/*! 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 36 Field at different radii \(*\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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Field at different radii \(*\) A current of 8000 amperes flows through an aluminum rod \(4 \mathrm{~cm}\) in diameter. Assuming the current density is uniform through the cross section, find the strength of the magnetic field at \(1 \mathrm{~cm}\), at \(2 \mathrm{~cm}\), and at \(3 \mathrm{~cm}\) from the axis of the rod.

Short Answer

Expert verified
The magnetic field at 1 cm is \(0.002 T\), at 2 cm is \(0.004 T\) (inside the rod), and at 3 cm is \(0.00267 T\) (outside the rod).

Step by step solution

01

Determining the type of problem

First, observe the problem and recognize it as pertaining to magnetic fields in current carrying conductors. Under the assumption that the current is evenly distributed and by knowing the total current, we can find the magnetic field at any radius r inside the conductor.
02

Calculate the radius of the conductor

In the given problem, the diameter of the aluminium rod is mentioned as 4 cm. Therefore, compute the radius, which is half the diameter, resulting in a radius, \(R\), of 2 cm or 0.02 m.
03

Compute the magnetic field within the conductor

Now, calculate the strength of the magnetic field at 1 cm and 2 cm from the axis of the rod, since those points are within the rod. The formula for the magnetic field inside the conductor is \(B=\frac{\mu I r}{2 \pi R^2}\), where \(\mu\) is the permeability of free space. Use these values: \(\mu = 4\pi × 10^{-7} T·m/A\), \(I = 8000 A\), \(r = 0.01 m\) for the first calculation and \(r = 0.02 m\) for the second calculation. This will give the magnetic field values inside the rod.
04

Compute the magnetic field outside the conductor

Next, compute the strength of the magnetic field at 3 cm from the axis of the rod, which is a point outside the rod. Use the formula \(B=\frac{\mu I}{2 \pi r}\). Substituting the values \(\mu = 4\pi × 10^{-7} T·m/A\), \(I = 8000 A\), and \(r = 0.03 m\), compute the magnetic field outside the rod.
05

Interpreting the Results

The strength of the magnetic field calculated at different distances from the axis of the conductor indicates that magnetic field is stronger closer to the conductor and weakens as we move further away in radial direction, as expected.

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Key Concepts

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

Current Density
Current density is a fundamental concept when dealing with current-carrying conductors. It describes how much electric current flows through a given area of a material. Imagine it as the concentration of the flow of electric charge.
  • If you have a large current flowing through a small cross-sectional area, the current density is high.
  • In this particular problem, the current density is uniform, meaning it's the same throughout the entire cross-sectional area of the rod.
The formula for current density, denoted as \( J \), is given by the ratio of the total current \( I \) flowing through the conductor to its cross-sectional area \( A \). Mathematically, it's expressed as:\[ J = \frac{I}{A} \]Understanding current density is crucial as it helps calculate other magnetic properties of the conductor, such as the magnetic field strength at various points.
Magnetic Field Strength
The magnetic field strength is a measure of the force exerted by a magnetic field at a given point. In a current-carrying conductor, the magnetic field is generated around the conductor as a result of the flow of electric current.
  • The strength of the magnetic field is directly proportional to the current in the conductor.
  • For points inside the conductor, the field strength also depends on the distance from the center of the wire.
In the exercise, different values of magnetic field strength were computed at 1 cm, 2 cm, and 3 cm from the axis using specific formulas. These formulas account for whether a point is within or outside the conductor:- Inside the conductor: \( B = \frac{\mu I r}{2 \pi R^2} \)- Outside the conductor: \( B = \frac{\mu I}{2 \pi r} \)Here, \( \mu \) is the permeability of free space, \( I \) is the current, \( R \) is the radius of the conductor, and \( r \) is the distance from the center of the conductor. Notice how the magnetic field weakens with increasing distance from the wire.
Ampere's Law
Ampere's Law is a pivotal part of understanding magnetic fields in conductors. It connects the electric current flowing through a loop to the magnetic field around that loop.The law is formulated as:\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu I \]Where:- \( \oint \mathbf{B} \cdot d\mathbf{l} \) is the line integral of the magnetic field \( \mathbf{B} \) along the path enclosing the current.- \( \mu \) is the permeability of free space.- \( I \) is the current through the loop.This mathematical expression shows that the magnetic field around a closed loop is proportional to the total current passing through the loop. In the context of the problem, Ampere's Law helps derive the formulae used to calculate the magnetic field both inside and outside the conductor by considering different paths in conjunction with the current.
Permeability of Free Space
The permeability of free space, denoted as \( \mu_0 \), is a physical constant that appears in several important equations in electromagnetism.
  • It's used to relate the magnetic field in a material to the magnetic field in a vacuum.
  • Its value is approximately \( 4\pi \times 10^{-7} \) T·m/A.
This constant is crucial when calculating the magnetic influence of a current-carrying conductor, as seen in the formula for magnetic field strength:\[ B = \frac{\mu_0 I}{2 \pi r} \]Here, \( \mu_0 \) facilitates the translation between current and magnetic field by acting as a proportionality factor, which amplifies the effect of current in generating magnetic fields. In the original exercise, \( \mu_0 \) was essential to compute the magnetic field at different radii from the aluminum rod.

