Answer to Question #176899 in Electricity and Magnetism for JohnDoe

Question

An electron moves at 2.50 X 10^6 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.40 X 10^-2 T. (a) what are the largest and smallest possible magnitudes of the velocity of the electron due to the magnetic field? (b) if the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

Answer 1

The Lorentzian force is $\vec{F} = q(\vec{V}\times\vec{B}), \; F = qVB\sin\phi,$ where $\phi$ is the angle between the initial velocity and the magnetic field.

Therefore, the force and the acceleration will be largest when the velocity and the magnetic field are perpendicular to each other and smallest (equal to 0) when the velocity and the magnetic field have the same direction.

So the smallest acceleration is 0 and the largest acceleration is $a = \dfrac{F}{m} = \dfrac{qVB}{m} = \dfrac{1.6\cdot10^{-19}\,\mathrm{C}\cdot2.5\cdot10^{6}\,\mathrm{m/s}\cdot7.4\cdot10^{-2}\,\mathrm{T}}{9.1\cdot10^{-31}\,\mathrm{kg}} = 3.3\cdot10^{16}\,\mathrm{m/s^2}.$

Worth noting, the acceleration is centripetal and the force is always perpendicular to the velocity, so this force does not produce work and does not change the kinetic energy of the electron. Therefore, the modulus of the total velocity will remain the same. When a = 0, the electron will move straightly, when a reaches its maximum, the electron moves in a circle.

b) the maximum acceleration is $a = \dfrac{F}{m} = \dfrac{qVB}{m}$ . The acceleration in question is $a_1 = \dfrac{qVB\sin\varphi}{m} = \dfrac14\cdot \dfrac{qVB}{m} \Rightarrow \sin\varphi = 0.25, \; \varphi =14.5^\circ.$