Question

An electron moving at 2.97×103 m/s in a 1.25 T magnetic field experiences a magnetic force of 1.40×10−16 N . What angle (in degrees) does the velocity of the electron make with the magnetic field? There are two possible answers. [Hint: Most calculators will only return the answer that falls in the first quadrant (0 - 90 degrees), but there is another angle in the second quadrant (90 - 180 degrees) that also satisfies the equation.]

Answer 1

To solve the problem it is necessary to take into account the concepts related to the Magnetic Force, which is given by,

$F= qvBsin\theta$

Where

F= Magnetic force

q= charge of proton

v= velocity

B Magnetic field

$\theta=$Angle between the velocity and the magnetic field.

Re-arrange the equation to find the angle we have,

$\theta = sin^{-1}(\frac{F}{qvB})$

Replacing our values we have,

$V= 2.97*10^3m/s\\B = 1.25T\\F = 1.4*10^{-16}N\\q = 1.6*10^{-19}C$

Then,

$\theta = sin^{-1}(\frac{1.4*10^{-16}}{(1.6*10^{-19})(1.25)(2.97*10^3)})\\\theta = 13.63\°$

The angle between 0 to 180 degrees would be,

$\theta' = 180-13.63\\\theta' = 166.36$

Therefore the two angles required are 13.63° and 166.36°