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Chapter 20: Problem 55

A circular metal loop is \(22 \mathrm{~cm}\) in diameter. (a) How large a current must flow through this metal so that the magnetic field at its center is equal to the earth's magnetic field of \(0.50 \times 10^{-4} \mathrm{~T} ?\) (b) Show how the loop should be oriented so that it can cancel the earth's magnetic field at its center.

Short Answer

Expert verified
(a) 8.75 A. (b) Orient the loop horizontally, opposing Earth's field direction.

Step by step solution

01

Understanding the Formula for Magnetic Field at the Center of a Loop

The magnetic field (B) at the center of a current-carrying loop is given by the formula \( B = \frac{\mu_0 I}{2R} \), where \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \, \text{T} \, \text{m/A}) \), \( I \) is the current, and \( R \) is the radius of the loop.
02

Calculating the Radius of the Loop

Given the diameter of the loop is 22 cm, we first convert it to meters: \( 22 \, \text{cm} = 0.22 \, \text{m} \). Therefore, the radius \( R = \frac{0.22}{2} = 0.11 \, \text{m} \).
03

Rearranging the Formula to Solve for Current

To find the current \( I \), rearrange the formula to \( I = \frac{2BR}{\mu_0} \).
04

Plugging in the Values and Solving for Current

Substitute the known values into the equation: \( B = 0.50 \times 10^{-4} \, \text{T} \), \( R = 0.11 \, \text{m} \), and \( \mu_0 = 4\pi \times 10^{-7} \, \text{T} \, \text{m/A} \): \[ I = \frac{2 \times 0.50 \times 10^{-4} \times 0.11}{4\pi \times 10^{-7}}. \] Evaluating gives \( I \approx 8.75 \, \text{A} \).
05

Orientation of the Loop to Cancel Earth's Magnetic Field

To cancel the Earth's magnetic field at the center of the loop, the loop should be oriented such that its magnetic field is in the opposite direction to the Earth's magnetic field. Place the loop horizontally with the north pole of its magnetic field pointing opposite to the Earth's magnetic field. This alignment ensures the net magnetic field at the center is zero.

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Key Concepts

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

Permeability of Free Space
The permeability of free space, also known as the magnetic constant, is a fundamental physical constant that describes the ability of a vacuum to sustain a magnetic field. The value of this constant is given by \(\mu_0 = 4\pi \times 10^{-7} \, \text{T} \, \text{m/A}\). This constant is crucial in equations describing magnetic fields, particularly when calculating the magnetic field created by electric currents in a wire or a loop.
  • Permeability implies how a material affects the magnetic fields inside it.
  • In a vacuum or free space, this permeability is standardized as \(\mu_0\).
By understanding this constant, we can better comprehend the magnetic interactions at play in different environments.
Current Calculation
Calculating the current required to create a specific magnetic field involves rearranging the formula for the magnetic field at the center of a loop. This process requires understanding and applying the formula: \[ B = \frac{\mu_0 I}{2R} \]where:
  • \(B\) is the magnetic field we aim for, such as the Earth's magnetic field strength \(0.50 \times 10^{-4} \, \text{T}\).
  • \(\mu_0\) is the permeability of free space.
  • \(I\) is the current, which we need to find.
  • \(R\) is the radius of the loop, half of the given diameter.
The formula can be rearranged to solve for the current:\[ I = \frac{2BR}{\mu_0} \]Understanding this step is crucial for finding the necessary current to achieve a certain magnetic field at the loop's center.
Magnetic Field Cancellation
Magnetic field cancellation is the act of orienting a magnetic loop to counteract an existing magnetic field, such as the Earth's. This means aligning the loop so that the magnetic field it generates is opposite to the field you are trying to cancel.
  • The loop's magnetic field must have the same magnitude but opposite direction to the targeted field.
  • Achieving zero net magnetic field at the loop's center requires precise orientation.
By aligning as described, the magnetic field from the loop effectively cancels the Earth's magnetic field, resulting in no net field at the loop's center.
Loop Orientation
The orientation of a loop has significant impacts on the magnetic field it produces. When trying to cancel the Earth's magnetic field, the loop should be placed in a specific manner so that its magnetic north points opposite to the Earth's.
  • Position the loop horizontally in a flat plane.
  • Ensure the north pole of the loop鈥檚 magnetic field faces in the opposite direction to the Earth's field.
This orientation will direct the loop's magnetic field against the Earth's field, cancelling out at the loop鈥檚 center and achieving the desired effect of magnetic neutrality.

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Most popular questions from this chapter

A particle having a mass of \(0.195 \mathrm{~g}\) carries a charge of \(-2.50 \times 10^{-8} \mathrm{C} .\) The particle is given an initial horizontal northward velocity of \(4.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\). What are the magnitude and direction of the minimum magnetic field that will balance the earth's gravitational pull on the particle?

You want to produce a magnetic field of magnitude \(5.50 \times 10^{-4} \mathrm{~T}\) at a distance of \(0.040 \mathrm{~m}\) from a long, straight wire's center. (a) What current is required to produce this field? (b) With the current found in part (a), how strong is the magnetic field \(8.00 \mathrm{~cm}\) from the wire's center?

A \(150 \mathrm{~g}\) ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a \(125 \mathrm{~m}\) vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal \(0.250 \mathrm{~T}\) magnetic field directed from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

A straight vertical wire carries a current of 1.20 A downward in a region between the poles of a large electromagnet where the field strength is \(0.588 \mathrm{~T}\) and is horizontal. What are the magnitude and direction of the magnetic force on a \(1.00 \mathrm{~cm}\) section of this wire if the magnetic- field direction is (a) toward the east, (b) toward the south, (c) \(30.0^{\circ}\) south of west?

An electron travels into a 0.3 T magnetic field perpendicular to its path, where it moves in a circular arc of diameter \(0.020 \mathrm{~m}\). What is the kinetic energy of the electron?
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