Question

A proton, traveling with a velocity of 3.7 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 5.4 × 10-14 n and direction of due south. what are the magnitude and direction of the magnetic field causing the force? if the field is up, then enter a number greater than zero. if the field is down, then enter a number less than zero.

Answer 1

The magnetic force (Lorentz force) experienced by the proton in the magnetic field is given by
$F=qvBsin\theta=qvB$
since $\theta = 90^{\circ}$, because the velocity v and the force F in this problem are perpendicular, and so also the angle $\theta$ between the velocity and the magnetic field B should be $90^{\circ}$.

Let's find the magnitude of the magnetic field; this is given by
$B= \frac{F}{qv}= \frac{5.4\cdot 10^{-14}N}{1.6\cdot 10^{-19}C \cdot 3.7\cdot 10^6 m/s}=0.091 T$

To understand the direction, let's use the right-hand rule:
-index finger: velocity
- middle finger: magnetic field
- thumb: force

Since the velocity (index) points east and the force (thumb) points south, then the magnetic field (middle finger) points downwards. So we write:
B = -0.091 T