The magnetic force acting on the proton is given by the following:
F = qv×B
F is the magnetic force vector
q is the charge
v is the velocity vector
B is the magnetic field vector
F is the cross product of qv and B, and the right-hand rule can be used to easily determine the direction of F. One way to utilize the right-hand rule is to:
- Point the thumb in the direction of v
- Point the other fingers in the direction of B
- The palm will face in the direction of F
v points in the +y direction and B points in the -x direction, therefore F points in the +z direction. Since v is perpendicular to B, the magnitude of F is given by:
|F| = q(|v|)(|B|)
Given values:
q = 1.60×10⁻¹⁹C
|v| = 2790m/s
|B| = 3.21×10⁻³T
|F| = 1.60×10⁻¹⁹(2790)(3.21×10⁻³)
|F| = 1.43×10⁻¹⁸N
We also have an E field acting on the proton. The magnitude of the electric force on the proton is:
|F| = q(|E|)
|F| is the magnitude of the electric force, q is the charge, and |E| is the magnitude of the E field.
For each case use q = 1.60×10⁻¹⁹C to calculate |F|
a) E is directed in +z and |E| = 5.97V/m, therefore F will point in +z with magnitude |F| = 9.55×10⁻¹⁹N. The electric force points in the same direction as the magnetic force, therefore the magnitude of the net force is just the sum of their magnitudes, 2.39×10⁻¹⁸N.
b) E is directed in -z and |E| = 5.97V/m, therefore F will point in -z with magnitude |F| = 9.55×10⁻¹⁹N. The electric force points in the opposite direction to the magnetic force, therefore the magnitude of the net force is the difference of their magnitudes, 4.75×10⁻¹⁹N.
c) E is directed in +x and |E| = 5.97V/m, therefore F will point in +x with magnitude |F| = 9.55×10⁻¹⁹N. The electric force is perpendicular to the magnetic force, therefore we will use the Pythagorean theorem to calculate the magnitude of the net force...
√((9.55×10⁻¹⁹)^{2} + (1.43×10⁻¹⁸)^{2})
= 1.72×10⁻¹⁸N