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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 \(0.60 \hat{\mathrm{i}}-0.80 \mathrm{j}\). (This unit vector gives the orientation of the magnetic dipole moment vector.) If the loop is located in a uniform magnetic field given by \(\vec{B}=(0.50 \mathrm{~T}) \hat{\mathrm{i}}+(0.20 \mathrm{~T}) \hat{\mathrm{k}}\), find (a) the torque on the loop (in unit-vector notation) and (b) the orientation energy of the loop.

Short Answer

Expert verified
Torque: \( \vec{\tau} = 0.000482 \hat{i} + 0.000644 \hat{k} \) N·m; Energy: \( U = -0.001205 \) J.

Step by step solution

01

Calculate the Magnetic Dipole Moment

The magnetic dipole moment \( \vec{\mu} \) of a loop is given by the formula \( \vec{\mu} = I \cdot A \cdot \hat{n} \), where \( I \) is the current, \( A \) is the area of the loop, and \( \hat{n} \) is the unit vector parallel to \( \vec{\mu} \). First, calculate the area \( A = \pi r^2 = \pi (0.08 \, \text{m})^2 \approx 0.0201 \, \text{m}^2 \). Then, \( \vec{\mu} = (0.20 \, \text{A}) \times 0.0201 \, \text{m}^2 \times (0.60 \hat{i} - 0.80 \hat{j}) = (0.00241 \hat{i} - 0.00322 \hat{j}) \, \text{A} \cdot \text{m}^2 \).
02

Calculate the Torque on the Loop

The torque \( \vec{\tau} \) on a current loop is given by \( \vec{\tau} = \vec{\mu} \times \vec{B} \). Using the cross product formula, \( \vec{\tau} = |\hat{i} \hat{j} \hat{k}| \begin{vmatrix} 0.00241 & -0.00322 & 0 \ 0.50 & 0 & 0.20 \end{vmatrix} = (-0.00322)(0.20)\hat{k} + (0.00241)(0.20)\hat{i} - (0.00241)(0)\hat{j} \). Simplifying, \( \vec{\tau} = 0.000482 \hat{i} + 0.000644 \hat{k} \, \text{N} \cdot \text{m} \).
03

Calculate the Orientation Energy

The potential energy \( U \) of a magnetic dipole in a magnetic field is given by \( U = -\vec{\mu} \cdot \vec{B} \). Calculate the dot product as \( \vec{\mu} \cdot \vec{B} = (0.00241 \hat{i} - 0.00322 \hat{j}) \cdot (0.50 \hat{i} + 0.20 \hat{k}) = (0.00241)(0.50) + (-0.00322)(0) \). Simplifying, \( U = -0.001205 \, \text{J} \).
04

Present Final Results

Combine the results: the torque on the loop is \( \vec{\tau} = 0.000482 \hat{i} + 0.000644 \hat{k} \, \text{N} \cdot \text{m} \), and the orientation energy is \( U = -0.001205 \, \text{J} \).

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Key Concepts

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

Torque on a Loop
When a circular loop of wire carrying an electric current is placed in a magnetic field, it experiences a torque. The torque \( \vec{\tau} \) is the rotational force that the magnetic field exerts on the loop. It can be calculated using the formula:
  • \( \vec{\tau} = \vec{\mu} \times \vec{B} \)
Here, \( \vec{\mu} \) is the magnetic dipole moment, and \( \vec{B} \) is the magnetic field vector. The cross product \( \times \) means that \( \vec{\tau} \) will be perpendicular to both \( \vec{\mu} \) and \( \vec{B} \).

In our exercise, we find the torque by calculating the cross product of the magnetic dipole moment \( \vec{\mu} = (0.00241\hat{i} - 0.00322\hat{j}) \) and the magnetic field \( \vec{B} = (0.50\hat{i} + 0.20\hat{k}) \). The resulting torque is \( \vec{\tau} = 0.000482 \hat{i} + 0.000644 \hat{k} \, \text{N} \cdot \text{m} \).

