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 anticlockwise? Draw the situation in your exercise book.

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 metres.

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 laboratory 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 for the current in the expression $B = \frac{\mu_0\times I}{2\pi\times r}$.
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 laboratory, 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 gradient 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 centre 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 standard form, rounded to two significant figures.

An anticlockwise 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 centre of the loop: upwards or downwards?
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 centre 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? Reason 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 centre 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 centimetres.

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 metre 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 metre.
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.

A long solenoid has a field $B = 8\times10^{-3}\ \text{T}$ at its interior centre.

1. a) By imagining the solenoid as the combination of two semi-infinite solenoids, reason that the field at the centre of one of its ends must be approximately half the interior field. What is the value of this field at the end?
2. b) What is the approximate value of the field at an exterior point close to the midpoint of the solenoid? Justify your answer by considering what happens to the field lines as they emerge.

A toroid is a coil wound in the shape of a ring. The field inside it, at a distance $r$ from the axis, is $B = \frac{\mu_0\times N\times I}{2\pi\times r}$, where $N$ is the total number of turns. A toroid with $N = 500$ turns carries a current $I = 2\ \text{A}$. Calculate the field at points $r = 10\ \text{cm}$ from the axis ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).

A toroid with $N = 1000$ turns has a mean radius of $15\ \text{cm}$.

1. a) Rearrange for the current in the expression $B = \frac{\mu_0\times N\times I}{2\pi\times r}$.
2. b) Calculate the current required for the field inside to be $B = 4\times10^{-3}\ \text{T}$ ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$).

A toroid with $N = 600$ turns has an inner radius of $8\ \text{cm}$ and an outer radius of $12\ \text{cm}$, and carries a current $I = 5\ \text{A}$. The field inside it, at a distance $r$ from the axis, is $B = \frac{\mu_0\times N\times I}{2\pi\times r}$ ($\mu_0 = 4\pi\times10^{-7}\ \text{T}\times\text{m/A}$). Observe the cross-section of the toroid in the figure.

1. a) Calculate the field next to the inner edge ($r = 8\ \text{cm}$) and next to the outer edge ($r = 12\ \text{cm}$).
2. b) Calculate the field at the mean radius ($r = 10\ \text{cm}$) and check that it lies between the two previous values.
3. c) In light of the results, is the field inside a toroid uniform? Why, then, is the mean radius approximation used?

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