion?

Answer 1

We can find the magnitude of the magnetic field as follows:

$$F_B=qvBsin\theta,$$

$B=\dfrac{F_B}{qvsin\theta}=\dfrac{8.0\cdot10^{-14}\ N}{1.6\cdot10^{-19}\ C\cdot6.8\cdot10^6\ \dfrac{m}{s}\cdot sin90^{\circ}}=0.073\ T.$

According to the right-hand rule, the magnetic field directed upward.