This torque determines how the loop will attempt to align itself within the magnetic field, rotating in the direction of the torque until equilibrium is reached.
Circular Loop of Wire
A circular loop of wire is a common configuration in electromagnetism due to its symmetry and simplicity in analysis. The characteristics of a circular loop allow us to study fundamental electromagnetic concepts like magnetic dipole moment and induced currents. For this exercise, we'll consider the circular loop of wire that has a 8.0 cm (radius).

The loop carries a steady current, which gives rise to a magnetic dipole moment. The dipole moment \( \vec{\mu} \) is a vector that reflects the magnetic strength and orientation of the loop. It's given by the product of the current \( I \), the area \( A \) of the loop, and the direction as specified by a unit vector.

The area \( A \) of this circular loop is computed using the formula \( A = \pi r^2 \), which results in approximately 0.0201 m² for a radius of 8.0 cm. This area is crucial in determining the magnitude of the magnetic dipole moment, showcasing the direct relationship between the loop's size and its magnetic properties.
Magnetic Field
Magnetic fields exert forces on moving charges and magnetic materials within them. In this exercise, we have a uniform magnetic field represented by the vector \( \vec{B} = (0.50 \mathrm{~T}) \hat{\mathrm{i}} + (0.20 \mathrm{~T}) \hat{\mathrm{k}} \).

The magnetic field is a vector field around a magnet where magnetic forces can be observed. It is defined by both magnitude and direction, thus influencing the torque on a loop of wire. The strength of a magnetic field is measured in Tesla (T).
  • The component \( 0.50 \mathrm{~T} \hat{\mathrm{i}} \) indicates the field's strength along the x-direction.
  • The component \( 0.20 \mathrm{~T} \hat{\mathrm{k}} \) indicates the field's strength along the z-direction.

Understanding these components helps in comprehensively determining how a magnetic loop will react within the field, calculating both the torque and potential energy interactions. Here, the torque calculated gives an insight into how the loop will behave and align due to the imposed magnetic field.

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Most popular questions from this chapter

A proton travels through uniform magnetic and electric fields. The magnetic field is \(\vec{B}=-3.25 \hat{\mathrm{i}} \mathrm{mT}\). At one instant the velocity of the proton is \(\vec{v}=2000 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}\). At that instant and in unit-vector notation, what is the net force acting on the proton if the electric field is (a) \(4.00 \mathrm{k} \mathrm{V} / \mathrm{m}\), (b) \(-4.00 \mathrm{k} \mathrm{V} / \mathrm{m}\), and (c) \(4.00 \hat{\mathrm{i}} \mathrm{V} / \mathrm{m}\) ?

An alpha particle travels at a velocity \(\vec{v}\) of magnitude \(620 \mathrm{~m} / \mathrm{s}\) 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 \(6.6 \times 10^{-27} \mathrm{~kg}\) ) 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 \(\vec{F}_{B} ?\) (c) Does the speed of the particle increase, decrease, or remain the same?

(a) 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 \(0.500 \mathrm{~m} ?(\mathrm{~b})\) What is the period of the motion?

Figure 28-22 shows a metallic block, with its faces parallel to coordinate axes. The block is in a uniform magnetic field of magnitude \(0.020 \mathrm{~T}\). One edge length of the block is \(32 \mathrm{~cm}\); the block is not drawn to scale. The block is moved at \(3.5 \mathrm{~m} / \mathrm{s}\) parallel to each axis, in turn, and the resulting potential difference \(V\) that appears across the block is measured. With the motion parallel to the \(y\) axis, \(V=12 \mathrm{mV}\); with the motion parallel to the \(z\) axis, \(V=18 \mathrm{mV}\); with the motion parallel to the \(x\) axis, \(V=0\). What are the block lengths (a) \(d_{x}\), (b) \(d_{y}\), and (c) \(d_{z} ?\)

A circular coil of 500 turns has a radius of \(1.90 \mathrm{~cm}\). (a) Calculate the current that results in a magnetic dipole moment of magnitude \(1.90 \mathrm{~A} \cdot \mathrm{m}^{2}\). (b) Find the maximum magnitude of the torque that the coil, carrying this current, can experience in a uniform \(35.0 \mathrm{mT}\) magnetic field.

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