/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 A conductor is made in the form ... [FREE SOLUTION] | 91Ó°ÊÓ

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A conductor is made in the form of a hollow cylinder with inner and outer radii a and \(b\) , respectively. It carries a current I uniformly distributed over its cross section. Derive expressions for the magnitude of the magnetic field in the regions (a) \(r < a\) ; (b) \(a< r b\).

Short Answer

Expert verified
The magnetic field is 0 for \( rb \).

Step by step solution

01

Analyze the Magnetic Field for r < a

Inside the hollow part of the cylinder (where \( r < a \)), there is no current enclosed, so the magnetic field is zero. This is due to the symmetry and the absence of current within this region. Thus, by Ampère's Law, \( B(r < a) = 0 \).
02

Determine the Current Enclosed for a < r < b

In the region \( a < r < b \), the current is uniformly distributed over the annular cross-section. We calculate the current enclosed within a circle of radius \( r \) by evaluating the current density \( J \), which is \( J = \frac{I}{\pi(b^2 - a^2)} \). The current through a circle with radius \( r \) is the current density multiplied by the area of the circle: \( I_{enc} = J \times \pi(r^2 - a^2) \).
03

Apply Ampère's Law for a < r < b

For \( a < r < b \), apply Ampère’s Law \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc} \), where \( \oint \mathbf{B} \cdot d\mathbf{l} = B\cdot 2\pi r \). Substitute the enclosed current from Step 2: \( B \cdot 2 \pi r = \mu_0 \cdot \frac{I}{\pi(b^2 - a^2)} \cdot \pi(r^2 - a^2) \). After simplifying, obtain \( B = \frac{\mu_0 I (r^2 - a^2)}{2\pi r(b^2-a^2)} \).
04

Analyze the Magnetic Field for r > b

For \( r > b \), the entire current \( I \) is enclosed. Applying Ampère’s Law gives \( B \cdot 2\pi r = \mu_0 I \). Solve for \( B \) to get \( B = \frac{\mu_0 I}{2\pi r} \).
05

Summarize the Magnetic Field Expressions for Different Regions

Compile the derived expressions for the magnetic field in each region: - For \( r < a \), \( B = 0 \). - For \( a < r < b \), \( B = \frac{\mu_0 I (r^2 - a^2)}{2\pi r(b^2-a^2)} \). - For \( r > b \), \( B = \frac{\mu_0 I}{2\pi r} \).

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Key Concepts

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

Magnetic field in conductors
The magnetic field in conductors is an important concept when dealing with currents and magnetism. For conductors carrying a current, the magnetic field is generated around them due to the movement of electric charges. This can be understood through Ampère's Law, which relates the magnetic field encircling a current to the current itself. Ampère's Law essentially says that the magnetic field around a closed loop is proportional to the total current passing through the loop.

In mathematical terms, Ampère's Law is expressed as \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I \), where \( \mathbf{B} \) is the magnetic field, \( d\mathbf{l} \) is a small segment of the loop, and \( I \) is the current enclosed by the loop. \( \mu_0 \) is the permeability of free space, a constant that characterizes the magnetic properties of vacuum.

For conductors like wires or cylinders, the distribution of the magnetic field depends on the shape of the conductor and how the current flows through it. Studying these distributions can help us understand fields in regions near conductors, which is crucial for applications ranging from electrical engineering to magnetic resonance imaging.
Current distribution in hollow cylinders
Understanding how current distributes in hollow cylinders helps in deriving the magnetic field in different regions. A hollow cylinder, or a tube, can carry current on its annular cross-section. This essentially means that there's current flowing through the material between the inner and outer radii, but not in the center core.

The current in these situations is considered to be uniformly distributed over the cross-sectional area between radii \( a \) and \( b \). To find the current at any point within this region, we calculate the current density, defined as the current per unit area. This is given by \( J = \frac{I}{\pi(b^2 - a^2)} \), where \( I \) is the total current, \( a \) is the inner radius, and \( b \) is the outer radius.

