Interaction with moving charges
A magnet does not attract a stationary charge. But as soon as that charge moves, a force appears that deflects it without slowing it down or speeding it up: this is the magnetic force, the same one that makes a motor turn or bends the paths of solar wind particles near the poles.
The Lorentz force: magnitude and direction
If you bring a magnet close to a beam of electrons at rest, nothing happens, but as soon as the electron gun fires them towards a screen, the beam suddenly bends to one side without slowing down: what do the direction and the size of this force, which only appears in motion, depend on?
The magnitude of the Lorentz force is $F = q\times v\times B\times\sen\alpha$ and its direction is given by the vector product $\vv{F} = q\times(\vv{v}\wedge\vv{B})$,…
The magnitude of the Lorentz force is $F = q\times v\times B\times\sen\alpha$ and its direction is given by the vector product $\vv{F} = q\times(\vv{v}\wedge\vv{B})$, which is determined using the right-hand rule. A proton moves horizontally to the east inside a uniform magnetic field also directed horizontally to the north. The situation, viewed from above, is shown in the figure.
- a) Determine the direction and sense of the force acting on the proton.
- b) What would the direction of the force be if the particle were an electron? Justify the difference.
From the expression for the magnitude of the Lorentz force, $F = q\times v\times B$ (perpendicular case): a) Rearrange for $B$ and deduce what combination of SI units…
From the expression for the magnitude of the Lorentz force, $F = q\times v\times B$ (perpendicular case):
- a) Rearrange for $B$ and deduce what combination of SI units one tesla is equivalent to.
- b) Using $1\ \text{C/s} = 1\ \text{A}$, express the tesla as $\text{T} = \frac{\text{N}}{\text{A}\times\text{m}}$ and explain in your own words what field is required so that a charge of $1\ \text{C}$ moving at $1\ \text{m/s}$ experiences a force of $1\ \text{N}$.
A proton ($q = 1.6\times10^{-19}\ \text{C}$) enters with $v = 3\times10^{6}\ \text{m/s}$ into a uniform magnetic field $B = 0.8\ \text{T}$, making an angle…
A proton ($q = 1.6\times10^{-19}\ \text{C}$) enters with $v = 3\times10^{6}\ \text{m/s}$ into a uniform magnetic field $B = 0.8\ \text{T}$, making an angle $\alpha = 30^\circ$ with it. Observe the diagram:
- a) Calculate the magnitude of the Lorentz force, $F = q\times v\times B\times\sen\alpha$ ($\sen 30^\circ = 0.5$).
- b) For what angle would the force be maximum? And for which would it be zero?
The magnetic force does no work: circular motion
When a charge enters a uniform field perpendicularly, the Lorentz force acts as the centripetal force and makes it describe a circle of constant radius.
The Lorentz force on a moving charge is always perpendicular to its velocity. a) Taking into account that work depends on the component of the force in the direction of…
The Lorentz force on a moving charge is always perpendicular to its velocity.
- a) Taking into account that work depends on the component of the force in the direction of displacement, explain why the magnetic force does no work on the charge.
- b) What then happens to the speed of the charge?
- c) And to its kinetic energy?
An electron ($m_e = 9.11\times10^{-31}\ \text{kg}$, $e = 1.6\times10^{-19}\ \text{C}$) enters perpendicularly with $v = 1\times10^{7}\ \text{m/s}$ into a uniform…
An electron ($m_e = 9.11\times10^{-31}\ \text{kg}$, $e = 1.6\times10^{-19}\ \text{C}$) enters perpendicularly with $v = 1\times10^{7}\ \text{m/s}$ into a uniform magnetic field $B = 1\times10^{-3}\ \text{T}$, as shown in the figure.
The radius of its circular trajectory is $r = \frac{m\times v}{q\times B}$. Calculate its value and express it in centimeters.
