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A circular loop of wire having a radius of \(8.0 \mathrm{~cm}\) carries a current of \(0.20 \mathrm{~A}\). A vector of unit length and parallel to the dipole moment \(\vec{\mu}\) of the loop is given by \(0.60 \hat{\mathrm{i}}-0.80 \hat{\mathrm{j}} .\) (This unit vector gives the orientation of the magnetic dipole moment vector.) If the loop is located in a uniform magnetic field given by \(\vec{B}=(0.25 \mathrm{~T}) \hat{\mathrm{i}}+(0.30 \mathrm{~T}) \hat{\mathrm{k}}\), find \((\mathrm{a})\) the torque on the loop (in unitvector notation) and (b) the orientation energy of the loop.

Short Answer

Expert verified
Torque on the loop: \((-0.000965 \hat{\mathrm{i}} + 0.000724 \hat{\mathrm{j}}) \; \mathrm{Nm}\), Orientation energy: \(-0.000603 \; \mathrm{J}\).

Step by step solution

01

Calculate the Magnetic Dipole Moment

The magnetic dipole moment \( \vec{\mu} \) of a loop of wire is given by the formula: \( \vec{\mu} = I \vec{A} \), where \( I \) is the current and \( \vec{A} \) is the area vector. The magnitude of \( \vec{A} \) can be calculated by \( \pi r^2 \), where \( r \) is the radius of the loop.Given that the current \( I = 0.20 \; \mathrm{A} \) and \( r = 0.08 \; \mathrm{m} \), we have:\[\|\vec{A}\| = \pi \times (0.08)^2 = \pi \times 0.0064 \approx 0.0201 \; \mathrm{m}^2\]Hence, the magnitude of \( \vec{\mu} \) is:\[\|\vec{\mu}\| = I \times \|\vec{A}\| = 0.20 \times 0.0201 = 0.00402 \; \mathrm{Am}^2\]
02

Determine the Direction of the Dipole Moment

The direction of \( \vec{\mu} \) is given by the vector \( 0.60 \hat{\mathrm{i}} - 0.80 \hat{\mathrm{j}} \). Therefore, we write the magnetic dipole moment vector as:\[\vec{\mu} = 0.00402 (0.60 \hat{\mathrm{i}} - 0.80 \hat{\mathrm{j}}) = (0.002412 \hat{\mathrm{i}} - 0.003216 \hat{\mathrm{j}}) \; \mathrm{Am}^2\]
03

Calculate the Torque on the Loop

The torque \( \vec{\tau} \) on a magnetic dipole in a magnetic field is calculated using the cross product \( \vec{\tau} = \vec{\mu} \times \vec{B} \). \( \vec{B} \) is given as \( (0.25 \hat{\mathrm{i}} + 0.30 \hat{\mathrm{k}}) \; \mathrm{T} \).Using the cross product:\[\vec{\tau} = (0.002412 \hat{\mathrm{i}} - 0.003216 \hat{\mathrm{j}}) \times (0.25 \hat{\mathrm{i}} + 0.30 \hat{\mathrm{k}}) \]Breaking it down using the vector identity \( \hat{\mathrm{i}} \times \hat{\mathrm{k}} = \hat{\mathrm{j}} \) and \( \hat{\mathrm{j}} \times \hat{\mathrm{k}} = -\hat{\mathrm{i}} \), we calculate:\[\vec{\tau} = (0.002412 \times 0.30) \hat{\mathrm{j}} - (0.003216 \times 0.30) \hat{\mathrm{i}} \]\[\vec{\tau} = (0.000724 \hat{\mathrm{j}} - 0.000965 \hat{\mathrm{i}}) \; \mathrm{Nm}\]
04

Calculate the Orientation Energy

The potential energy \( U \) of a magnetic dipole in a magnetic field is given by \( U = - \vec{\mu} \cdot \vec{B} \), the dot product of \( \vec{\mu} \) and \( \vec{B} \).\[U = - (0.002412 \hat{\mathrm{i}} - 0.003216 \hat{\mathrm{j}}) \cdot (0.25 \hat{\mathrm{i}} + 0.30 \hat{\mathrm{k}})\]Using the dot product identity \( \hat{\mathrm{i}} \cdot \hat{\mathrm{i}} = 1 \) and \( \hat{\mathrm{j}} \cdot \hat{\mathrm{k}} = 0 \), we calculate:\[U = - (0.002412 \times 0.25)\]\[U = -0.000603 \; \mathrm{J}\]

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Key Concepts

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

Torque on a Magnetic Dipole
Understanding how torque affects a magnetic dipole begins with the concept of magnetic dipole moments. When a loop of wire carrying current is placed within a magnetic field, it experiences a force that causes it to rotate, known as torque. This is similar to how a compass needle rotates to align with the Earth's magnetic field.

