Magnetism

Magnetic Fields

Magnetic fields are a modification of electric fields to account for Einstein's theory of special relativity. Like electric fields, magnetic fields are produced by charged particles, but only when the charged particle is moving. A magnetic field applies a force on a moving charged particle. Watch Veritasium explain.

Moving charges produce magnetic fields

Electric current is a source of moving charges. Electric current in a straight wire produces a magnetic field that curves around the wire.

The right hand rule can be used to find the direction the field turns. Place your right hand's thumb in the direction of the current. Your finger will curl in the direction the field turns.

Current moving in a ring shape has the same magnetic field as a permanent magnet. The magnetic field from permanent magnets comes form spinning charges.

Spinning charges produce magnetic fields

Electrons spin. Normally the electrons in atoms spin in opposing directions, but in ferromagnetic materials some spins remain unbalanced. The unbalanced spinning electrons produce a net magnetic field. Watch Minute Physics explain.

Permanent magnets always have unbalanced spins.
Example: a refrigerator magnet.

Paramagnetic magnets only have unbalanced spins when exposed to a magnetic field.
Example: the refrigerator door.

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Earth's Magnetic Field

The earth produces a magnetic field that is closely aligned with its axis of rotation. The source of the field is probably from internal convection of conductive materials in earth's core, but the process is complex and has only been recently modeled with computer simulations.

We also have strong evidence that the North and South poles experience periodic reversals. This means that a compass a million years ago would have pointed in the opposite direction it points now. The reversals are unpredictable, but generally there is several thousand years of stability between flips.

The earth's magnetic field slows down charged particles. This shields the earth from the solar wind's ionizing radiation. This slowing down can be seen at the north and south poles as an aurora.

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Magnetic Force on a Particle

A charged particle will experience a force when it moves in a magnetic field. The magnetic field is pointed out of the page.

B v F

$$ F=qvB sin \theta $$

\(F\) = magnetic force [N, Newton, kg m s⁻²] vector
\(q\) = charge [C, Coulomb]
\(c\) = velocity [m s⁻¹] vector
\(B\) = magnetic field [T, tesla, kg C⁻¹ s⁻¹] vector
\(\theta\) = angle between v and B
orders of magnitude of teslas

The ⊙ symbol indicates a vector pointed out of the page.
The ⊕ symbol indicates a vector pointed into the page.

Example: A +2C particle is moving at 10m/s through a 5mT field from a refrigerator magnet. What magnitude force will be applied to the particle?

solution $$ F=qvB $$ $$ F = (2)(10)(0.005) $$ $$ F = 0.1 N $$

Example: In order to learn about the strength of a magnetic field electrons are fired into the field at 100 m/s. The electrons accelerate at 1m/s². What is the magnitude of the magnetic field?

solution $$ F = ma $$ $$ F = (9.1 \times 10^{-31})(1)$$ $$ F = 9.1 \times 10^{-31}N$$
$$ F = q v B $$ $$ B = \frac{F}{qv} $$ $$ B = \frac{9.1 \times 10^{-31}}{(1.6 \times 10^{-19})(100)} $$ $$ B = 5.688 \times 10^{-14} T $$

Magnetic Force on a Wire

A wire in a magnetic field will experience a force when current runs through it. The length of the wire is multiplied by the current in the wire times the magnetic field.

B I F l

$$ F=lIB sin \theta $$

\(F\) = magnetic force [N, Newton, kg m s⁻²] vector
\(l\) = length [m, meter]
\(I\) = current [A, Amps] vector
\(B\) = magnetic field [T, tesla, kg C⁻¹ s⁻¹] vector
\(\theta\) = angle between the magnetic field and the current

Example: A 15cm wire has 20mA flowing through it east to west. The wire is in a magnetic field of 10T directed up. Describe the force on the wire.

solution $$ F=lIB$$ $$ F = (0.15)(0.02)(10) $$ $$ F = 0.03 N \quad North$$

Example: How long is a wire that has a force of 0.050N caused by the 0.21A current interacting with a 0.45T magnetic field?

solution $$ F=lIB $$ $$ l= \frac{F}{IB} $$ $$ l= \frac{0.050}{(0.21)(0.45)} $$ $$ l= 0.53m $$

Direction of Magnetic Forces

The 3-d nature of the magnetic force comes from the vector operation called a cross product. The direction of a cross product can be found with the right hand rule.

