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
