Lorentz Force Law (Magnetic and Mixed Fields)
The magnetic force on a particle of charge $q$ moving with a velocity $\vec{v}$ through a region with a magnetic field $\vec{B}$ is
$\vec{F} =q\vec{v}\times\vec{B}.$
If an electric field $\vec{E}$ is also present in the region, then the net force due to the two fields is called the Lorentz force.
$\vec{F}_{\text{Lorentz}} = q(\vec{E} + \vec{v}\times\vec{B})$
Contents
Physical effects on the motion of the particle
By definition of the cross product, the magnetic force must be perpendicular to both the velocity and the magnetic field. As a result of the perpendicularity with the velocity, the magnetic field cannot change the speed, but only the direction of the velocity. Hence it is always a centripetal force.
Particle circulating in a uniform magnetic field
A particle that enters a uniform magnetic field, $\vec{B},$ experiences a force of magnitude $F_B = qvB.$ Since this force is perpendicular to the direction the object travels, it will undergo centripetal acceleration.
$F_B = ma_{\text{centripetal}}$
$qvB = mv^2/r$
$r=\frac{mv}{qB}$
Find the radius of orbit for a particle of charge $q=5 \text{ C}$ and mass $m=20\text{ kg}$ moving with speed $v = 16 \frac{\text{m}}{\text{s}}$ through a uniform magnetic field $B = 2\text{ T}.$
$r=\frac{mv}{qB}=\frac{(20\text{ kg})(16 \frac{\text{m}}{\text{s}})}{(5 \text{ C})(2\text{ T})}=32\text{ m}$
Mass spectrometer
Many devices leverage the circular motion of a charge particle moving through a uniform field. One such device is a mass spectrometer, which sorts the ions in a sample according to their mass to charge ratio.
Find the charge to mass ratio of a particle that travels with speed $v=4\text{m/s}$ through a mass spectrometer set to $B=100\text{ mT}$ along an orbit of radius $r = 20\text{ cm}.$
From $r=\frac{mv}{qB},$ the charge to mass ratio is $\frac{q}{m} = \frac{v}{rB} = \frac{4}{(0.20)(0.1)} = 200\frac{\text{C}}{\text{kg}}.$
A rudimentary way to enrich uranium for use in nuclear energy or weaponry is to pass an ionized beam of vaporized uranium through a modified mass spectrometer known as a calutron. Assuming an electronmagnet that produces a magnetic field of magnitude $B = 1\text{ T}$ and a voltage oven rated at a potential difference of $V = 2\text{ kV}$ are available, what is the radius of the path traveled by the uranium-235 being isolated? Assume the beam of vaporized uranium is ionized to a charge of $q = 1.6\times10^{-19}\text{C}.$
Lorentz force due to electric and magnetic fields
The Lorentz force is the force felt by a particle of charge $q$ moving with a velocity $\vec{v}$ through a region with both an electric field $\vec{E}$ and a magnetic field $\vec{B}.$
$\vec{F}_{Lorentz} = q(\vec{E} + \vec{v}\times\vec{B})$
The net force on a particle moving through a region with both electric and magnetic fields is just the vector sum of the fields. $\vec{F}_{Lorentz} = \vec{F}_E + \vec{F}_B=q\vec{E} + q\vec{v}\times\vec{B}=q(\vec{E} + \vec{v}\times\vec{B})$
Find the net force on a $3 \text{ C}$ particle moving with velocity $\vec{v} = 6\hat{x}\frac{\text{m}}{\text{s}}$ in a region with electric field $\vec{E} = 5\hat{y}\frac{\text{N}}{\text{C}}$ and magnetic field $\vec{B} = 20\hat{z}\text{ T}.$
It is best to start with evaluating the cross product.
$\vec{v}\times\vec{B} = (6\hat{x})\times(5\hat{z}) = -30\hat{y}$
Now the Lorentz force is
$\vec{F}_{Lorentz} = q(\vec{E} + \vec{v}\times\vec{B}) = (3)(5\hat{y} - 30\hat{y}) = -75\hat{y}\text{ N}.$
A positron moves through a region in which the electric fields is uniform in the $x$-direction and the magnetic field is uniform in the $y$-direction. What is the direction of the terminal velocity of the positron?
References
- IndianFace, ., & Primefac, . Action of the Lorentz force bending the path of an electron in a magnetic field. Retrieved May 4, 2016, from https://commons.wikimedia.org/wiki/File:Action_of_the_Lorentz_force_bending_the_path_of_an_electron_in_a_magnetic_field.gif