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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?

Short Answer

Expert verified
The current density \(J\) inside the wire must depend on \(r\) such that \(J\propto1/r\) in order for the magnetic field to have constant magnitude inside the wire.

Step by step solution

01

Apply Ampere's law

Ampere's law relates the circulation of magnetic field around a closed loop to the current through it. For a cylindrical wire, we choose a circular Amperean loop of radius \(r\) inside the wire. By symmetry, the magnetic field \(B\) is always tangent to this loop and it has the same magnitude at every point. The field and the direction of integration are both in azimuthal direction, so the angle between them is zero. Ampere's law becomes \(\oint B \cdot dl = \mu_0 I_{enc}\), where \(I_{enc}\) is the enclose current within the loop. We can rewrite this equation as \(B \cdot 2 \pi r = \mu_0 I_{enc}\).
02

Express the enclosed current in terms of current density

The current density \(J\) is defined as the current per unit area. For the cylindrical wire, the current enclosed within the circular loop of \(r\) is given by \(I_{enc} = \int J \cdot da\), where \(da = r \cdot d \theta \cdot dr\) is the area element in cylindrical coordinates. Substituting this into the Ampere's law equation gives \(B \cdot 2 \pi r = \mu_0 \int J \cdot r d \theta dr\). In order to derive a constant magnetic field \(B\), it is necessary for \(J\) to be a function of \(r\).
03

Solve for current density

By symmetry, \(J\) does not depend on the angle \(\theta\) and the integration over \(\theta\) yields \(2 \pi\). The equation then becomes \(B \cdot r = \mu_0 \int J \cdot r dr\). If \(B\) is to be constant, the right hand side must also be constant w.r.t \(r\). It follows that \(J\propto1/r\), i.e., the current density decreases inversely with the radius.

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Key Concepts

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

Magnetic Field Inside a Conductor
The concept of the magnetic field inside a conductor can be intriguing, as it involves imagining how invisible lines of magnetic force permeate a material carrying electric current. In simple terms, when a conductor such as a wire carries a current, it generates a magnetic field around it. The direction of this magnetic field can be determined by the right-hand rule, which states that if you point the thumb of your right hand in the direction of the current, the direction in which your fingers curl will indicate the magnetic field direction.

For a long, straight conductor, the magnetic field lines form circular loops around the wire. However, in a cylindrical wire, the situation becomes more complex due to the wire's geometry. Using Ampere's law, \( \oint B \cdot dl = \mu_0 I_{enc} \), we understand that the magnetic field inside a wire is directly related to the current enclosed by the path upon which we are applying the law. A key insight for having a constant magnetic field within a cylindrical conductor is ensuring that the relationship between the magnetic field and the current density is carefully managed.
Current Density Dependence on Radius
Getting to grips with how current density varies with the radius in a cylindrical wire requires a bit of spatial thinking. Current density, often denoted by \(J\), is the amount of electric current flowing per unit area within the conductor. Thinking about a thick wire, rather than current flowing uniformly across the entire cross-section, we realize that the distribution of current can vary.

In our problem, we are asked to determine how \(J\) should depend on the radial position \(r\) so that the magnetic field inside the wire remains constant. Since the magnetic field \(B\) is directly proportional to the enclosed current \(I_{enc}\) in Ampere's law, and \(I_{enc}\) depends on the current density and area over which it is flowing, having a constant \(B\) implies a very specific radial dependence for \(J\). The solution unveils that this dependence is inversely proportional, meaning \(J \propto 1/r\). This indicates that closer to the center of the wire, where \(r\) is smaller, the current density must be higher to maintain a uniform magnetic field throughout the wire's interior.
Application of Cylindrical Coordinates in Electromagnetism
Cylindrical coordinates become exceedingly helpful when dealing with problems in electromagnetism involving cylindrical symmetry. Unlike Cartesian coordinates, which are best suited for rectangular shapes, cylindrical coordinates are aligned with circular or cylindrical shapes, making them more intuitive for solving such problems.

In the context of our exercise, we applied cylindrical coordinates to describe the area element \(da\), which is essentially a small piece of the wire's cross-section at a certain radius \(r\) and angle \(\theta\). This allows us to integrate over the cross-section to find the enclosed current that is contributing to the magnetic field at that radius. The beauty of using cylindrical coordinates lies in their alignment with the problem's symmetry, simplifying the integration process and guiding us to the correct dependence of current density on radius to achieve the desired uniform magnetic field inside the cylindrical conductor.

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Most popular questions from this chapter

Fields from two rings \(*\) A ring with radius \(r\) and linear charge density \(\lambda\) spins with frequency \(\omega\). A second ring with radius \(2 r\) has the same density \(\lambda\) and frequency \(\omega\). Each ring produces a magnetic field at its center. How do the magnitudes of these fields compare?

Far field from a square loop ** Consider a square loop with current \(I\) and side length \(a\). The goal of this problem is to determine the magnetic field at a point a large, distance \(r\) (with \(r \gg a\) ) from the loop. (a) At the distant point \(P\) in Fig. 6.36, the two vertical sides give essentially zero Biot-Savart contributions to the field, because they are essentially parallel to the radius vector to \(P\). What are the Biot-Savart contributions from the two horizontal sides? These are easy to calculate because every little interval in these sides is essentially perpendicular to the radius vector to \(P\). Show that the sum (or difference) of these contributions equals \(\mu_{0} I a^{2} / 2 \pi r^{3}\), to leading order in \(a\). (b) This result of \(\mu_{0} I a^{2} / 2 \pi r^{3}\) is not the correct field from the loop at point \(P\). The correct field is half of this, or \(\mu_{0} I a^{2} / 4 \pi r^{3} .\) We will eventually derive this in Chapter 11, where we will show that the general result is \(\mu_{0} I A / 4 \pi r^{3}\), where \(A\) is the area of a loop with arbitrary shape. But we should be able to calculate it via the Biot-Savart law. Where is the error in the reasoning in part (a), and how do you go about fixing it? This is a nice one - don't peek at the answer too soon!

Helmholtz coils One way to produce a very uniform magnetic field is to use a very long solenoid and work only in the middle section of its interior. This is often inconvenient, wasteful of space and power. Can you suggest ways in which two short coils or current rings might be arranged to achieve good uniformity over a limited region? Hint: Consider two coaxial current rings of radius \(a\), separated axially by a distance \(b\). Investigate the uniformity of the field in the vicinity of the point on the axis midway between the two coils. Determine the magnitude of the coil separation \(b\) that for given coil radius \(a\) will make the field in this region as nearly uniform as possible.

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.

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.

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