Question

An electron is moving through a magnetic field whose magnitude is 8.70x 10^-4T. The electron experiences only a magnetic force and has an acceleration of magnitude 3.50 x 10^14m/s^2. At a certain instant, it has a speed of 6.80 X 10^6 m/s^2. Determine the angle (less than 90 degrees) between the electrons velocity and the magnetic field.

Answer 1

Magnetic field= 8.70 X 10^-4 T=B

The electron experiences only a magnetic force and has an acceleration of magnitude

       = 3.50 * 10¹⁴ m/s ².

At a certain instant, it has a speed = 6.80 * 10 ⁶ m/s.

We have to determine the angle less than right angle between electron velocity and magnetic field which can be given as--

    F = q*v*B Sin θ

sin θ = F / q*v*B

    θ =sin ^-1 (F / q*v*B   )

         θ = sin ^-1 ( m*a / q*v*B   )   -------------(1) (Since F=m*a)

m is mass of the electron =9.11*10^-31 Kg

The acceleration of electron is given in the question as--

a = 3.50 X 10¹⁴ m/s ²

q is charge of the electron = 1.60*10^-19 C

Speed of electron = v = 6.80 *10 ⁶ m/s

B is the magnetic field = 8.70 X 10^-4 T.

Substituting the values in equation (1) we get-

θ =sin ^-1 (3.5 * (10^14)*9.1*10^(-31)/(1.6*10^(-19))*(6.8*10^6)*(8.7*10^-4))

θ=sin ^-1 (0.336)

θ=19.633 degrees. (Since we need angle less than the right angle)