When a charged particle enters a magnetic field with a velocity vector perpendicular to the magnetic field lines, it experiences a magnetic force perpendicular to both the magnetic field lines and the velocity vector (i.e., if the magnetic field lines run on the x-axis and the particle moves along the y-axis, then the force will be on the z-axis). This force can be quantified by the formula:
If the particle enters perfectly perpendicular to the magnetic field, it follows a perfectly circular path.
If the particle enters the magnetic field at an angle 0 < θ < 90, the particle will move in a helical path (a path resembling the curly hair strand).
If the particle enters parallel to the magnetic field, it experiences no force and continues moving straight.
To find the radius of the circle in the case it enters perfectly perpendicular, the following formula can be used:
A proton of mass \(1.67 \times 10^{-27}\) kg, and charge \(1.6 \times 10^{-19}\) C is moving at a speed of \(2.0 \times 10^6\) m/s perpendicular to a uniform magnetic field of strength 0.50T. What is the magnitude of the magnetic force acting on the proton? What is the shape of the path of the particle?
Solution:
\(F = qv \times B\)
\(F = (1.6 \times 10^{-19})(2 \times 10^6)(0.5)\)
\(F = 1.6 \times 10 ^{-13} \: N\), since the particle enters the magnetic field perfectly perpendicular, the shape of the path is a circle.
An electron enters a uniform magnetic field with a velocity perpendicular to said magnetic field. Derive a formula for the period of the rotation of said particle in terms of q, B, m and any constants needed.
Solution:
\(T = \frac{\text{Circumference}}{\text{velcoity}}\)
\(T = \frac{2πr}{v}\)
\(T = (\frac{2π}{v})(\frac{mv}{qB})\)
\(T = \frac{2πm}{qB}\)