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Two concentric, circu- endlar wire loops, of radii \(r_{1}=20.0 \mathrm{~cm}\) and \(r_{2}=30.0 \mathrm{~cm}\), are located in an \(x y\) plane; each carries a clockwise current of \(7.00 \mathrm{~A}\) (Fig. \(28-47\) ). (a) Find the magnitude of the net magnetic dipole moment of the system. (b) Repeat for reversed current in the inner loop.

Short Answer

Expert verified
(a) 0.91\pi Am²; (b) 0.35\pi Am².

Step by step solution

01

Understanding Magnetic Dipole Moment

The magnetic moment of a current-carrying loop is given by the product of the current and the area enclosed by the loop. Mathematically, the magnetic moment \( \mu \) is expressed as \( \mu = I \times A \), where \( I \) is the current and \( A \) is the area of the loop.
02

Calculating the Area of Each Loop

The area \( A \) of a circle is given by \( A = \pi r^2 \). For the inner loop with radius \( r_1 = 20.0 \text{ cm} = 0.2 \text{ m} \), the area is \( A_1 = \pi (0.2)^2 = 0.04\pi \text{ m}^2 \). For the outer loop with radius \( r_2 = 30.0 \text{ cm} = 0.3 \text{ m} \), the area is \( A_2 = \pi (0.3)^2 = 0.09\pi \text{ m}^2 \).
03

Calculating the Magnetic Dipole Moment for Each Loop

Using \( \mu = I \times A \), for the inner loop, \( \mu_1 = 7.00 \text{ A} \times 0.04\pi \text{ m}^2 = 0.28\pi \text{ Am}^2 \). For the outer loop, \( \mu_2 = 7.00 \text{ A} \times 0.09\pi \text{ m}^2 = 0.63\pi \text{ Am}^2 \).
04

Finding the Net Magnetic Dipole Moment

Since both loops have clockwise current, their magnetic moments add when considering direction. So, the net magnetic dipole moment of the system is \( \mu_{net} = \mu_1 + \mu_2 = 0.28\pi + 0.63\pi = 0.91\pi \text{ Am}^2 \).
05

Reversed Current in the Inner Loop

If the current in the inner loop is reversed, \( \mu_1 \) becomes negative, i.e., \( -0.28\pi \text{ Am}^2 \). Thus, the net magnetic dipole moment is \( \mu_{net} = \mu_2 - \mu_1 = 0.63\pi - 0.28\pi = 0.35\pi \text{ Am}^2 \).

