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 f…

Question

asked Aug 1, 2020 144k views

1 Answer

Ademar · Answer 1 · 2020-08-08T05:10:04+0000

Answer:

$8.1^(\circ), 171.9^(\circ)$

Step-by-step 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:

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

which gives the following two angles:

$\theta = 8.1^(\circ)\\\theta = 180^(\circ)-8.1^(\circ) = 171.9^(\circ)$