An electron ($m_e = 9.11\times10^{-31}\ \text{kg}$, $e = 1.6\times10^{-19}\ \text{C}$) orbits in a uniform magnetic field $B = 2\times10^{-3}\ \text{T}$, with its…
An electron ($m_e = 9.11\times10^{-31}\ \text{kg}$, $e = 1.6\times10^{-19}\ \text{C}$) orbits in a uniform magnetic field $B = 2\times10^{-3}\ \text{T}$, with its velocity perpendicular to the field.
- a) Equate the Lorentz force to the centripetal force and, using $v = \frac{2\pi\times r}{T}$, show that the orbital period is $T = \frac{2\pi\times m}{q\times B}$.
- b) Calculate its value.
- c) The result does not contain $v$ or $r$: explain what this means physically.
Force between currents and on a conductor
When the circuit of two flexible wires hanging in parallel is closed, they are seen to move together if their currents flow in the same direction and to move apart if they flow in opposite directions, even though the wires are neutral: how can one current exert a mechanical force on another?
Two straight parallel conductors carry currents in the same direction. a) Draw in your notebook the magnetic field created by the first conductor at the position of the…
Two straight parallel conductors carry currents in the same direction.
- a) Draw in your notebook the magnetic field created by the first conductor at the position of the second (right-hand rule) and the force exerted on the second conductor.
- b) Do they attract or repel each other?
- c) What changes if the direction of one of the currents is reversed?
You can do it on paper and send a photo with your answer.
A straight conductor of length $L = 40\ \text{cm}$, placed perpendicular to a uniform magnetic field $B = 0.1\ \text{T}$, carries a current $I = 5\ \text{A}$. The…
A straight conductor of length $L = 40\ \text{cm}$, placed perpendicular to a uniform magnetic field $B = 0.1\ \text{T}$, carries a current $I = 5\ \text{A}$. The magnetic force on it is $F = B\times I\times L$.
- a) Calculate its magnitude.
- b) Draw the situation in your notebook and determine, using the right-hand rule, the direction and sense of the force relative to the current and the field.
- c) If the conductor made an angle of $30^\circ$ with the field, the force would be $F = B\times I\times L\times\sen 30^\circ$; calculate it.
A horizontal rod of length $L = 50\ \text{cm}$ and mass $m = 25\ \text{g}$ is in a uniform horizontal magnetic field $B = 0.5\ \text{T}$, perpendicular to the rod. It is…
A horizontal rod of length $L = 50\ \text{cm}$ and mass $m = 25\ \text{g}$ is in a uniform horizontal magnetic field $B = 0.5\ \text{T}$, perpendicular to the rod. It is desired that the magnetic force on the rod, $F = B\times I\times L$, exactly balances its weight (take $g = 9.8\ \text{m/s}^2$). The situation is shown in the figure.
- a) Set up the condition for equilibrium between the magnetic force and the weight, and calculate the required current.
- b) Determine, using the right-hand rule, what relative orientation the current and field must have for the force to be vertically upwards.
- c) Reason what would happen if the current were increased by $10\ \%$: would the rod move? In which direction and with what initial net force?
Applications: selector, spectrometer and cyclotron
When a filament is heated, ions are released with a continuous mixture of disordered velocities, but to measure their masses accurately we need all of them to move at exactly the same speed: how can we filter a beam so that only particles with a given velocity get through?
In the next region, a second magnetic field bends the path into a semicircle of radius $r = \dfrac{m\cdot v}{|q|\cdot B}$: since $v$ is already the same for all the selected particles, the radius is determined by the mass. This is how isotopes are separated in a mass spectrometer. In a cyclotron, a fixed magnetic field keeps the radius increasing while an alternating voltage accelerates the particle on each turn, until it reaches the radius of the device.
In a mass spectrometer, an ion of $\text{He}^+$ ($m = 6.64\times10^{-27}\ \text{kg}$, $q = 1.6\times10^{-19}\ \text{C}$) describes a circular arc of radius…
In a mass spectrometer, an ion of $\text{He}^+$ ($m = 6.64\times10^{-27}\ \text{kg}$, $q = 1.6\times10^{-19}\ \text{C}$) describes a circular arc of radius $r = 12\ \text{cm}$ in a field $B = 0.5\ \text{T}$ perpendicular to its velocity. The situation is shown in the figure.
