An electron moves with a speed of 8.0 × 10^{6} m/s along the +x-axis. It enters a region where there is a magnetic field of 2.5 T, directed at an angle of 60° to the +x-axis and lying in the xy-plane. What is the magnitude of the magnetic force of the electron?

The magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N

Explanation:

Given;

speed of the electron, v =  8.0 × 10⁶ m/s

magnetic field strength, B = 2.5 T

angle of inclination of the field, θ = 60°

The magnetic force experienced by the electron in the magnetic field is given as;

F = qvBsinθ

where;

q is charge of electron = 1.6 x 10⁻¹⁹ C

B is strength of magnetic field

v is speed of the electron

Substitute the given values and solve for F

F = (1.6 x 10⁻¹⁹)( 8.0 × 10⁶)(2.5)sin60

F = 2.77 x 10⁻¹² N

Thus, the magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form