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Zero field outside a solenoid \(*\) * We showed in the solution to Problem \(6.19\) that the magnetic field is zero outside an infinite solenoid with arbitrary (uniform) crosssectional shape. We can demonstrate this fact in another way, similar in spirit to Problem 1.17.

Short Answer

Expert verified
The magnetic field is zero outside the solenoid based on Ampere's Law and the dot product of magnetic field vector and vector element of the path outside the solenoid.

Step by step solution

01

- Understanding Ampere's Law

Ampere's Law is expressed as \(\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}\), where B is the magnetic field, \(d\mathbf{l}\) is a vector element of a closed path, and \(I_{\text{enc}}\) is the current enclosed by the path.
02

- Applying Ampere's Law Outside the Solenoid

If we apply Ampere's Law to a closed loop path that is outside the solenoid, we have \(I_{\text{enc}}=0\) because no current passes through the closed loop path, hence, \(\oint \mathbf{B} \cdot d\mathbf{l} = 0 \mu_0\).
03

- Understanding the Dot Product of Magnetic Field and Path Element

The integral \(\oint \mathbf{B} \cdot d\mathbf{l}\) signifies that the magnetic field is being integrated around a closed path. That is to say, it is a dot product between the magnetic field and an element of the path. The dot product between two vectors is zero if the vectors are orthogonal, meaning at 90 degrees to each other. In this case, if the magnetic field is zero outside the solenoid, then \(\mathbf{B}\) must be perpendicular to \(d\mathbf{l}\).
04

- Concluding the Problem

Therefore, given the orthogonal relationship (at 90 degrees) outside the solenoid, and recalling that the dot product of 90 degrees angle will be zero, hence \(\mathbf{B}=0\). So, it can be concluded that the magnetic field is zero outside the solenoid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ampere's Law

At the heart of understanding magnetic fields in physics lies Ampere's Law. This fundamental principle offers a way to relate magnetic fields to the electric currents that generate them. It can be pretty tricky, but think of it like a helpful guide to finding out where magnetic forces are hiding and how strong they are. Ampere's Law is mathematically expressed as \(\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}\), where \(\mathbf{B}\) represents the magnetic field, \(d\mathbf{l}\) is a small piece of an imaginary loop around the current, and \(I_{\text{enc}}\) is the electric current trapped inside that loop. It's like taking a walk around a path and adding up all the little magnetic pushes along the way; if there's current inside your path, there will be a total magnetic push to tally up at the end.

Solenoid

Imagine a solenoid as a tightly wound spiral staircase for electrons – it's a long coil of wire that, when electrified, becomes an electromagnet. It's a super handy gadget that can be found everywhere from car starters to physics labs. A solenoid's superpower is its ability to create a pretty uniform magnetic field on the inside – kind of like a personal bubble of magnetic energy.

A handy thing to remember about this is that all the magic mostly happens on the inside. The magnetic field outside isn't quite so orderly and, in the case of an ideal, infinitely long solenoid, it's actually zero – the outside world isn't privy to the magnetic dance happening within those coiled wires.

Magnetic Field

Picture the magnetic field as the personal space of a magnet – it's an invisible area around a magnet where magnetic forces can play tug-of-war with iron things and other magnets. This magnetic field can be visualized using little imaginary arrows that show the direction and strength of these invisible forces.

The idea that magnetic fields fill the space around magnets or currents might remind you of the force fields in sci-fi movies. They have lines of force, which you can think of as the paths that the north pole of a teeny-tiny magnet would follow if it could explore the field on its own.

Integral Calculus in Physics

Now, think of integral calculus as the mathematical version of a net. It's used to scoop up an infinite number of small quantities to find the total. In physics, we use integrals to do things like find the total twist of a magnetic field along a path. Integrals grab all those little magnetic nudges you find using Ampere's Law and add them up into something that makes sense.

Let's apply it to our earlier concept: when we talk about \(\oint \mathbf{B} \cdot d\mathbf{l}\), what we're really doing is adding up all the little \(d\mathbf{l}\) vectors (small steps along the path) after considering how much they're aligned with the magnetic field \(\mathbf{B}\). If the field is zero in a certain area (like outside an ideal solenoid), the integral calculus confirms that – no small nudges to net, no total magnetic twist. It’s like having a net with nothing to catch!

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Most popular questions from this chapter

Field from an infinite wire \(* *\) Use the Biot-Savart law to calculate the magnetic field at a distance \(b\) from an infinite straight wire carrying current \(I\).

Maximum field in a cyclotron ** For some purposes it is useful to accelerate negative hydrogen ions in a cyclotron. A negative hydrogen ion, \(\mathrm{H}^{-}\), is a hydrogen atom to which an extra electron has become attached. The attachment is fairly weak; an electric field of only \(4.5 \cdot 10^{8} \mathrm{~V} / \mathrm{m}\) in the frame of the ion (a rather small field by atomic standards) will pull an electron loose, leaving a hydrogen atom. If we want to accelerate \(\mathrm{H}^{-}\)ions up to a kinetic energy of \(1 \mathrm{GeV}\left(10^{9} \mathrm{eV}\right)\), what is the highest magnetic field we dare use to keep them on a circular orbit up to final energy? (To find \(\gamma\) for this problem you only need the rest energy of the \(\mathrm{H}^{-}\)ion, which is of course practically the same as that of the proton, approximately \(1 \mathrm{GeV}\).)

Constant magnitude of \(B\) ** How should the current density inside a thick cylindrical wire depend on \(r\) so that the magnetic field has constant magnitude inside the wire?

Integral of \(A\), flux of \(B\) Show that the line integral of the vector potential \(\mathbf{A}\) around a closed curve \(C\) equals the magnetic flux \(\Phi\) through a surface \(S\) bounded by the curve. This result is very similar to Ampère's law, which says that the line integral of the magnetic field \(\mathbf{B}\) around a closed curve \(C\) equals (up to a factor of \(\mu_{0}\) ) the current flux \(I\) through a surface \(S\) bounded by the curve.

Hall voltage \(* *\) A Hall probe for measuring magnetic fields is made from arsenic-doped silicon, which has \(2 \cdot 10^{21}\) conduction electrons per \(\mathrm{m}^{3}\) and a resistivity of \(0.016 \mathrm{ohm}-\mathrm{m}\). The Hall voltage is measured across a ribbon of this \(n\)-type silicon that is \(0.2 \mathrm{~cm}\) wide, \(0.005\) \(\mathrm{cm}\) thick, and \(0.5 \mathrm{~cm}\) long between thicker ends at which it is connected into a \(1 \mathrm{~V}\) battery circuit. What voltage will be measured across the \(0.2 \mathrm{~cm}\) dimension of the ribbon when the probe is inserted into a field of 1 kilogauss?

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