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Key Concepts

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

Current-Carrying Loop
A current-carrying loop acts like a tiny magnet. It creates a magnetic dipole moment, which is a measure of its magnetic strength and orientation. Imagine you have a loop of wire, like the ones in the exercise. When current flows through it, a magnetic field is generated around it. This is much like how electric flow creates a magnetic field in a straight wire but, in the case of a loop, the effect is more pronounced.
The magnetic moment of the loop is a result of both the current flowing through the wire and the area the loop covers. In simple terms, the bigger the loop or the stronger the current, the larger the magnetic moment. This concept is essential in understanding how we calculate the net magnetic moment in the given exercise.
  • The direction of the magnetic moment depends on the direction of the current. Clockwise and counterclockwise currents will produce magnetic moments pointing in opposite directions.
  • This principle is very useful in technologies like electric motors and generators.
Area of a Circle
The area of a circle is a fundamental concept in geometry and physics, especially when dealing with circular loops. It is determined using the formula \( A = \pi r^2 \), where \( A \) represents the area, \( \pi \) is a constant (approximately 3.14159), and \( r \) is the radius of the circle.
In the exercise, two different loops are considered, each with its own radius.
  • For the inner loop, with a radius of 20 cm, we convert it to meters (0.2 m) for calculations. The resulting area is \( A_1 = \pi (0.2)^2 = 0.04\pi \text{ m}^2 \).
  • For the outer loop, with a radius of 30 cm, the converted radius is 0.3 m, and the area becomes \( A_2 = \pi (0.3)^2 = 0.09\pi \text{ m}^2 \).
The area is crucial because it directly influences the magnetic dipole moment. Larger areas result in larger magnetic moments when the current is constant.
Net Magnetic Moment
The net magnetic moment refers to the combined magnetic effect of a system of current-carrying loops. To understand this, we consider both the direction and magnitude of each loop's magnetic dipole moment.
In the exercise, both loops carry a clockwise current. This means their magnetic moments aid each other, contributing to a larger net magnetic moment. Mathematically, this is computed as:
  • For the inner loop, \( \mu_1 = 7.00 \text{ A} \times 0.04\pi \text{ m}^2 = 0.28\pi \text{ Am}^2 \).
  • For the outer loop, \( \mu_2 = 7.00 \text{ A} \times 0.09\pi \text{ m}^2 = 0.63\pi \text{ Am}^2 \).
  • The total, or net, magnetic moment when currents are aligned is \( \mu_{net} = 0.28\pi + 0.63\pi = 0.91\pi \text{ Am}^2 \).
However, if one loop's current is reversed, its magnetic moment becomes negative, partially cancelling the effect of the other loop. Understanding how to calculate and appreciate this balance helps in many applications, from magnetic resonance imaging (MRI) to studying magnetic fields in physics.

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Most popular questions from this chapter

In a nuclear experiment a proton with kinetic energy \(1.0 \mathrm{MeV}\) moves in a circular path in a uniform magnetic field. What energy must (a) an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) and (b) a deuteron \((q=+e, m=2.0 \mathrm{u})\) have if they are to circulate in the same circular path?

A proton moves through a uniform magnetic field given by \(\vec{B}=(10 \hat{\mathrm{i}}-20 \hat{\mathrm{j}}+30 \hat{\mathrm{k}}) \mathrm{mT}\). At time \(t_{1}\), the proton has a velocity given by \(\vec{v}=v_{x} \hat{\mathrm{i}}+v_{y} \hat{\mathrm{j}}+(2.0 \mathrm{~km} / \mathrm{s}) \hat{\mathrm{k}}\) and the magnetic force on the proton is \(\vec{F}_{B}=\left(4.0 \times 10^{-17} \mathrm{~N}\right) \hat{\mathrm{i}}+\left(2.0 \times 10^{-17} \mathrm{~N}\right) \hat{\mathrm{j}}\). At that instant, what are (a) \(v_{x}\) and (b) \(v_{y} ?\)

A wire of length \(25.0 \mathrm{~cm}\) carrying a current of \(4.51 \mathrm{~mA}\) is to be formed into a circular coil and placed in a uniform magnetic field \(\vec{B}\) of magnitude \(5.71 \mathrm{mT}\). If the torque on the coil from the field is maximized, what are (a) the angle between \(\vec{B}\) and the coil's magnetic dipole moment and (b) the number of turns in the coil? (c) What is the magnitude of that maximum torque?

A circular wire loop of radius \(15.0 \mathrm{~cm}\) carries a current of \(2.60 \mathrm{~A}\). It is placed so that the normal to its plane makes an angle of \(41.0^{\circ}\) with a uniform magnetic field of magnitude \(12.0 \mathrm{~T}\). (a) Calculate the magnitude of the magnetic dipole moment of the loop. (b) What is the magnitude of the torque acting on the loop?

An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The particle has a charge of \(q=+2 e\) and a mass of \(4.00 \mathrm{u}\), where \(\mathrm{u}\) is the atomic mass unit, with \(1 \mathrm{u}=1.661 \times 10^{-27} \mathrm{~kg} .\) Suppose an alpha particle travels in a circular path of radius \(4.50 \mathrm{~cm}\) in a uniform magnetic field with \(B=1.20 \mathrm{~T}\). Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.

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