. An electron moving at 4.00×103m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40×10−16N . What angle does the velocity of the electron make with the magnetic field? There are two answers.

9.59° and 350.41°

Explanation: The formulae that relates the force F exerted on a moving charge q with velocity v in a magnetic field of strength B is given as

F =qvB sin x

Where x is the angle between the strength of magnetic field and velocity of the charge.

q = 1.609×10^-19 C

v = 4×10³ m/s

B = 1.25 T

F = 1.40×10^-16 N

By substituting the parameters, we have that

1.40×10^-16 = 1.609×10^-19 × 4×10³ × 1.25 × sinx

sin x = 1.40×10^-16/ 1.609×10^-19 × 4×10³ × 1.25

sin x = 1.40×10^-16 /8.045*10^(-16)

sin x = 0.1666

x = 9.59°

The value of sin x is positive in first and fourth quadrant.

Hence to get the second value of x, we move to the 4th quadrant of the trigonometric quadrant which is 360 - x

Hence = 360 - 9.59 = 350.41°

good looking response