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A coil with magnetic moment 1.45 \(\mathrm{A} \cdot \mathrm{m}^{2}\) is oriented initially with its magnetic moment antiparallel to a uniform \(0.835-\mathrm{T}\) magnetic field. What is the change in potential energy of the coil when it is rotated \(180^{\circ}\) so that its magnetic moment is parallel to the field?

Short Answer

Expert verified
The change in potential energy is \(-2.4205\,\text{J}\).

Step by step solution

01

Understand the Concept of Potential Energy in a Magnetic Field

The potential energy (U) of a magnetic dipole moment (\(\mu\)) in a magnetic field (\(B\)) is given by \[ U = -\mu \cdot B \cdot \cos(\theta) \] where \(\theta\) is the angle between the magnetic moment and the magnetic field. A rotation from antiparallel to parallel changes \(\theta\) from \(180^\circ\) to \(0^\circ\).
02

Calculate Initial Potential Energy

Initially, the coil's magnetic moment is antiparallel to the field, which means \(\theta = 180^\circ\). The initial potential energy is \[ U_\text{initial} = -\mu \cdot B \cdot \cos(180^\circ) \] Since \(\cos(180^\circ) = -1\), it simplifies to: \[ U_\text{initial} = \mu \cdot B \]
03

Calculate Final Potential Energy

After rotating the coil \(180^\circ\), the magnetic moment is parallel to the field, so \(\theta = 0^\circ\). The final potential energy is \[ U_\text{final} = -\mu \cdot B \cdot \cos(0^\circ) \] Since \(\cos(0^\circ) = 1\), it simplifies to: \[ U_\text{final} = -\mu \cdot B \]
04

Determine the Change in Potential Energy

The change in potential energy \(\Delta U\) as the coil rotates is calculated by the formula: \[ \Delta U = U_\text{final} - U_\text{initial} \] Substituting the expressions for \(U_\text{final}\) and \(U_\text{initial}\): \[ \Delta U = (-\mu \cdot B) - (\mu \cdot B) = -2 \cdot \mu \cdot B \]
05

Plug in Known Values

Given \(\mu = 1.45\,\mathrm{A} \cdot \mathrm{m}^2\) and \(B = 0.835\,\mathrm{T}\), substitute into the expression for \(\Delta U\):\[ \Delta U = -2 \cdot 1.45 \cdot 0.835 = -2.4205 \,\text{J} \] Thus, the change in potential energy is \(-2.4205\,\text{J}\).

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Key Concepts

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

Magnetic Dipole Moment
A magnetic dipole moment is like a tiny bar magnet inside an object or particle, which has a north and south pole. This characteristic of an object, often a current loop such as a coil, describes how strongly it can get magnetized in the presence of a magnetic field. The magnetic dipole moment (\(\mu\)) is a vector quantity, which means it has both a magnitude and direction.
Its unit is \(\mathrm{A} \cdot \mathrm{m}^2\), which represents an ampere-loop of area. Think of it as a measure of how effective an object is at creating a magnetic field in its surrounding space.
For example, in this exercise, a coil with a magnetic dipole moment of \(1.45 \mathrm{A} \cdot \mathrm{m}^2\) is being analyzed to understand the change in energy when aligned with a magnetic field.
Some important points about magnetic dipole moments:
  • Magnetic dipole moments arise from either the spin of charged particles or the movement of electrons in atoms.
  • Larger dipole moments imply stronger magnetic behavior in an external field.
  • The direction of the magnetic dipole moment vector is perpendicular to the plane of the loop in the coil.
Understanding this concept is essential for analyzing the effects of magnetic fields on electrical components.
Magnetic Field Strength
Magnetic field strength, often denoted as \(B\), represents the force exerted by a magnetic field on a moving charged particle or magnetic dipole. It is a vector field, which means it has both a direction and a magnitude. The unit of magnetic field strength is the tesla (\(\mathrm{T}\)), which is equivalent to one newton per ampere-meter.
When a magnetic field interacts with a magnetic dipole moment, it can exert forces and torques on the dipole. This is highly relevant in scenarios like our coil rotating in the magnetic field because the interaction changes its potential energy.
Key aspects of magnetic field strength include:
  • The direction of \(B\) is defined as the direction a north magnetic pole would be pushed.
  • It varies in strength, determined by the source of the magnetic field, such as permanent magnets or electric currents.
  • Uniform magnetic fields, such as the one described in our exercise (\(0.835 \mathrm{T}\)), apply a consistent force throughout a defined region.
Understanding magnetic field strength is crucial for grasping how magnetic fields interact with objects possessing magnetic dipole moments.
Potential Energy Calculation
The potential energy of a magnetic dipole in a magnetic field is influenced by its orientation relative to the field. This energy is given by the formula: \[ U = -\mu \cdot B \cdot \cos(\theta) \] where \(\mu\) is the magnetic dipole moment, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the magnetic dipole and the magnetic field direction.
The formula indicates how energy changes based on orientation. When a coil is rotated from anti-parallel (\(180^\circ\)) to parallel (\(0^\circ\)) with the field, the change in angle results in a shift in potential energy.
Let's break down the process:
  • Initially, when the dipole is antiparallel (\(\cos(180^\circ) = -1\)), the potential energy is maximized as \(\mu \times B\).
  • When aligned parallel to the field (\(\cos(0^\circ) = 1\)), the potential energy is minimized at \(-\mu \times B\).
  • The change, given by \(-2 \mu B\), signifies that energy decreases as the coil aligns with the field.
This understanding explains the calculated energy shift of \(-2.4205\) Joules for the coil, highlighting the impact of position within a magnetic field.

