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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 \(28.81 .\) )

Short Answer

Expert verified
The magnitude is \(B = \frac{\mu_0}{2} \sigma v\); the direction is perpendicular to the surface.

Step by step solution

01

Identify the Equivalent Current Distribution

The moving charge on the surface of the belt can be considered an equivalent current. Since the charge density is \(\sigma\) and it moves with a speed \(v\), the equivalent surface current density \(K\) can be expressed as \(K = \sigma \times v\). This represents the amount of charge per unit time crossing a unit width orthogonal to the motion.
02

Apply the Biot-Savart Law for a Current Sheet

For an infinite current sheet with surface current density \(K\), the magnetic field \(B\) at a point just above the sheet can be determined using the modified form of Ampère's law or Biot-Savart Law. The magnetic field is constant and is given by \(B = \frac{\mu_0}{2} K\), where \(\mu_0\) is the permeability of free space.
03

Calculate the Magnitude of the Magnetic Field

Substitute \(K = \sigma v\) into the expression for the magnetic field. This gives \(B = \frac{\mu_0}{2} \sigma v\). This expression represents the magnitude of the magnetic field produced just above the surface of the moving belt.
04

Determine the Direction of the Magnetic Field

According to the right-hand rule, the direction of the magnetic field due to a current is perpendicular to both the current direction and the axis from which the field is measured. Given that the belt moves to the right, the magnetic field is directed upwards or downwards, depending on your reference point.

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Key Concepts

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

Current Distribution
In the context of the moving insulating belt with a uniform charge, current distribution is a fundamental concept necessary for magnetic field calculations. Current distribution refers to how current is spread over a surface or through a volume. Here, the belt, moving with speed \(v\), carries a positive surface charge per unit area denoted by \(\sigma\).

This moving charge is treated as an equivalent current, even though it's not a wire carrying electrons. Instead of point charges moving around, the entire surface of the belt participates, contributing to a total effective current. This distribution is quantified as the surface current density \(K\), calculated with the formula:

\[K = \sigma \times v\]

This equation tells us that the faster the belt moves, or the higher the charge per area, the greater the equivalent current passing over the belt's surface.
Biot-Savart Law
The Biot-Savart Law is central for determining the magnetic field resulting from a current distribution. While typically used for line currents, in this scenario, it is adapted for an infinite current sheet. For our case, the current sheet is the belt with surface current density \(K\).

To calculate the magnetic field \(B\), the simplified version of the Biot-Savart Law or modified Ampère's law for infinite sheets is used:

\[B = \frac{\mu_0}{2} K\]

Where \(\mu_0\) is the permeability of free space. This relationship demonstrates how the magnetic field is directly proportional to the surface current density. The surface charge density and speed of the belt dictate the current density, leading to a constant magnetic field regardless of the position over the belt, as long as it's near the surface and away from the edges.
Surface Current Density
Surface current density is a crucial concept when dealing with objects like our charged belt. Unlike a regular current through a wire (where current is considered along a line), surface current density \(K\) is all about how current is dispersed over a surface area.

In our problem, surface current density \(K\) was determined by multiplying the surface charge density \(\sigma\) by the velocity \(v\) of the belt. This relationship emerges because each unit of area on the surface contributes to the flow of current proportional to its charge and velocity:
  • \( \sigma \) - Represents how much charge is placed per unit area on the belt surface.
  • \( v \) - Indicates the speed at which this charge moves.
  • \( K = \sigma \times v \) - Provides a measure of current flow per unit width.
Surface current densities are particularly significant because they allow us to calculate resultant fields from extended surfaces, making them invaluable in electromagnetism.
Right-Hand Rule
The right-hand rule is an indispensable tool for determining the direction of vector quantities resulting from cross-products, such as magnetic fields due to currents. In our scenario, knowing the magnetic field's direction is vital after calculating its magnitude using the Biot-Savart Law.

To apply the right-hand rule in this context, consider the belt's direction: moving to the right. Point your right thumb in the direction of the current (same as the direction the belt moves). Then, curl your fingers following the direction from a point on the surface above the belt.

Your fingers now point along the direction of the magnetic field, which is perpendicular to both the current's direction and the surface plane. Depending on the observer's frame, this can mean the magnetic field is directed upwards or downwards. This step ensures an intuitive grasp of spatial orientations in electromagnetic scenarios.

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Most popular questions from this chapter

A negative point charge \(q=-7.20 \mathrm{mC}\) is moving in a reference frame. When the point charge is at the origin, the magnetic field it produces at the point \(x=25.0 \mathrm{cm}, y=0, z=0\) is \(\vec{B}=(6.00 \mu \mathrm{T}) \hat{j},\) and its speed is 800 \(\mathrm{km} / \mathrm{s}\) . (a) What are the \(x\) - \(y\) .and \(z\) -components of the velocity \(\vec{v}_{0}\) of the charge? (b) At this same instant, what is the magnitude of the magnetic field that the charge produces at the point \(x=0, y=25.0 \mathrm{cm}, z=0 ?\)

A closed curve encircles several conductors. The line integral \(\phi \overrightarrow{\boldsymbol{B}} \cdot d \vec{l}\) around this curve is \(3.83 \times 10^{-4} \mathrm{T} \cdot \mathrm{m} .\) (a) What is the net current in the conductors? (b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral? Explain.

A toroidal solenoid (see Example 28.10 ) has inner radius \(r_{1}=15.0 \mathrm{cm}\) and ourer radius \(r_{2}=18.0 \mathrm{cm} .\) The solenoid has 250 turns and carries a current of 8.50 A. What is the magnitude of the magnetic field at the following distances from the center of the torus: (a) \(12.0 \mathrm{cm} ;(\mathrm{b}) 16.0 \mathrm{cm} ;(\mathrm{c}) 20.0 \mathrm{cm} ?\)

A long, straight wire lies along the \(y\) -axis and carries a current \(I=8.00\) A in the \(-y\) -direction (Fig. 28.39\()\) . In addition to the magnetic field due to the current in the wire, a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) with magnitude \(1.50 \times 10^{-6} \mathrm{T}\) is in the \(+x\) -direction \(\mathrm{What}\) is the total field (magnitude and direction) at the following points in the \(x z\) -plane: \((a) x=0, z=1.00 \mathrm{m}\) (b) \(x=1.00 \mathrm{m}, z=0 ;(\mathrm{c}) x=0\) \(z=-0.25 \mathrm{m} ?\)

Along, straight wire lies along the \(z\) -axis and carries a \(4.00-\mathrm{A}\) current in the \(+z\) -direction. Find the magnetic field (magnitude and direction) produced at the following points by a \(0.500-\mathrm{mm}\) segment of the wire centered at the origin: (a) \(x=2.00 \mathrm{m}, y=0\) , \(z=0 ;(b) x=0, y=2.00 \mathrm{m}, z=0 ;(\mathrm{c}) x=2.00 \mathrm{m}, y=2.00 \mathrm{m}\) \(z=0 ;(\mathrm{d}) x=0, y=0, z=2.00 \mathrm{m}\)

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