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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 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 of the coil when it is rotated 180 degrees is -2.42 J.

Step by step solution

01

Compute Initial Potential Energy

Calculate the initial potential energy when the magnetic moment is antiparallel to the magnetic field using the formula for potential energy in a magnetic field, \(-\mathbf{M} \cdot \mathbf{B}\). Here, \(M = 1.45 \mathrm{~A} \cdot \mathrm{m}^{2}\) and \(B = 0.835 \mathrm{~T}\), hence the potential energy is \(-1.45 \mathrm{~A} \cdot \mathrm{m}^{2} \times -0.835 \mathrm{~T} = 1.21 \mathrm{~J}\).
02

Compute Final Potential Energy

Calculate the final potential energy when the magnetic moment is parallel to the magnetic field. The magnetic field direction is in line with magnetic moment. Therefore, \(M = 1.45 \mathrm{~A} \cdot \mathrm{m}^{2}\) and \(B = 0.835 \mathrm{~T}\), the potential energy is \(-1.45 \mathrm{~A} \cdot \mathrm{m}^{2} \times 0.835 \mathrm{~T} = -1.21 \mathrm{~J}\).
03

Change in Potential Energy

Finally, calculate the change in potential energy, which is the difference between the final and initial potential energy. Therefore, \(-1.21 \mathrm{~J} - 1.21 \mathrm{~J} = -2.42 \mathrm{~J}\).

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Key Concepts

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

Magnetic Moment
The magnetic moment is a vector quantity that represents the magnetic strength and orientation of a magnet or other object that produces a magnetic field. Specifically, it's the torque that a magnetic object experiences in the presence of an external magnetic field. Imagine holding a compass; the needle, when freely suspended, aligns with Earth's magnetic field because it has a magnetic moment.

In our exercise, the coil's magnetic moment of \(1.45 \text{ A} \text{ m}^2\) tells us about its magnetic 'power' and which way it points. When it is antiparallel to the magnetic field, it's like the compass needle pointing south while Earth's field points north; it's in a high-energy state, storing potential energy that can be released if the coil is allowed to flip. This flipping to a parallel state with the field would then take the system to a lower energy state.
Uniform Magnetic Field
A uniform magnetic field is constant in magnitude and direction at all points in the area it covers. Such consistency is ideal for examining basic magnetic interactions without the complexity of varying field strength or direction. Take a flashlight beam shining evenly on a wall as an analogy; it illuminates the surface uniformly, just as a uniform magnetic field 'covers' an area consistently in terms of magnetic influence.

In our example, the coil is within a uniform 0.835 T (Tesla) magnetic field, offering us a stable environment to analyze how the coil's potential energy changes when its magnetic moment flips direction. Understanding how objects behave in uniform fields is often helpful before moving to non-uniform, more complex fields.
Potential Energy Calculation
Potential energy represents stored energy, which can be calculated in different contexts like gravity, spring systems, and, as in our case, magnetic systems. For a magnet in a uniform magnetic field, potential energy depends on the strength of both the magnetic moment and the field, and their relative orientations.

We calculate it using the formula \(-\textbf{M} \times \textbf{B}\), where \(M\) is the magnetic moment, and \(B\) is the magnetic field. The negative sign indicates that the lowest potential energy (most stable state) occurs when \(M\) and \(B\) are in the same direction. When we compute changes in this energy, we assess the system's energetic 'journey' from one configuration to another, as seen in the exercise when flipping the coil affects its energy by \(-2.42 \text{ J}\). Understanding this change helps us predict how the system might behave dynamically under different conditions.

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Most popular questions from this chapter

A straight, \(2.5 \mathrm{~m}\) wire carries a typical household current of \(1.5 \mathrm{~A}\) (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

An electron in the beam of a cathode-ray tube is accelerated by a potential difference of \(2.00 \mathrm{kV}\). Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius \(0.180 \mathrm{~m} .\) What is the magnitude of the field?

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at \(1.50 \mathrm{~km} / \mathrm{s}\) in the \(+x\) -direction experiences a force of \(2.25 \times 10^{-16} \mathrm{~N}\) in the \(+y\) -direction, and an electron moving at \(4.75 \mathrm{~km} / \mathrm{s}\) in the \(\begin{array}{llll}-z \text { -direction } & \text { experiences } & \text { a } & \text { force } & \text { of } & 8.50 \times 10^{-16} \mathrm{~N} & \text { in the }\end{array}\) \(+y\) -direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the \(-y\) -direction at \(3.20 \mathrm{~km} / \mathrm{s} ?\)

The magnetic poles of a small cyclotron produce a magnetic field with magnitude \(0.85 \mathrm{~T}\). The poles have a radius of \(0.40 \mathrm{~m},\) which is the maximum radius of the orbits of the accelerated particles. (a) What is the maximum energy to which protons \(\left(q=1.60 \times 10^{-19} \mathrm{C}\right.\), \(m=1.67 \times 10^{-27} \mathrm{~kg}\) ) can be accelerated by this cyclotron? Give your answer in electron volts and in joules. (b) What is the time for one revolution of a proton orbiting at this maximum radius? (c) What would the magnetic-field magnitude have to be for the maximum energy to which a proton can be accelerated to be twice that calculated in part (a)? (d) For \(B=0.85 \mathrm{~T}\), what is the maximum energy to which alpha particles \(\left(q=3.20 \times 10^{-19} \mathrm{C}, m=6.64 \times 10^{-27} \mathrm{~kg}\right)\) can be accelerated by this cyclotron? How does this compare to the maximum energy for protons?

An electron is moving in the \(x y\) -plane. If at time \(t\) a magnetic field \(B=0.200 \mathrm{~T}\) in the \(+z\) -direction exerts a force on the electron equal to \(F=5.50 \times 10^{-18} \mathrm{~N}\) in the \(-y\) -direction, what is the velocity (magnitude and direction) of the electron at this instant?

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