- a) Rearrange for velocity in the expression $r = \frac{m\times v}{q\times B}$.
- b) Calculate its value in scientific notation.
Helical motion: the Van Allen belts
When the velocity is neither perpendicular nor parallel to the field, but makes some angle $\alpha$ with it, it is convenient to split it into a component parallel to $\vv{B}$ (which experiences no force, and therefore does not change) and a perpendicular component (which goes round in a circle). The result is a helix: the charge moves at constant velocity in the direction of the field while spiralling around it.
The Van Allen belts trap charged particles from the solar wind in the Earth's magnetic field: as they approach the poles the field increases, the radius of the path ($r = m v_\perp/(|q|B)$) decreases and the helix "winds" tighter and tighter, until the particle bounces back. Particles that spiral down toward the poles produce the auroras seen in Alaska and northern Minnesota.
If two particles enter with the same kinetic energy instead of the same speed, write $v = \sqrt{\dfrac{2\,KE}{m}}$ in the radius formula: $r = \dfrac{\sqrt{2m\cdot KE}}{|q|B}$, so $r$ is proportional to $\sqrt{m}/|q|$.
A proton ($m_p = 1.67\times10^{-27}\ \text{kg}$) and an electron ($m_e = 9.11\times10^{-31}\ \text{kg}$) enter with the same speed, perpendicularly, into the same…
A proton ($m_p = 1.67\times10^{-27}\ \text{kg}$) and an electron ($m_e = 9.11\times10^{-31}\ \text{kg}$) enter with the same speed, perpendicularly, into the same uniform magnetic field.
- a) Write down the ratio $\frac{r_p}{r_e}$ of the radii of their trajectories and calculate its value.
- b) Taking into account that $T = \frac{2\pi\times m}{q\times B}$, calculate also the ratio of their periods and comment on which particle orbits faster.
The Van Allen belts trap charged particles in the Earth's magnetic field, which is much weaker than laboratory fields. An electron (…
The Van Allen belts trap charged particles in the Earth's magnetic field, which is much weaker than laboratory fields. An electron ($m_e = 9.11\times10^{-31}\ \text{kg}$, $e = 1.6\times10^{-19}\ \text{C}$) moves at $v = 3\times10^{7}\ \text{m/s}$ in the equatorial region, where $B = 3\times10^{-5}\ \text{T}$, with its velocity perpendicular to the field.
- a) Calculate the radius of its trajectory, $r = \frac{m\times v}{q\times B}$.
- b) At latitudes near the poles, the field is twice as strong, $B = 6\times10^{-5}\ \text{T}$; calculate the radius there.
- c) Relate the results to the fact that the trajectories of the trapped particles become more tightly coiled as they approach the poles.
A proton ($m_p$, $q_p = e$) and an alpha particle ($m_\alpha = 4\times m_p$, $q_\alpha = 2\times e$) enter with the same velocity, perpendicularly, into the same uniform…
A proton ($m_p$, $q_p = e$) and an alpha particle ($m_\alpha = 4\times m_p$, $q_\alpha = 2\times e$) enter with the same velocity, perpendicularly, into the same uniform magnetic field.
- a) Calculate the ratio $\frac{r_\alpha}{r_p}$ of the radii of their trajectories.
- b) Calculate the ratio $\frac{T_\alpha}{T_p}$ of their orbital periods.
- c) Repeat the ratio of radii assuming that both particles enter with the same kinetic energy (use $v = \sqrt{\frac{2\times E_c}{m}}$ in the expression for the radius) and comment on whether a mass spectrometer could separate them in each of the two cases.
Back to the start: An electric motor turns because the opposite currents flowing along the two sides of a loop experience magnetic forces in opposite directions, producing a torque that makes it rotate continuously.