To know the current enclosed within a circle of radius \( r \) (where \( a < r < b \)), we multiply this density by the area of the circle from \( a \) to \( r \), which is \( \pi(r^2 - a^2) \). This understanding of current distribution is crucial for determining magnetic fields in these regions.
Magnetic field regions
Different regions around a hollow cylindrical conductor have varying magnetic fields based on their distance from the center.

  • **Region 1: Inside the hollow part (\( r < a \))**
    Here, there is no current, and by symmetry and Ampère's Law, the magnetic field is zero.

  • **Region 2: Within the conducting material (\( a < r < b \))**
    The magnetic field is found by applying Ampère’s Law, using the enclosed current within this area. The expression is \( B = \frac{\mu_0 I (r^2 - a^2)}{2\pi r(b^2-a^2)} \). This shows how the field varies with radius \( r \).

  • **Region 3: Outside the conductor (\( r > b \))**
    In this region, the entire current \( I \) is enclosed, and the magnetic field follows an inverse relationship with \( r \), similar to a long straight wire, given as \( B = \frac{\mu_0 I}{2\pi r} \).

These expressions show how the magnetic field's strength changes based on the position around the hollow cylindrical conductor. Understanding these regions is important for designing systems that rely on magnetic fields, like motors and transformers.

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Most popular questions from this chapter

Helmholtz Coils. Fig. 28.59 is a sectional view of two circular coils with radius \(a\) , each wound with \(N\) turns of wire carrying a current \(I,\) circulating in the same direction in both coils. The coils are separated by a distance \(a\) equal to their radii. In this configuration the coils are called Helmholtz coils; they produce a very uniform magnetic field in the region between them. (a) Derive the expression for the magnitude \(B\) of the magnetic field at a point on the axis a distance \(x\) to the right of point \(P,\) which is midway between the coils. (b) Graph \(B\) versus \(x\) for \(x=0\) to \(x=a / 2\) . Compare this graph to one for the magnetic field due to the right-hand coil alone. (c) From part (a), obtain an expression for the magnitude of the magnetic field at point \(P .\) (d) Calculate the magnitude of the magnetic field at \(P\) if \(N=300\) turns, \(I=6.00 \mathrm{A},\) and \(a=8.00 \mathrm{cm} .\) ( e) Calculate \(d B / d x\) and \(d^{2} B / d x^{2}\) at \(P(x=0)\) . Discuss how your results show that the field is very uniform in the vicinity of \(P .\)

Effect of Transmission Lines, Two hikers are reading a compass under an overhead transmission line that is 5.50 \(\mathrm{m}\) above the ground and carries a current of 800 \(\mathrm{A}\) in a horizontal direction from north to south. (a) Find the magnitude and direction of the magnetic field at a point on the ground directiy under the conductor. (b) One hiker suggests they walk on another 50 \(\mathrm{m}\) to avoid inaccurate compass readings caused by the current. Considering that the magnitude of the earth's field is of the order of \(0.5 \times 10^{-4} \mathrm{T},\) is the current really a problem?

A wide, long, insulating belt has a uniform positive charge per unit area \(\sigma\) on its upper surface. Rollers at each end move the belt to the right at a constant speed \(v\) . Calculate the magnitude and direction of the magnetic field produced by the moving belt at a point just above its surface. (Hint . At points near the surface and far from its edges or ends, the moving belt can be considered to be an infinite current sheet like that in Problem \(28.81 .\) )

Two identical circular, wire loops 40.0 \(\mathrm{cm}\) in diameter each carry a current of 1.50 \(\mathrm{A}\) in the same direction. These loops are parallel to each other and are 25.0 \(\mathrm{cm}\) apart. Line \(a b\) is normal to the plane of the loops and passes through their centers. A proton is fired at 2400 \(\mathrm{km} / \mathrm{sperpendicular}\) to line \(a b\) from a point midway between the centers of the loops. Find the magnitude and direction of the magnetic force these loops exert on the proton just after it is fired.

An alpha particle (charge + 2e) and an electron move in opposite directions from the same point, each with the speed of \(2.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) (Fig. \(28.32 ) .\) Find the magnitude and direction of the total magnetic field these charges produce at point \(P,\) which is 1.75 \(\mathrm{nm}\) from each of them.

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