Interaction of Charges and Currents with a Magnetic Field

An electric charge is at rest inside a uniform magnetic field. The magnitude of the Lorentz force is $F = q\times v\times B\times\sen\alpha$.

1. a) Does any magnetic force act on it? Justify your answer using the formula.
2. b) The charge is set in motion in the exact direction of the field ($\alpha = 0^\circ$). Does the answer change? Why?

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.

1. a) Determine the direction and sense of the force acting on the proton.
2. b) What would the direction of the force be if the particle were an electron? Justify the difference.

A proton ($q = 1.6\times10^{-19}\ \text{C}$) moves at $v = 2\times10^{6}\ \text{m/s}$ perpendicular to a magnetic field $B = 0.5\ \text{T}$. Calculate the magnitude of the Lorentz force acting on it, $F = q\times v\times B$.

From the expression for the magnitude of the Lorentz force, $F = q\times v\times B$ (perpendicular case):

1. a) Rearrange for $B$ and deduce what combination of SI units one tesla is equivalent to.
2. 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 $1\ \text{N}$.

The Lorentz force on a moving charge is always perpendicular to its velocity.

1. 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.
2. b) What then happens to the speed of the charge?
3. 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 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.

A charged particle enters with velocity $\vv{v}$ perpendicular to a uniform magnetic field.

1. a) Explain why the Lorentz force acts as a centripetal force and write down the equality $q\times v\times B = \frac{m\times v^2}{r}$.
2. b) Rearrange for the radius $r$ and comment on which quantities it depends on: does it increase or decrease if the particle moves faster? What if the field is stronger?

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:

1. a) Calculate the magnitude of the Lorentz force, $F = q\times v\times B\times\sen\alpha$ ($\sen 30^\circ = 0.5$).
2. b) For what angle would the force be maximum? And for which would it be zero?

A proton ($m_p = 1.67\times10^{-27}\ \text{kg}$, $e = 1.6\times10^{-19}\ \text{C}$) enters perpendicularly with $v = 5\times10^{6}\ \text{m/s}$ into a uniform magnetic field $B = 0.2\ \text{T}$.

1. a) Calculate the radius of its circular trajectory, $r = \frac{m\times v}{q\times B}$, and express it in centimeters.
2. b) What would happen to the radius if the proton entered at half the speed? Reason using the formula.

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.

1. a) Equate the Lorentz force to the centripetal force and, using $v = \frac{2\pi\times r}{T}$, show that the orbital period is $T = \frac{2\pi\times m}{q\times B}$.
2. b) Calculate its value.
3. c) The result does not contain $v$ or $r$: explain what this means physically.

Two straight parallel conductors carry currents in the same direction.

1. 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.
2. b) Do they attract or repel each other?
3. 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.

In a laboratory demonstration, two parallel wires separated by $d = 10\ \text{cm}$ carry currents in the same direction, $I_1 = 20\ \text{A}$ and $I_2 = 30\ \text{A}$. The force per unit length between them is $\frac{F}{L} = \frac{\mu_0\times I_1\times I_2}{2\pi\times d}$ ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$). Observe the diagram:

1. a) Calculate the force per unit length.
2. b) Do the wires attract or repel each other? Justify your answer.
3. c) Calculate the total force on a $1\ \text{m}$ section of each wire.

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 $F = B\times I\times L$.

1. a) Calculate its magnitude.
2. 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.
3. 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.

The old definition of the ampere (in force until 2019) stated: one ampere is the constant current that, if maintained in two straight parallel conductors of infinite length and separated by $1\ \text{m}$ in a vacuum, produces between them a force of $2\times10^{-7}\ \text{N}$ per meter of conductor ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).

1. a) Write down the expression for the force per unit length when both conductors carry the same current, $\frac{F}{L} = \frac{\mu_0\times I^2}{2\pi\times d}$.
2. b) Substitute $d = 1\ \text{m}$ and $\frac{F}{L} = 2\times10^{-7}\ \text{N/m}$ and check that $I = 1\ \text{A}$ is obtained.
3. c) Comment on the practical difficulties of this definition ('infinite' conductors, measuring such small forces) compared to the current definition, which fixes the value of the elementary charge.

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.

1. a) Set up the condition for equilibrium between the magnetic force and the weight, and calculate the required current.
2. b) Determine, using the right-hand rule, what relative orientation the current and field must have for the force to be vertically upwards.
3. 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?