Using your right hand only place your thumb in the direction the charge is moving and place your index/fore finger in the direction of the magnetic field. The palm of your hand will face in the correct direction of the force.

Example: A magnetic field is pointed North. A positively charged particle is moving down through the field. What direction is the magnetic force on the particle?

solution
Using the right hand rule place your thumb down and your index finger north. $$East$$

Example: What direction magnetic field would produce a downward force on a negatively charged particle that is moving west?

solution
A negatively charged particle will go in the opposite direction of a positively charged particle. Using the right hand rule place your thumb west east and the palm of your hand down. $$South$$

Example: A 1C charged particle is moving west at 10 m/s through a 2T magnetic field directed east. What is the magnitude and direction of the force on the particle?

solution

Since the velocity and magnetic field are in opposite directions there is no force.

$$F=qvB sin \theta $$ $$sin 180 = 0$$ $$F=0$$

Example: Find the magnitude and direction of the force on a 24μC charge moving down at 13m/s through an 8.5T magnetic field directed South.

solution $$ F = q v B $$ $$ F = (24 \times 10^{-6})(13)(8.5) $$ $$ F = 0.00265 N \quad West$$

Path of a Charged Particle

The magnetic force on a charged particle is always at a 90 degree angle to the velocity. This satisfies the requirement for circular motion and it means we can set the equation for magnetic force equal to the centripetal force. $$ F = qvB \quad \quad F = \frac{mv^{2}}{r}$$ $$qvB = \frac{mv^{2}}{r}$$ $$qB = \frac{mv}{r}$$

r v F

$$r = \frac{mv}{qB}$$

\(r\) = radius of circular motion [m]
\(m\) = mass [kg, kilograms]
\(v\) = velocity [m s⁻¹] vector
\(q\) = charge [C, Coulomb]
\(B\) = magnetic field [T, tesla, kg C⁻¹ s⁻¹] vector




See the Pen charged particle motion in a uniform magnetic field by Landgreen (@lilgreenland) on CodePen.

Example: The earth's magnetic field protects us from solar wind. What is the radius for the circular motion of an electron in the earth's magnetic field? Wikipedia says that the field is up to 70μT, and particles from the solar wind travel at up to 500 km/s.

solution $$r = \frac{mv}{qB}$$ $$r = \frac{(9.1 \times 10^{-31})(500,000)}{(1.6 \times 10^{-19})(70 \times 10^{-6})}$$ $$r = 0.0406 m $$

Example: Find the radius for a proton in the same conditions.

solution $$r = \frac{mv}{qB}$$ $$r = \frac{(1.67 \times 10^{-27})(500,000)}{(1.6 \times 10^{-19})(70 \times 10^{-6})}$$ $$r = 74.55 m $$

Example: Find the radius of the circular path of an electron moving at 111m/s in a 2T magnetic field.

solution $$r = \frac{mv}{qB}$$ $$r = \frac{(9.1 \times 10^{-31})(111)}{(1.6 \times 10^{-19})(2)}$$ $$r = 3.1 \times 10^{-10}m $$

Example: You fire a 1kg object charged with +1C at 1m/s. The object arcs into a circle with a radius of 1m. What magnetic field could have produced this arc.

solution $$r = \frac{mv}{qB}$$ $$B = \frac{mv}{qr}$$ $$B = \frac{(1)(1)}{(1)(1)}$$ $$B = 1 T$$

Electromagnetic Induction

Electromagnetic induction is the production of voltage from a conductor moving past a magnetic field. The process is used to make electricity in a generator at a power plant. In reverse it can also turn energy into motion in an electric motor.

You can see a simulation of induction in the Phet simulation below.

Faraday's Law
Click to Run

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