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Most popular questions from this chapter

A slab and a sheet \(* *\) A volume current density \(\mathbf{J}=J \hat{\mathbf{z}}\) exists in a slab between the infinite planes at \(x=-b\) and \(x=b\). (So the current is coming out of the page in Fig. 6.37.) Additionally, a surface current density \(\mathcal{J}=2 b J\) points in the \(-\hat{\mathbf{z}}\) direction on the plane at \(x=b\) (a) Find the magnetic field as a function of \(x\), both inside and outside the slab. (b) Verify that \(\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}\) inside the slab. (Don't worry about the boundaries.)

Zero force in any frame \(* *\) A neutral wire carries current \(I\). A stationary charge is nearby. There is no electric field from the neutral wire, so the electric force on the charge is zero. And although there is a magnetic field, the charge isn't moving, so the magnetic force is also zero. The total force on the charge is therefore zero. Hence it must be zero in every other frame. Verify this, in a particular case, by using the Lorentz transformations to find the \(\mathbf{E}\) and \(\mathbf{B}\) fields in a frame moving parallel to the wire with velocity \(\mathbf{v}\).

Field at the center of a disk * A disk with radius \(R\) and surface charge density \(\sigma\) spins with angular frequency \(\omega\). What is the magnetic field at the center?

Proton beam \(* *\) A high-energy accelerator produces a beam of protons with kinetic energy \(2 \mathrm{GeV}\) (that is, \(2 \cdot 10^{9} \mathrm{eV}\) per proton). You may assume that the rest energy of a proton is \(1 \mathrm{GeV}\). The current is 1 milliamp, and the beam diameter is \(2 \mathrm{~mm}\). As measured in the laboratory frame: (a) what is the strength of the electric field caused by the beam \(1 \mathrm{~cm}\) from the central axis of the beam? (b) What is the strength of the magnetic field at the same distance? (c) Now consider a frame \(F^{\prime}\) that is moving along with the protons. What fields would be measured in \(F^{\prime}\) ?

Creating a uniform field *** For a delicate magnetic experiment, a physicist wants to cancel the earth's field over a volume roughly \(30 \times 30 \times 30 \mathrm{~cm}\) in size, so that the residual field in this region will not be greater than 10 milligauss at any point. The strength of the earth's field in this location is \(0.55\) gauss, making an angle of \(30^{\circ}\) with the vertical. It may be assumed constant to a milligauss or so over the volume in question. (The earth's field itself would hardly vary that much over a foot or so, but in a laboratory there are often local perturbations.) Determine roughly what solenoid dimensions would be suitable for the task, and estimate the number of ampere turns (that is, the current \(I\) multiplied by the number of turns \(N\) ) required in your compensating system.

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