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Chapter 21: Problem 75

The maximum torque experienced by a coil in a 0.75-T magnetic field is \(8.4 \times 10^{-4} \mathrm{N} \cdot \mathrm{m} .\) The coil is circular and consists of only one turn. The current in the coil is 3.7 A. What is the length of the wire from which the coil is made?

Short Answer

Expert verified
The length of the wire is approximately 0.0617 m or 6.17 cm.

Step by step solution

01

Understanding Torque in a Magnetic Field

The torque (\( \tau \)) experienced by a coil in a magnetic field can be calculated using the formula \( \tau = B \cdot I \cdot A \cdot \sin(\theta) \), where \( B \) is the magnetic field, \( I \) is the current, \( A \) is the area of the coil, and \( \theta \) is the angle between the magnetic field direction and the normal to the coil. The maximum torque occurs when \( \sin(\theta) = 1 \), i.e., when \( \theta \) is 90 degrees.
02

Calculating the Area of the Coil

We can rearrange the formula to find the area of the coil: \( A = \frac{\tau}{B \cdot I} \). Substituting the given values, \( \tau = 8.4 \times 10^{-4} \, \mathrm{N \cdot m} \), \( B = 0.75 \, \mathrm{T} \), and \( I = 3.7 \, \mathrm{A} \), we find \( A = \frac{8.4 \times 10^{-4}}{0.75 \cdot 3.7} \approx 3.04 \times 10^{-4} \, \mathrm{m^2} \).
03

Expressing Area in Terms of Radius

Since the coil is circular and has one turn, the area \( A \) can also be expressed as \( \pi r^2 \), where \( r \) is the radius. So, \( \pi r^2 = 3.04 \times 10^{-4} \, \mathrm{m^2} \).
04

Solving for the Radius

To solve for \( r \), rearrange the formula: \( r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{3.04 \times 10^{-4}}{\pi}} \approx 9.82 \times 10^{-3} \, \mathrm{m} \).
05

Calculating the Circumference

The length of the wire used to make the coil is equal to the circumference of the circle, which is \( 2\pi r \). Thus, substituting the value of \( r \), we find the length \( L = 2\pi (9.82 \times 10^{-3}) \approx 6.17 \times 10^{-2} \, \mathrm{m} \).

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Key Concepts

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

Torque
Torque is a measure of how much a force acting on an object causes that object to rotate. When it comes to a coil in a magnetic field, torque plays a crucial role. The formula for the torque (\( \tau \)) experienced by a coil in a magnetic field is given by \( \tau = B \cdot I \cdot A \cdot \sin(\theta) \). Here:
  • \( B \) represents the magnetic field strength.
  • \( I \) denotes the current flowing through the coil.
  • \( A \) is the area of the coil.
  • \( \theta \) is the angle between the magnetic field and the normal to the coil's plane.
The torque is maximized when \( \theta \) is 90 degrees since \( \sin(90^\circ) = 1 \). This means the plane of the coil is perpendicular to the magnetic field direction. Understanding this helps in scenarios where maximizing rotational effect is desired, like in electric motors.
Circular Coil
A circular coil is a loop made of conducting material, shaped into a circle, through which electric current can flow. This particular problem involves a coil made from a single turn of wire. The area of the circular coil, which significantly affects the calculation, is given by the formula \( A = \pi r^2 \). Knowing the area is essential as it determines the coil's ability to produce torque when subjected to a magnetic field.
  • The coil's radius \( r \) can be calculated from its area using the formula \( r = \sqrt{\frac{A}{\pi}} \).
  • The wire length, or circumference, of the coil is \( 2\pi r \).
Thus, understanding the properties of a circular coil helps solve problems related to magnetic forces and electricity, as it engages both the physical shape and geometrical properties in any application.
Current
Current is the flow of electric charge through a conductor, typically measured in amperes (A). In this context, the current flowing through the coil is a vital factor in determining how much torque the coil experiences when placed in a magnetic field. By combining the current \( I \) with the magnetic field and the area of the coil, one can compute the torque using \( \tau = B \cdot I \cdot A \cdot \sin(\theta) \).
  • Current affects the magnitude of the torque directly, meaning the more current that flows through the coil, the greater the exerted torque for a given magnetic field strength and coil area.
  • It represents the availability of moving charges that interact with the magnetic field, thus contributing to the force exerted on the coil.
Controlling the current, therefore, becomes a method of adjusting the torque output in practical applications like electromagnetic devices.
Magnetic Field Strength
Magnetic field strength, denoted as \( B \) and measured in Tesla (T), describes the intensity of the magnetic field at a given point. In this problem, the strength of the magnetic field affects the torque produced by the coil. The torque calculation is directly proportional to the magnetic field strength, given by \( \tau = B \cdot I \cdot A \cdot \sin(\theta) \).
  • The magnetic field strength determines how strongly the charged particles in the current-carrying coil will be pushed or pulled, thereby influencing the torque.
  • A stronger magnetic field results in a higher torque for the same coil area and current.
Understanding the concept of magnetic field strength is key to analyzing how magnetic forces affect current-carrying conductors, crucial in the design and operation of electrical systems and devices.

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Most popular questions from this chapter

A magnetic field has a magnitude of \(1.2 \times 10^{-3} \mathrm{T}\), and an electric field has a magnitude of \(4.6 \times 10^{3} \mathrm{N} / \mathrm{C}\). Both fields point in the same direction. A positive \(1.8 \mu \mathrm{C}\) charge moves at a speed of \(3.1 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.

A small compass is held horizontally, the center of its needle a distance of \(0.280 \mathrm{m}\) directly north of a long wire that is perpendicular to the earth's surface. When there is no current in the wire, the compass needle points due north, which is the direction of the horizontal component of the earth's magnetic field at that location. This component is parallel to the earth's surface. When the current in the wire is \(25.0 \mathrm{A}\), the needle points \(23.0^{\circ}\) east of north. (a) Does the current in the wire flow toward or away from the earth's surface? (b) What is the magnitude of the horizontal component of the earth's magnetic field at the location of the compass?

A wire carries a current of 0.66 A. This wire makes an angle of \(58^{\circ}\) with respect to a magnetic field of magnitude \(4.7 \times 10^{-5} \mathrm{T}\). The wire experiences a magnetic force of magnitude \(7.1 \times 10^{-5} \mathrm{N}\). What is the length of the wire?

A long, cylindrical conductor is solid throughout and has a radius \(R\). Electric charges flow parallel to the axis of the cylinder and pass uniformly through the entire cross section. The arrangement is, in effect, a solid tube of current \(I_{0} .\) The current per unit cross-sectional area (i.e., the current density) is \(I_{0} /\left(\pi R^{2}\right) .\) Use Amp猫re's law to show that the magnetic field inside the conductor at a distance \(r\) from the axis is \(\mu_{0} I_{0} r /\left(2 \pi R^{2}\right)\). (Hint: For a closed path, use a circle of radius r perpendicular to and centered on the axis. Note that the current through any surface is the area of the surface times the current density.)

Two charged particles move in the same direction with respect to the same magnetic field. Particle 1 travels three times faster than particle 2. However, each particle experiences a magnetic force of the same magnitude. Find the ratio \(\left|q_{1}\right| / q_{2} \mid\) of the magnitudes of the charges.
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