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)((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)((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°