Fields of wires, loops, solenoids and toroids

An upward electric current flows through a vertical wire.

1. a) State in your own words the right-hand rule that relates the direction of the current to the direction of the magnetic field lines it creates.
2. b) Apply it: viewed from above, do the field lines rotate clockwise or counterclockwise? Draw the situation in your notebook.

You can do it on paper and send a photo with your answer.

A current $I = 10\ \text{A}$ flows through a very long straight wire. The magnetic field it creates at a distance $r$ is $B = \frac{\mu_0\times I}{2\pi\times r}$, with $\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$. Calculate the field at $r = 5\ \text{cm}$ from the wire. Remember to convert the distance to meters.

The field created by a very long straight wire is $B = \frac{\mu_0\times I}{2\pi\times r}$. Without doing numerical calculations, reason using the formula:

1. a) What happens to $B$ if the current is doubled while keeping the distance constant?
2. b) What if the distance is doubled while keeping the current constant?
3. c) Where is the field stronger, near or far from the wire? Relate this to how its field lines spread apart with distance.

A lab probe measures the magnetic field at $r = 4\ \text{cm}$ from a very long straight wire and obtains $B = 2\times10^{-4}\ \text{T}$.

1. a) Rearrange the expression $B = \frac{\mu_0\times I}{2\pi\times r}$ to solve for the current.
2. b) Calculate its value ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).

The pantograph cable of an electric train can carry currents of $I = 500\ \text{A}$.

1. a) Rearrange for the distance $r$ in the formula for the field of a wire, $B = \frac{\mu_0\times I}{2\pi\times r}$.
2. b) Calculate at what distance from the cable the field it creates equals the Earth's magnetic field, $B_T \approx 5\times10^{-5}\ \text{T}$ ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).
3. c) Calculate the field at $10\ \text{m}$ from the cable and comment on whether it could affect a compass placed at that distance.

For wire A, which carries a current $I_A = 5\ \text{A}$, we measure the field at a distance of $10\ \text{cm}$; for wire B, which carries $I_B = 20\ \text{A}$, we measure it at $40\ \text{cm}$.

1. a) Calculate both fields using $B = \frac{\mu_0\times I}{2\pi\times r}$ ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).
2. b) Explain, using the dependence $B \propto \frac{I}{r}$, why the two results are equal.

In the lab, the magnetic field created by a very long wire is measured at several distances, with the following results: at $r = 20\ \text{cm}$ we obtain $B = 2\times10^{-5}\ \text{T}$; at $r = 10\ \text{cm}$, $B = 4\times10^{-5}\ \text{T}$; at $r = 5\ \text{cm}$, $B = 8\times10^{-5}\ \text{T}$.

1. a) Plot $B$ against $\frac{1}{r}$ in your notebook and explain what type of relationship appears.
2. b) The slope of the above line is $\frac{\mu_0\times I}{2\pi}$; estimate it from the data and deduce the value of the current $I$ flowing through the wire ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).

A circular loop of radius $R = 10\ \text{cm}$ carries a current $I = 2\ \text{A}$. The magnetic field at its center is $B = \frac{\mu_0\times I}{2\times R}$, with $\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$.

1. a) Calculate its value.
2. b) Express the result in scientific notation, rounded to two significant figures.

A counterclockwise current, viewed from above, flows through a circular loop placed horizontally. The loop, viewed from above, is shown in the figure.

1. a) Apply the right-hand rule and state the direction of the magnetic field at the center of the loop: upward or downward?
2. b) Reason which face of the loop (the top face or the bottom face) acts as a north pole.

A flat coil with $N = 50$ turns has a radius $R = 8\ \text{cm}$ and carries a current $I = 0.3\ \text{A}$. The field at its center is $B = \frac{\mu_0\times N\times I}{2\times R}$.

1. a) Calculate its value ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).
2. b) How many times greater is it than the field of a single loop of the same radius carrying the same current? Explain your answer using the formula.

What radius should a circular loop carrying a current of $I = 1\ \text{A}$ have so that the field at its center equals the Earth's magnetic field, $B_T \approx 5\times10^{-5}\ \text{T}$?

1. a) Rearrange for the radius in the expression $B = \frac{\mu_0\times I}{2\times R}$.
2. b) Calculate its value ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$) and express it in centimeters.

