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Chapter 27: Problem 38

An electromagnet produces a magnetic field of 0.550 \(\mathrm{T}\) in a cylindrical region of radius 2.50 \(\mathrm{cm}\) between its poles. A straight wire carrying a current of 10.8 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force is exerted on the wire?

Short Answer

Expert verified
The magnitude of the force exerted on the wire is 0.297 N.

Step by step solution

01

Identify the Relevant Formula

The force exerted on a current-carrying wire in a magnetic field is given by the formula \( F = I L B \sin \theta \), where \( F \) is the force, \( I \) is the current, \( L \) is the length of the wire in the magnetic field, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the wire and the magnetic field. In this scenario, \( \theta = 90^\circ \), so \( \sin \theta = 1 \).
02

Determine the Wire Length Affected

Since the wire passes through the center of the cylindrical region, and the force is only exerted within this region, we consider the diameter of the region as the length of the wire affected by the magnetic field. The diameter \( L \) of the region is twice the radius: \( L = 2 \times 2.50 \text{ cm} = 5.00 \text{ cm} = 0.050 \text{ m} \).
03

Substitute Known Values into the Formula

Now that we have all the values, we substitute them into the formula: \( F = 10.8 \text{ A} \times 0.050 \text{ m} \times 0.550 \text{ T} \times 1 = 0.297 \text{ N} \).
04

Calculate the Magnitude of the Force

Perform the multiplication to find the force: \( F = 0.297 \text{ N} \). This is the magnitude of the force exerted on the wire.

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Key Concepts

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

Electromagnetism
Electromagnetism is a fundamental force of nature that encompasses the forces of electricity and magnetism. These two forces are intimately connected and are explained through the theory of electromagnetism. The interplay between electricity and magnetism is visibly demonstrated in scenarios involving current-carrying wires and magnetic fields. A moving electric charge, like current in a wire, generates its own magnetic field. This field interacts with external magnetic fields, resulting in observable physical forces. This fundamental principle is the cornerstone of technologies like motors and generators. In layman's terms, if you have an electric current close to a magnet, you鈥檝e got both an electric and a magnetic aspect forming a single force we call electromagnetism.
Magnetic Field Strength
Magnetic field strength, also known as magnetic flux density, refers to the concentration of magnetic field lines in a given area. It is a vector quantity, measured in Tesla (T), and signifies the force that a magnetic field exerts on moving charges or other magnetic materials within its influence.
  • A stronger magnetic field means that magnetic field lines are more densely packed, leading to a greater force on current-carrying wires.
  • In the problem at hand, a magnetic field strength of 0.550 T is crucial since it determines the force applied to the wire.
Understanding magnetic field strength is essential, especially when calculating the force exerted on current-carrying wires within a magnetic field. To grasp this fully, picture magnetic field lines as invisible lines of force around magnets: the closer these lines are, the stronger the field is in that area.
Current-Carrying Wire
A current-carrying wire is simply a wire through which electrical current is flowing. When placed in a magnetic field, this wire experiences a magnetic force. This force is the result of interaction between the magnetic field generated by the current and the external magnetic field.
  • The direction and magnitude of this force are dictated by the well-known right-hand rule and the formula: \( F = I L B \sin \theta \).
  • In our example, the wire is affected by a magnetic field when the current passes through, creating a perpendicular orientation to the field (\(\theta = 90^\circ\)).
The force on the wire is directly proportional to the current and the length of the wire within the field. Hence, increasing either the current or the wire length interacting with the field will amplify the force. A practical understanding of these interactions allows for effective design and operation of electrical devices like transformers and inductors.

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Most popular questions from this chapter

A particle with a charge of \(-1.24 \times 10^{-8} \mathrm{C}\) is moving with instantaneous velocity \(\vec{v}=\left(4.19 \times 10^{4} \mathrm{m} / \mathrm{s}\right) \hat{\imath}+(-3.85 \times\) \(10^{4} \mathrm{m} / \mathrm{s} ) \hat{\boldsymbol{J}}\) . What is the force exerted on this particle by a mag- netic field (a) \(\vec{\boldsymbol{B}}=(1.40 \mathrm{T}) \hat{\boldsymbol{i}}\) and \((\mathrm{b}) \vec{\boldsymbol{B}}=(1.40 \mathrm{T}) \hat{\boldsymbol{k}} ?\)

An electron moves at \(2.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) through a region in which there is a magnetic field of unspecified direction and magnitude \(7.40 \times 10^{-2} \mathrm{T}\) (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

In a 1.25 -T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50\(\mu \mathrm{C}\) and initially moving northward at 4.75 \(\mathrm{km} / \mathrm{s}\) is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle.

Paleoclimate. Climatologists can determine the past temperature of the earth by comparing the ratio of the isotope oxygen-18 to the isotope oxygen-16 in air trapped in ancient ice sheets, such as those in Greenland. In one method for separating these isotopes, a sample containing both of them is first singly ionized (one electron is removed) and then accelerated from rest through a potential difference \(V\) . This beam then enters a magnetic field \(B\) at right angles to the field and is bent into a quarter-circle. A particle detector at the end of the path measures the amount of each isotope. (a) Show that the separation \(\Delta r\) of the two isotopes at the detector is given by $$\Delta r = \frac { \sqrt { 2 e V } } { e B } \left( \sqrt { m _ { 18 } } - \sqrt { m _ { 16 } } \right)$$ where \(m _ { 16 }\) and \(m _ { 18 }\) are the masses of the two oxygen isotopes, (b) The measured masses of the two isotopes are \(2.66 \times\) \(10 ^ { - 26 } \mathrm { kg } \left( ^ { 16 } \mathrm { O } \right)\) and \(2.99 \times 10 ^ { - 26 } \mathrm { kg } ( 8 \mathrm { O } ) .\) If the magnetic field is \(0.050 \mathrm { T } ,\) what must be the accelerating potential \(V\) so that these two isotopes will be separated by 4.00\(\mathrm { cm }\) at the detector?

A singly charged ion of \(^{7} \mathrm{Li}\) (an isotope of lithium) has a mass of \(1.16 \times 10^{-26} \mathrm{kg}\) . It is accelerated through a potential dif- ference of 220 \(\mathrm{V}\) and then enters a magnetic field with magnitude 0.723 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?
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