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Chapter 29: Problem 63

At a certain location in the Philippines, Earth's magnetic field of \(39 \mu \mathrm{T}\) is horizontal and directed due north. Suppose the net field is zero exactly \(2.0 \mathrm{~cm}\) above a long, straight, horizontal wire that carries a constant current. What are the (a) magnitude and (b) direction of the current?

Short Answer

Expert verified
The current is 3.9 A, flowing east.

Step by step solution

01

Understand the Problem

We need to find the magnitude and direction of the electric current in a wire that will result in the net magnetic field being zero at a point above the wire. The Earth's magnetic field is given, and it acts to the north horizontally.
02

Use Biot-Savart Law

The magnetic field due to a long straight wire with current is given by Biot-Savart Law as \( B = \frac{\mu_0 I}{2\pi r} \), where \( I \) is the current, \( r \) is the distance from the wire, and \( \mu_0 \) is the permeability of free space \(4\pi \times 10^{-7} \; \text{Tm/A} \).
03

Set Up Equation for Zero Net Field

The net magnetic field at the point must be zero. This means the magnetic field due to the wire, \( B = \frac{\mu_0 I}{2\pi r} \), must be equal and opposite to the Earth's magnetic field of \(39 \mu\mathrm{T}\). Therefore, we set up the equation: \[ \frac{\mu_0 I}{2\pi r} = 39 \times 10^{-6} \; \text{T} \]
04

Solve for the Current \( I \)

We rearrange the equation to find \( I \): \[ I = \frac{2\pi r \times 39 \times 10^{-6}}{\mu_0} \]Substitute \( r = 0.02 \; \text{m} \) and \( \mu_0 = 4\pi \times 10^{-7} \; \text{T m/A} \) into the equation:\[ I = \frac{2\pi \times 0.02 \times 39 \times 10^{-6}}{4\pi \times 10^{-7}} \]
05

Calculate the Current

Simplifying the equation:\[ I = \frac{0.02 \times 39 \times 10^{-6}}{2 \times 10^{-7}} = \frac{0.78 \times 10^{-6}}{2 \times 10^{-7}} = 3.9 \; \text{A} \]
06

Determine the Direction of the Current

To ensure the magnetic fields cancel out, the direction of the magnetic field created by the wire must be opposite to the Earth's field. By the right-hand rule, if the field of the Earth is north, the current must flow towards the east to create a southward magnetic field.

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Key Concepts

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

Biot-Savart Law
The Biot-Savart Law is a fundamental principle used to calculate the magnetic field generated by a current-carrying wire. It is especially useful for determining the magnetic field around long straight wires. According to this law, the magnetic field \( B \) created by a current \( I \) at a certain distance \( r \) from a wire is given by the formula: \[ B = \frac{\mu_0 I}{2\pi r} \] Where:
  • \( \mu_0 \) is the permeability of free space, valued at \( 4\pi \times 10^{-7} \; \text{Tm/A} \).
  • \( I \) is the current flowing through the wire.
  • \( r \) is the perpendicular distance from the wire to the point of interest.
This equation shows that the strength of the magnetic field decreases with increasing distance from the wire. This inverse relationship with distance is an essential feature of the Biot-Savart Law and helps understand how magnetic fields disperse around current-carrying wires.
Right-Hand Rule
The Right-Hand Rule is an intuitive method for determining the direction of a magnetic field created by a current. It can be applied easily with your hand:
  • Point your thumb in the direction of the current flow.
  • Curl your fingers naturally around the wire.
  • Your fingers now point in the direction of the magnetic field loops circling the wire.
This visualization helps in finding the orientation of the magnetic field. For the given exercise, the Earth's magnetic field is directed north, hence using the right-hand rule, the current must be directed east to oppose the northern Earth's magnetic field. This rule is a crucial tool when considering how different currents interact to influence magnetic field directions.
Earth's Magnetic Field
Earth's magnetic field is a natural phenomenon that extends from the Earth's interior out into space, where it interacts with the solar wind. In the context of the exercise, the Earth's magnetic field is horizontal and directed due north at the given location.
  • The magnetic field strength in the exercise is given as \(39 \mu \mathrm{T}\).
  • It is important to note that this field affects all magnetic and electromagnetic activities on Earth.
The Earth's magnetic field is an essential aspect when measuring and calculating other magnetic fields, like that of a wire. It acts as a baseline or reference point for observing how additional magnetic forces might influence the total magnetic environment around us. Understanding the Earth's magnetic field helps in comprehending phenomena such as compass directions and auroras.

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Most popular questions from this chapter

In a particular region there is a uniform current density of \(18 \mathrm{~A} / \mathrm{m}^{2}\) in the positive \(z\) direction. What is the value of \(\oint \vec{B} \cdot d \vec{s}\) when that line integral is calculated along a closed path consisting of the three straight-line segments from \((x, y, z)\) coordinates \((4 d, 0,0)\) to \((4 d, 3 d, 0)\) to \((0,0,0)\) to \((4 d, 0,0)\), where \(d=20 \mathrm{~cm} ?\)

A long solenoid has 123 turns/cm and carries current \(i\). An electron moves within the solenoid in a circle of radius \(2.30 \mathrm{~cm}\) perpendicular to the solenoid axis. The speed of the electron is \(0.0187 c\) ( \(c=\) speed of light). Find the current \(i\) in the solenoid.

Figure 29-25a shows a length of wire carrying a current \(i\) and bent into a circular coil of one turn. In Fig. 29-25b the same length of wire has been bent to give a coil of two turns, each of half the original radius. (a) If \(B_{a}\) and \(B_{b}\) are the magnitudes of the magnetic fields at the centers of the two coils, what is the ratio \(B_{b} / B_{a}\) ? (b) What is the ratio \(\mu_{b} / \mu_{a}\) of the dipole moment magnitudes of the coils?

A solenoid that is \(95.0 \mathrm{~cm}\) long has a radius of \(2.00 \mathrm{~cm}\) and a winding of 1500 turns; it carries a current of \(3.60 \mathrm{~A}\). Calculate the magnitude of the magnetic field inside the solenoid.

An electron is shot into one end of a solenoid. As it enters the uniform magnetic field within the solenoid, its speed is \(500 \mathrm{~m} / \mathrm{s}\) and its velocity vector makes an angle of \(30^{\circ}\) with the central axis of the solenoid. The solenoid carries \(4.0 \mathrm{~A}\) and has 8000 turns along its length. How many revolutions does the electron make along its helical path within the solenoid by the time it emerges from the solenoid's opposite end? (In a real solenoid, where the field is not uniform at the two ends, the number of revolutions would be slightly less than the answer here.)
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