Question

An electron in a TV camera tube is moving at 7.20×10⁶ m/s in a magnetic field of strength 59 mT. Without knowing the direction of the field, what can you say about the greatest and least magnitude of the force acting on the electron due to the field? Maximum force?

Minimum force?

At one point the acceleration of the electron is 6.327×10¹⁶ m/s². What is the angle between the electron velocity and the magnetic field? (deg)

Answer 1

Maximum force acts on electron when angle between field and velocity of electron is 90 degree

Here magnitude of maximum force = qvB sin 90 = (1.6 x 10^-19)( 7.20×10⁶ )(59 x 10^-3)(1) = 6.8 x 10^-14 N

Minimum force will act on electron when angle between field and velocity of electron is 0 degree

Here magnitude of minimum force = qvB sin 0 = 0

If acceleration a= 6.327×10¹⁶ m/s²

Then Force = mass x acceleration

qvB sin( angle) = ma

(1.6 x 10^-19)( 7.20×10⁶ )(59 x 10^-3)sin (angle) = (9.1 x 10^-31)( 6.327×10¹⁶)

angle = 57.9 degrees

Maximum force acts on electron when angle between field and velocity of electron is 90 degree

Here magnitude of maximum force = qvB sin 90 = (1.6 x 10^-19)( 7.20×10⁶ )(59 x 10^-3)(1) = 6.8 x 10^-14 N

Minimum force will act on electron when angle between field and velocity of electron is 0 degree

Here magnitude of minimum force = qvB sin 0 = 0

If acceleration a= 6.327×10¹⁶ m/s²

Then Force = mass x acceleration

qvB sin( angle) = ma

(1.6 x 10^-19)( 7.20×10⁶ )(59 x 10^-3)sin (angle) = (9.1 x 10^-31)( 6.327×10¹⁶)

angle = 57.9 degrees