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Chapter 20: Problem 35

\(\cdot\) The plane of a 5.0 \(\mathrm{cm}\) by 8.0 \(\mathrm{cm}\) rectangular loop of wire is parallel to a 0.19 T magnetic field, and the loop carries a cur- rent of 6.2 \(\mathrm{A}\) . (a) What torque acts on the loop? (b) What is the magnetic moment of the loop?

Short Answer

Expert verified
Torque is 0.004712 Nm. Magnetic moment is 0.0248 Am sup{2}.

Step by step solution

01

Understand the Problem

We need to find the torque acting on a rectangular loop of wire in a magnetic field and the magnetic moment of the loop. The given values are: Width of the loop = 5.0 cm, Length of the loop = 8.0 cm, Magnetic field strength ( B ) = 0.19 T, Current ( I ) = 6.2 A.
02

Calculate the Area of the Loop

The area (A) of the rectangular loop is calculated as:\[ A = ext{Length} \times ext{Width} \]Convert dimensions to meters, 5.0 cm = 0.05 m and 8.0 cm = 0.08 m. Thus,\[ A = 0.08 \, m \times 0.05 \, m = 0.004 \, m^2 \]
03

Find the Magnetic Moment

The magnetic moment (\mu) of the loop is given by:\[ \mu = I \times A \]Substitute the known values:\[ \mu = 6.2 \, A \times 0.004 \, m^2 = 0.0248 \, Am^2 \]
04

Calculate the Torque on the Loop

The torque (\tau) acting on the loop is given by:\[ \tau = \mu \times B \times \sin(\theta) \]Since the plane of the loop is parallel to the magnetic field, \(\theta = 90^\circ\) and \( \sin(90^\circ) = 1 \). Thus,\[ \tau = 0.0248 \, Am^2 \times 0.19 \, T \times 1 = 0.004712 \, Nm \]

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Key Concepts

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

Torque on a Loop
The concept of torque on a current-carrying loop in a magnetic field is essential to grasp when dealing with electromagnetic systems. Torque, in this context, is the tendency of a force to rotate an object about an axis. For a rectangular loop with sides of lengths 5.0 cm and 8.0 cm, placed parallel to a magnetic field of 0.19 Tesla, we must consider how the magnetic field interacts with the current flowing through the wire loop.

Torque (\( \tau \)) on a loop is derived from the equation:\[ \tau = \mu \times B \times \sin(\theta) \]where \( \mu \) is the magnetic moment, \( B \) is the magnetic field, and \( \theta \) is the angle between \( \mu \) and \( B \).
  • If the loop is parallel to the magnetic field, as given, \( \theta \) is \( 90^\circ \)
  • This makes \( \sin(90^\circ) = 1 \), simplifying the torque calculation
Ultimately, this gives us an intuitive explanation. The torque attempts to align the loop's magnetic moment with the applied magnetic field. This is a critical factor in devices like electric motors, where torque translates into mechanical rotation.
Rectangular Loop in Magnetic Field
A rectangular loop in a magnetic field presents an intriguing scenario for examining concepts such as magnetic flux, torque, and loop behavior under electromagnetic forces. A loop, with a width of 5.0 cm and a length of 8.0 cm, is placed in a uniform magnetic field of 0.19 Tesla. The orientation becomes crucial here.

Since the loop is perfectly parallel to the field, it experiences the maximum interaction as governed by its physical dimensions and current.
  • Area Calculation: The area \( A \) of the loop can be calculated using \( A = \text{Length} \times \text{Width} \).
  • Convert this to meters for consistency, leading to an area of \( 0.004 \, m^2 \).
  • The loop's effective exposure to the magnetic field influences its electromagnetic properties like torque and magnetic moment.
This setup is foundational in electromagnetic applications, as it highlights how altering angles or dimensions affects the resulting forces on the loop.
Current-Carrying Loop
An understanding of a current-carrying loop in a magnetic field is pivotal to analyzing systems such as generators and electric motors. Consider a loop that carries a 6.2 A current as given in the problem. This current influences the magnetic moment and, consequently, the electromagnetic interactions with the magnetic field.

To calculate the magnetic moment \( \mu \), we utilize the formula:\[ \mu = I \times A \]
  • Here, \( I \) is the current (6.2 A), and \( A \) is the area (0.004 \, m^2).
  • This yields a magnetic moment of \( 0.0248 \, Am^2 \).
The magnetic moment serves as a measure of the strength and orientation of the loop's magnetic field. When placed in an external magnetic field, this moment interacts dynamically, seeking alignment with the external field.
This principle is what enables motors to convert electrical energy into mechanical movement, manifesting the practical importance of understanding current-carrying loops in magnetic fields.

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Most popular questions from this chapter

\(\bullet\) A beam of protons traveling at 1.20 \(\mathrm{km} / \mathrm{s}\) enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field in a direction perpendicular to its original direction (Fig. \(20.60 ) .\) The beam travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field?

A solenoid having 165 turns and a cross-sectional area of 6.75 \(\mathrm{cm}^{2}\) carries a current of 1.20 A. If it is placed in a uniform 1.12 T magnetic field, find the torque this field exerts on the solenoid if its axis is oriented (a) perpendicular to the field, (b) parallel to the field, (c) at 35. \(0^{\circ}\) with the field.

\bullet A 150 g ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a 125 vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal 0.250 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.

\bullet A single circular current loop 10.0 \(\mathrm{cm}\) in diameter carries a 2.00 A current. (a) What is the magnetic field at the center of this loop? (b) Suppose that we now connect 1000 of these loops in series within a 500 \(\mathrm{cm}\) length to make a solenoid 500 \(\mathrm{cm}\) long. What is the magnetic field at the center of this solenoid? Is it 1000 times the field at the center of the loop in part (a)? Why or why not?

\(\bullet\) A 3.25 g bullet picks up an electric charge of 1.65\(\mu C\) as it travels down the barrel of a rifle. It leaves the barrel at a speed of 425 \(\mathrm{m} / \mathrm{s}\) , traveling perpendicular to the earth's magnetic field, which has a magnitude of \(5.50 \times 10^{-4} \mathrm{T} .\) Calculate (a) the magnitude of the magnetic force on the bullet and (b) the magnitude of the bullet's acceleration due to the magnetic force at the instant it leaves the rifle barrel.
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