A solenoid has $n = 1000\ \text{espiras/m}$ and carries a current $I = 0.5\ \text{A}$. The magnetic field inside it is $B = \mu_0\times n\times I$, with $\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$.

1. a) Calculate its value.
2. b) Express it in milliteslas ($1\ \text{mT} = 10^{-3}\ \text{T}$).

Describe the magnetic field inside a long solenoid carrying a current:

1. a) What are the field lines like in the central area of the interior: curved, parallel, or converging?
2. b) As we move through the interior, does the field increase, decrease, or remain practically constant?
3. c) What happens to the field on the outside compared to the inside?

We want to construct a solenoid that generates a field of $B = 2.5\times10^{-3}\ \text{T}$ inside it with a maximum current of $I = 2\ \text{A}$.

1. a) Rearrange for the turn density $n$ in the expression $B = \mu_0\times n\times I$.
2. b) Calculate how many turns per meter are required ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).
3. c) If the tube on which it is wound measures $40\ \text{cm}$, how many turns will need to be wound in total?

In the laboratory, $N = 400$ turns are wound around a tube of length $L = 20\ \text{cm}$ and a current $I = 1.5\ \text{A}$ is passed through it.

1. a) Calculate the turn density $n = \frac{N}{L}$, in turns per meter.
2. b) Calculate the field inside the solenoid, $B = \mu_0\times n\times I$ ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).

The magnets in a magnetic resonance imaging (MRI) scanner are solenoids carrying large currents. Consider one with $n = 2000\ \text{espiras/m}$ and $I = 15\ \text{A}$ ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$; Earth's magnetic field $B_T \approx 5\times10^{-5}\ \text{T}$).

1. a) Calculate the field inside the solenoid.
2. b) How many times larger is it than the Earth's magnetic field?
3. c) Real scanners use superconducting coils instead of normal coils. Comment on the advantage provided by the superconductor having no electrical resistance when large, permanent currents are required.

Two straight parallel conductors, separated by $20\ \text{cm}$, carry currents in the same direction $I_1 = I_2 = 10\ \text{A}$ ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$). The arrangement, in the plane perpendicular to the wires, is shown in the figure.

1. a) Calculate the magnitude of the field created by each wire at the midpoint of the line segment joining them.
2. b) Using the right-hand rule, determine the direction of each of these fields at the midpoint and justify why the total field there is zero.
3. c) If the currents were different, would the field at the midpoint still be zero? Reason your answer.

Two straight, very long, parallel conductors, perpendicular to the plane of the paper and $20\ \text{cm}$ apart, carry currents in opposite directions, $I_1 = I_2 = 10\ \text{A}$ ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$). The situation is shown in the figure.

1. a) Using the right-hand rule, determine the direction of the magnetic field created by each wire at the midpoint and check that, since the currents are opposite, the fields add up there.
2. b) Calculate the total magnetic field at the midpoint.
3. c) Calculate the total magnetic field at point $P$ on the line joining the wires, located outside the segment, $10\ \text{cm}$ from wire 2 and $30\ \text{cm}$ from wire 1, indicating whether the fields add or subtract there.

Two straight parallel conductors, separated by $d = 40\ \text{cm}$, carry currents in the same direction, $I_1 = 30\ \text{A}$ and $I_2 = 10\ \text{A}$ ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$). Observe the diagram:

1. a) Using the right-hand rule, explain in which region (between the wires or outside them) their magnetic fields have opposite directions.
2. b) Set up the condition $B_1 = B_2$ for a point located at a distance $x$ from the first wire, within the segment joining them, and solve the resulting equation.
3. c) Check the consistency of the result: is the point closer to the wire carrying the smaller current? Why must this be the case?

A very long straight wire, lying in the plane of a circular loop of radius $R = 10\ \text{cm}$, passes at $d = 5\ \text{cm}$ from its center. A current of $I_1 = 10\ \text{A}$ flows through the wire and $I_2 = 4\ \text{A}$ through the loop, with directions chosen such that their magnetic fields at the center of the loop have opposite directions ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).

1. a) Calculate the magnitude of the wire's magnetic field at the center of the loop.
2. b) Calculate the magnitude of the loop's magnetic field at its center.
3. c) Justify why both fields are perpendicular to the plane of the loop and calculate the total magnetic field at the center.