Question

An electron moves in a region where the magnetic field is constant and has a magnitude of 3.0 x 10-4 T. If the magnitude of the acceleration of the electron is 8.0x1012m/s2 at an instant when the speed is 200 km/s, what is the angle between the velocity and the field. Restrict your answer to the first quadrant values (from 0 to 90o). Do not write the symbol for degrees. Round off your answer to one decimal place. Show Work.

Answer 1

  In absence of electrical field the Lorentz force on the particle is given by
F = q · (v × B)
v × B denotes cross product of velocity vector and magnetic field vector
The magnitude of the Lorentz force is
|F| = |q| · ( |v| · |B| ·sin(θ) )
where θ is the angle between velocity vector and magnetic field vector

For the problem solve equation above for the angle
θ = arcsin[ |F| / ( |q| · |v| · |B| ) ]
= arcsin[ m·|a| / ( |q| · |v| · |B| ) ]
= arcsin[ 9.11×10^-31kg * 8 ×10¹² m/s² / ( 1.60×10^-19C * 2 ×10⁵ m/s * 3×10^-4 T ) ]
= arcsin[ 0.759167 ]
= 49.39°