Question

An electron moving at 5.06 103 m/s in a 1.23 T magnetic field experiences a magnetic force of 1.40 10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers between 0° and 180°. (Enter your answers from smallest to largest.)

Answer 1

Answer:

$8.1^{\circ}, 171.9^{\circ}$

Explanation:

The magnitude of the magnetic force exerted on the moving electron is:

$F=qvB sin \theta$

where here we have

$F=1.40\cdot 10^{-16} N$ is the magnitude of the force

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

B = 1.23 T is the magnetic field intensity

$\theta$ is the angle between the direction of the electron's velocity and the magnetic field

Solving the equation for $\theta$, we find:

$sin \theta = \frac{F}{qvB}=\frac{1.40\cdot 10^{-16}N}{(1.6\cdot 10^{-19} C)(5.06\cdot 10^3 m/s)(1.23 T)}=0.141$

which gives the following two angles:

$\theta = 8.1^{\circ}\\\theta = 180^{\circ}-8.1^{\circ} = 171.9^{\circ}$