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Most popular questions from this chapter

A circular area with a radius of 6.50 \(\mathrm{cm}\) lies in the \(x y\) -plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field \(B=0.230 \mathrm{T}\) (a) in the \(+z\) -direction; \((b)\) at an angle of \(53.1^{\circ}\) from the \(+z\) -direction; \((\mathrm{c})\) in the \(+y\) -direction?

An Electromagnetic Rail Gun. A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{\boldsymbol{B}}\) fills the region between the rails (Fig. 27.63\()\) . (a) Find the magnitude and direction of the net force on the conducting bar Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\) , find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\) . (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \((11.2 \mathrm{km} / \mathrm{s}) .\) Let \(B=0.50 \mathrm{T}, \quad I=2.0 \times 10^{3} \mathrm{A}\) \(m=25 \mathrm{kg},\) and \(L=50 \mathrm{cm} .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts \((a)\) and \((b)\) , even though gravity plays an important role in an actual launch in space.

(a) An 16 nucleus (charge \(+8 e )\) moving horizontally from west to east with a speed of 500 \(\mathrm{km} / \mathrm{s}\) experiences a magnetic force of 0.00320 \(\mathrm{nN}\) vertically downward. Find the magnitude and direction of the weakest magnetic field required to produce this force. Explain how this same force could be caused by a larger magnetic field. (b) An electron moves in a uniform, horizontal, 2.10 - T magnetic field that is toward the west. What must the magnitude and direction of the minimum velocity of the electron be so that the magnetic force on it will be 4.60 \(\mathrm{pN}\) , vertically upward? Explain how the velocity could be greater than this minimum value and the force still have this same magnitude and direction.

A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2\(\Omega\) . When the motor is running at full load on a \(120-\mathrm{V}\) line, the emf in the rotor is 105 \(\mathrm{V}\) . (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

A particle of charge \(q>0\) is moving at speed \(v\) in the \(+z\) -direction through a region of uniform magnetic field \(\overrightarrow{\boldsymbol{B}} .\) The magnetic force on the particle is \(\overrightarrow{\boldsymbol{F}}=F_{0}(3 \hat{\imath}+4 \hat{\jmath}),\) where \(\boldsymbol{F}_{0}\) is a positive constant. (a) Determine the components \(B_{x}, B_{y},\) and \(B_{z},\) or at least as many of the three components as is possible from the information given. (b) If it is given in addition that the magnetic field has magnitude \(6 F_{0} / q v,\) determine as much as you can about the remaining components of \(\overrightarrow{\boldsymbol{B}}\) .

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