Question

A proton moving with a speed of 200 m/s enters perpendicular to the direction of the magnetic field and experiences a force of 2 * 10 ^ -18 N. What is the magnitude of the magnetic field?

Answer 1

Answer:

0.0625 T

Explanation:

When a charged particle is moving in a region with a magnetic field, the particle experiences a force given by the equation

$F=qvB sin \theta$

where

q is the charge of the particle

v is its velocity

B is the magnitude of the magnetic field

$\theta$ is the angle between the direction of the velocity and of the field

For the proton in this problem, we have:

$F=2.0\cdot 10^{-18}N$ is the force experienced

$q=1.6\cdot 10^{-19}C$ is the charge of the proton

$v=200 m/s$ is its velocity

$\theta=90^{\circ}$, since the proton enters the region perpendicular to the magnetic field

Therefore, we can re-arrange the equation to find B:

$B=\frac{F}{qvsin \theta}=\frac{2\cdot 10^{-18}}{(1.6\cdot 10^{-19})(200)(sin 90^{\circ})}=0.0625 T$