High-voltage power lines transport current back and forth along parallel cables. Two of them, separated by $d = 2\ \text{m}$, carry $I = 800\ \text{A}$ in opposite directions ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).

1. a) State, justifying your answer using the right-hand rule, whether the force between the cables is attractive or repulsive.
2. b) Calculate the force per unit length between them, $\frac{F}{L} = \frac{\mu_0\times I_1\times I_2}{2\pi\times d}$.
3. c) In a short circuit, the current can reach $10^{4}\ \text{A}$; calculate the new force per unit length and explain why the cables must be securely anchored to the pylons.

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.

1. a) Rearrange for velocity in the expression $r = \frac{m\times v}{q\times B}$.
2. b) Calculate its value in standard form.

A velocity selector has an electric field $E = 2\times10^{4}\ \text{N/C}$ and a magnetic field $B = 0.05\ \text{T}$, perpendicular to each other and to the velocity of the ions. The electric and magnetic forces balance each other when $q\times E = q\times v\times B$.

1. a) Solve for the velocity of the ions that pass through the selector without being deflected and calculate it.
2. b) Does this velocity depend on the charge or mass of the ion? Comment on what this device is used for before a deflection region.

A mass spectrometer consists of a velocity selector (perpendicular electric field $E = 5\times10^{4}\ \text{N/C}$ and magnetic field $B_1 = 0.1\ \text{T}$) followed by a region with another magnetic field $B_2 = 0.2\ \text{T}$, perpendicular to the velocity, where the ions move along semicircles. Carbon-12 and carbon-14 ions enter, both with charge $q = e$ ($e = 1.6\times10^{-19}\ \text{C}$; atomic mass unit $u = 1.66\times10^{-27}\ \text{kg}$). The diagram of the apparatus is shown in the figure.

1. a) Calculate the speed at which the ions emerge from the selector ($q\times E = q\times v\times B_1$).
2. b) Calculate the radius of the trajectory of each isotope in the deflection region (take the masses as $12\times u$ and $14\times u$).
3. c) The ions strike a photographic plate at a distance from the entry point equal to the diameter of their semicircle: calculate the separation between the two marks they leave.

A cyclotron accelerates deuterons ($m = 3.34\times10^{-27}\ \text{kg}$, $q = 1.6\times10^{-19}\ \text{C}$) with a magnetic field $B = 1.2\ \text{T}$ perpendicular to the plane of the trajectories. The diagram of the cyclotron is shown in the figure.

1. a) The alternating generator must be synchronised with the orbital frequency, $f = \frac{q\times B}{2\pi\times m}$; calculate it and express it in megahertz.
2. b) The particles emerge when their radius reaches the radius of the device, $R = 0.5\ \text{m}$; calculate their maximum speed using $r = \frac{m\times v}{q\times B}$.
3. c) Calculate the kinetic energy with which the deuterons emerge.

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.

1. a) Write down the ratio $\frac{r_p}{r_e}$ of the radii of their trajectories and calculate its value.
2. 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.

A proton ($m_p = 1.67\times10^{-27}\ \text{kg}$, $e = 1.6\times10^{-19}\ \text{C}$) enters with $v = 2\times10^{6}\ \text{m/s}$ into a uniform magnetic field $B = 0.1\ \text{T}$, making an angle of $60^\circ$ with it, so that it describes a helical trajectory ($\sen 60^\circ \approx 0.87$; $\cos 60^\circ = 0.5$). Observe the diagram:

1. a) Resolve the velocity into its components parallel and perpendicular to the field.
2. b) Calculate the radius of the helix, determined by the perpendicular component.
3. c) Calculate the orbital period, $T = \frac{2\pi\times m}{q\times B}$, and the pitch (the distance it advances in the direction of the field per revolution).

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.

1. a) Calculate the radius of its trajectory, $r = \frac{m\times v}{q\times B}$.
2. b) At latitudes near the poles, the field is twice as strong, $B = 6\times10^{-5}\ \text{T}$; calculate the radius there.
3. 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 magnetic field.

1. a) Calculate the ratio $\frac{r_\alpha}{r_p}$ of the radii of their trajectories.
2. b) Calculate the ratio $\frac{T_\alpha}{T_p}$ of their orbital periods.
3. 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.