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.

$$\vec{F}_{\text{Lorentz}} = q(\vec{E} + \vec{v}\times\vec{B})$$

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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.

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 $B = 2\text{ T}.$

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$$r=\frac{mv}{qB}=\frac{(20\text{ kg})(16 \frac{\text{m}}{\text{s}})}{(5 \text{ C})(2\text{ T})}=32\text{ m}$$

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 $r = 20\text{ cm}.$

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From $r=\frac{mv}{qB},$ the charge to mass ratio is $$\frac{q}{m} = \frac{v}{rB} = \frac{4}{(0.20)(0.1)} = 200\frac{\text{C}}{\text{kg}}.$$

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 $\vec{B} = 20\hat{z}\text{ T}.$

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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

$$\vec{F}_{Lorentz} = q(\vec{E} + \vec{v}\times\vec{B}) = (3)(5\hat{y} - 30\hat{y}) = -75\hat{y}\text{ N}.$$

1. 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