The torque \( \vec{\tau} \) acting on the magnetic dipole can be described by the equation:
  • \( \vec{\tau} = \vec{\mu} \times \vec{B} \)
Here, \( \vec{\mu} \) represents the magnetic dipole moment of the loop, and \( \vec{B} \) is the magnetic field vector.
The cross product in this equation means that the torque is perpendicular to both \( \vec{\mu} \) and \( \vec{B} \). This results in a twisting effect that attempts to align the magnetic moment with the field direction.

For this exercise, the dipole moment vector \( \vec{\mu} \) was calculated using the direction \( 0.60 \hat{\mathrm{i}} - 0.80 \hat{\mathrm{j}} \) and multiplied by the scalar magnitude of the dipole moment \( 0.00402 \, \mathrm{Am}^2 \). The magnetic field \( \vec{B} \) was given as \( 0.25 \hat{\mathrm{i}} + 0.30 \hat{\mathrm{k}} \) T, leading to a torque \( \vec{\tau} \) calculated via the vector cross product, resulting in a new vector that dictates how the loop will rotate in response to the magnetic field.
Magnetic Field Interaction
Magnetic field interactions can be understood by examining how magnetic dipoles, like our loop of wire, interact with external fields. Each part of the dipole—positive and negative magnetic poles—experiences force in a magnetic field, affecting how the dipole aligns.

In this exercise, the wire loop's magnetic dipole interacts with a uniform magnetic field \( \vec{B} \) given by its vector components, creating complex but interesting dynamics.
  • The field tries to align \( \vec{\mu} \) parallel to itself.
  • The torque calculated before is a result of these field interactions.
These forces aid in explaining phenomena in electromagnetism, like how electric motors function, by discussing how arrangement in a field creates mechanical energy output. Understanding this concept is essential in many technological applications where magnetic fields are used to perform work, such as generators and maglev trains.

The vector nature of magnetic fields means they have both direction and magnitude, and this dual property is what causes the interesting re-alignment and movement of dipoles placed within them. Overall, the strength and orientation of \( \vec{B} \) significantly dictate how objects like our loop respond when inside such fields.
Orientation Energy in Magnetism
When a magnetic dipole, such as a loop of current, is positioned in a magnetic field, it possesses a certain amount of potential energy known as orientation energy. This energy measures the likely tendency of the dipole to align either with or against the direction of the field.

Mathematically, the potential energy \( U \) associated with a magnetic dipole in a magnetic field is denoted by:
  • \( U = - \vec{\mu} \cdot \vec{B} \)
This expression involves the dot product of the magnetic moment \( \vec{\mu} \) and the magnetic field \( \vec{B} \).
The negative sign indicates that the system is in its lowest potential energy state when the dipole aligns with the magnetic field. Unlike the torque, which indicates the tendency to rotate, the orientation energy signifies the stability of this arrangement.

In practical terms, it implies that minimal external energy will align the dipole with the magnetic direction, while more energy is required for it to align perpendicularly or against the field. Consequently, aligning a dipole opposite to a magnetic field requires energy input and results in a positive potential energy state, indicating lower stability and a tendency for the system to revert to its minimal energy state.

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Most popular questions from this chapter

What uniform magnetic field, applied perpendicular to a beam of electrons moving at \(1.30 \times 10^{6} \mathrm{~m} / \mathrm{s}\), is required to make the electrons travel in a circular arc of radius \(0.350 \mathrm{~m} ?\)

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A stationary circular wall clock has a face with a radius of 15 \(\mathrm{cm}\). Six turns of wire are wound around its perimeter; the wire carries a current of \(2.0 \mathrm{~A}\) in the clockwise direction. The clock is located where there is a constant, uniform external magnetic field of magnitude \(70 \mathrm{mT}\) (but the clock still keeps perfect time). At exactly 1:00 P.M., the hour hand of the clock points in the direction of the external magnetic field. (a) After how many minutes will the minute hand point in the direction of the torque on the winding due to the magnetic field? (b) Find the torque magnitude.

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An alpha particle travels at a velocity \(\vec{v}\) of magnitude \(550 \mathrm{~m} / \mathrm{s}\) through a uniform magnetic field \(\vec{B}\) of magnitude \(0.045 \mathrm{~T}\). (An alpha particle has a charge of \(+3.2 \times 10^{-19} \mathrm{C}\) and a mass of \(6.6 \times\) \(\left.10^{-27} \mathrm{~kg} .\right)\) The angle between \(\vec{v}\) and \(\vec{B}\) is \(52^{\circ} .\) What is the magnitude of (a) the force \(\vec{F}_{B}\) acting on the particle due to the field and(b) the acceleration of the particle due to \(\vec{F}_{B} ?(\mathrm{c})\) Does the speed of the particle increase, decrease, or remain the same?

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