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A very large flat conducting sheet of thickness \(t\) carries a uniform current density \(\overrightarrow{\mathbf{j}}\) throughout (Fig. 28-56). Determine the magnetic field (magnitude and direction) at a distance \(y\) above the plane. (Assume the plane is infinitely long and wide.)

Short Answer

Expert verified
The magnetic field magnitude is \(B = \frac{\mu_0 j t}{2}\) and directed in the positive \(z\)-direction.

Step by step solution

01

Understanding the Problem

We need to find the magnetic field at a distance \( y \) above a very large flat conducting sheet carrying a uniform current density \( \overrightarrow{\mathbf{j}} \). The sheet is infinite in extent and has a thickness \( t \).
02

Conceptualize the Magnetic Field Due to a Current Sheet

For an infinite current sheet, the magnetic field at a point above (or below) is uniform and parallel to the plane of the sheet. If the current density points in the \( x \)-direction, the magnetic field will be directed in the \( z \)-direction above the sheet.
03

Applying Ampere’s Law

Use Ampere’s Law: \( \oint \overrightarrow{\mathbf{B}} \cdot d\overrightarrow{\mathbf{l}} = \mu_0 I_{enc} \). Construct an Amperian loop in the shape of a rectangle that extends above and below the sheet, ensuring one side is perpendicular to the plane.
04

Calculate the Enclosed Current

The enclosed current, \( I_{enc} \), through the Amperian loop is equal to the product of the current density \( j \) and the thickness \( t \). Thus, \( I_{enc} = jtL \), where \( L \) is the length of the side of the loop parallel to the current flow.
05

Solve for Magnetic Field

Since the magnetic field is uniform on either side of the sheet, and zero along the sides perpendicular to the sheet, the integration reduces to: \( 2BL = \mu_0 jtL \). Hence, the magnitude of the magnetic field is \( B = \frac{\mu_0 j t}{2} \).
06

Determine Direction of Magnetic Field

Using the right-hand rule, with the thumb pointing in the direction of current density \( \overrightarrow{\mathbf{j}} \), the fingers curl towards the direction above the sheet. Therefore, for current density along the \( x \)-axis, the magnetic field above the sheet is directed in the positive \( z \)-direction.

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Key Concepts

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

Ampere's Law
Ampere's Law is a fundamental principle in electromagnetism. It relates the magnetic field around a closed loop to the electric current passing through the loop. Mathematically, it's expressed as: \[ \oint \overrightarrow{\mathbf{B}} \cdot d\overrightarrow{\mathbf{l}} = \mu_0 I_{enc} \]Here, \( \oint \overrightarrow{\mathbf{B}} \cdot d\overrightarrow{\mathbf{l}} \) is the line integral of the magnetic field \( \overrightarrow{\mathbf{B}} \) around a closed path, \( \mu_0 \) is the permeability of free space, and \( I_{enc} \) is the enclosed current within the loop.

In simple terms, it means the total magnetic field along a closed loop is determined by how much current passes through that loop. Ampere's Law is especially useful in symmetric situations like infinite wires or sheets, where it simplifies the process of finding the magnetic field.
Current Sheet
A current sheet is a two-dimensional plane through which electric current flows uniformly. Think of it as a thin, flat conductor with current spread across its entire surface. In our problem, the sheet is infinite and has a uniform current density, denoted as \( \overrightarrow{\mathbf{j}} \), which means the current per unit area is constant.

This property of uniform current distribution helps in determining the magnetic field. For a very large sheet, the effects at a distance are similar no matter where you are above the sheet, as long as you stay reasonably close. Thus, the magnetic field is parallel to the sheet and has a uniform strength above and below.
  • Uniform current distributes evenly across the sheet.
  • The sheet's infinite extent simplifies calculations, as edge effects are negligible.
  • Current density direction (e.g., x-axis) affects the resulting magnetic field direction.
Magnetic Field Direction
Understanding the direction of the magnetic field is crucial in solving electromagnetic problems. For an infinite conducting current sheet with a current density \( \overrightarrow{\mathbf{j}} \) pointing in a specific direction (say the \( x \)-axis), the magnetic field direction above the sheet will be perpendicular to both the current density and the sheet.

Based on the geometric setup, if the current flows along the \( x \)-axis, the magnetic field generated will emerge in the positive \( z \)-direction above the sheet and the negative \( z \)-direction below the sheet.
  • Magnetic field lines are continuous and loop around the current flow.
  • The field direction can be determined using the right-hand rule.
  • For uniform sheets, the direction is consistent above and below the sheet but opposite in orientation.
Right-Hand Rule
The right-hand rule is a simple tool to determine the direction of the magnetic field relative to the current flow. Here's how it works:
  • Extend your right hand's thumb in the direction of the current density vector \( \overrightarrow{\mathbf{j}} \).
  • Your fingers, when curled into a fist, will point in the direction of the magnetic field.
This visualization helps us easily find the magnetic field direction. If the current is flowing along the \( x \)-axis, the right-hand thumb points along \( x \), and your fingers will curl towards the \( z \)-direction above the current sheet. This visualization aids in understanding electromagnetic setups by connecting a physical gesture with field orientations.

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Most popular questions from this chapter

A 550 -turn solenoid is \(15 \mathrm{~cm}\) long. The current into it is 33 A. A 3.0-cm-long straight wire cuts through the center of the solenoid, along a diameter. This wire carries a 22 -A current downward (and is connected by other wires that don't concern us). What is the force on this wire assuming the solenoid's field points due east?

3\. (1II) A very long flat conducting strip of width \(d\) and negli- gible thickness lies in a horizontal plane and carries a uniform current \(I\) across its cross section. \((a)\) Show that at points a distance \(y\) directly above its center, the field is given by $$B=\frac{\mu_{0} I}{\pi d} \tan ^{-1} \frac{d}{2 y}$$ assuming the strip is infinitely long. [Hint. Divide the strip into many thin wires and sum (integrate) over these. (b) What value does \(B\) approach for \(y \gg d ?\) Does this make sense? Explain.

(II) A toroid (Fig. \(28-17\) ) has a 50.0 -cm inner diameter and a 54.0-cm outer diameter. It carries a \(25.0 \mathrm{~A}\) current in its 687 coils. Determine the range of values for \(B\) inside the toroid.

You have 1.0 \(\mathrm{kg}\) of copper and want to make a practical solenoid that produces the greatest possible magnetic field for a given voltage. Should you make your copper wire long and thin, short and fat, or something else? Consider other variables, such as solenoid diameter, length, and so on.

(II) A 20.0 -m-long copper wire, \(2.00 \mathrm{~mm}\) in diameter including insulation, is tightly wrapped in a single layer with adjacent coils touching, to form a solenoid of diameter \(2.50 \mathrm{~cm}\) (outer edge). What is \((a)\) the length of the solenoid and \((b)\) the field at the center when the current in the wire is \(16.7 \mathrm{~A}\) ?

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