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Chapter 28: Problem 80

A circular wire of radius \(5.0 \mathrm{~cm}\) has a current of \(3.0 \mathrm{~A}\) flowing in it. The wire is placed in a uniform magnetic field of \(5.0 \mathrm{mT}.\) a) Determine the maximum torque on the wire. b) Determine the range of the magnetic potential energy of the wire.

Short Answer

Expert verified
Answer: The maximum torque exerted on the circular wire is \(0.00011781 \mathrm{~N\cdot m}\). The range of the magnetic potential energy of the wire is from \(-0.00011781 \mathrm{~J}\) to \(0.00011781 \mathrm{~J}\).

Step by step solution

01

Calculate the area of the circular wire

The area of a circle is given by the formula: \(A = \pi * r^2\) where \(r\) is the radius of the circle. In this case, the radius is \(5.0 \mathrm{~cm} = 0.05 \mathrm{~m}\). So, \(A = \pi * (0.05)^2 = 0.007854 \mathrm{~m^2}\).
02

Calculate the maximum torque on the wire

The maximum torque on a closed current loop in a magnetic field is given by the formula: \(\tau = I * A * B * sin(\theta)\) where \(I\) is the current, \(A\) is the loop area, \(B\) is the magnetic field, and \(\theta\) is the angle between the normal to the plane of the loop and the magnetic field. The maximum torque occurs when the angle \(\theta = 90掳\), so \(sin(\theta) = 1\). Substituting the given values and the calculated area, \(\tau = (3.0 \mathrm{~A})(0.007854 \mathrm{~m^2})(5.0 * 10^{-3}\mathrm{~T})(1)= 0.00011781 \mathrm{~N\cdot m}\) So, the maximum torque on the wire is \(0.00011781 \mathrm{~N\cdot m}\).
03

Calculate the minimum and maximum magnetic potential energy of the wire

The magnetic potential energy of a closed current loop in a magnetic field is given by the formula: \(U = -\tau * cos(\theta)\) where \(\tau\) is the torque and \(\theta\) is the angle between the normal to the plane of the loop and the magnetic field. For minimum potential energy, \(\theta = 0掳\) (loop normal is parallel to the field), so \(cos(\theta) = 1\). Therefore, \(U_\text{min} = -\tau * cos(0掳)\) \(U_\text{min} = -(0.00011781 \mathrm{~N\cdot m})(1) = -0.00011781 \mathrm{~J}\) For maximum potential energy, \(\theta = 180掳\) (loop normal is anti-parallel to the field), so \(cos(\theta) = -1\). Therefore, \(U_\text{max} = -\tau * cos(180掳)\) \(U_\text{max} = -(0.00011781 \mathrm{~N\cdot m})(-1) = 0.00011781 \mathrm{~J}\) The range of the magnetic potential energy of the wire is from \(-0.00011781 \mathrm{~J}\) to \(0.00011781 \mathrm{~J}\).

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Key Concepts

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

Magnetic Field
A magnetic field is an invisible force that surrounds magnetic materials and electric currents. It is a vector field, meaning it has both magnitude and direction.
In the context of a wire loop, the magnetic field can exert a force on the current flowing through the loop, which in turn affects the torque and potential energy of the system.
A uniform magnetic field, like the one in our circular wire problem, has the same strength and direction at every point in space, making calculations more straightforward.
Current Loop
A current loop refers to current flowing through a closed loop of wire, which can interact with a magnetic field.
A circular current loop, like the one in our exercise, is a common configuration where the circle allows for symmetrical calculations.
The flow of electric current in the loop creates its own magnetic field, influencing how it interacts with external fields.
  • In this exercise, a current of 3.0 A flows through the circular loop.
  • This interaction is what allows us to calculate properties like torque and potential energy.
Magnetic Potential Energy
Magnetic potential energy (\( U \)) is the energy a current loop possesses due to its orientation in a magnetic field.
This energy depends on the torque (\( \tau \)) and the cosine of the angle (\( \theta \)) between the loop's normal and the magnetic field.
Mathematically:
  • \( U = -\tau \cdot \cos(\theta) \)
When the loop is aligned with the field (\( \theta = 0^\circ \)), the potential energy is at its minimum. Conversely, when it is anti-aligned (\( \theta = 180^\circ \)), the energy reaches its maximum.
This concept is essential for understanding the range of energy in a magnetic field.
Circular Wire
A circular wire loop is a common configuration for studying magnetic effects due to its simplicity and symmetry.
In this exercise, the circular loop has a radius of 5.0 cm.
  • The area of the loop is calculated using \( A = \pi r^2 \), where \( r = 0.05 \text{ m} \).
  • The resulting area is \( 0.007854 \text{ m}^2 \).
The circular shape ensures that any changes in its properties, like torque and potential energy, are uniformly distributed across the loop.
Maximum Torque
Maximum torque (\( \tau \)) on a current loop occurs when the loop is oriented such that the sinusoidal component of the interaction is at its peak.
This happens at \( \theta = 90^\circ \), where \( \sin(\theta) = 1 \).
The torque is given by:
  • \( \tau = I \cdot A \cdot B \cdot \sin(\theta) \)
where \( I \) is the current, \( A \) is the area of the loop, and \( B \) is the magnetic field.
  • In this exercise, the maximum torque is calculated as \( 0.00011781 \text{ N}\cdot\text{m} \).
Understanding maximum torque is crucial for designing and optimizing systems like electric motors that rely on magnetic forces.

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Most popular questions from this chapter

A long, straight wire carrying a 2.00-A current lies along the \(x\) -axis. A particle with charge \(q=-3.00 \mu \mathrm{C}\) passes parallel to the \(z\) -axis through the point \((x, y, z)=(0,2,0)\). Where in the \(x y\) -plane should another long, straight wire be placed so that there is no magnetic force on the particle at the point where it crosses the plane?

Two long, parallel wires separated by a distance, \(d\), carry currents in opposite directions. If the left-hand wire carries a current \(i / 2,\) and the right-hand wire carries a current \(i\), determine where the magnetic field is zero.

The magnetic force cannot do work on a charged particle since the force is always perpendicular to the velocity. How then can magnets pick up nails? Consider two parallel current-carrying wires. The magnetic fields cause attractive forces between the wires, so it appears that the magnetic field due to one wire is doing work on the other wire. How is this explained? a) The magnetic force can do no work on isolated charges; this says nothing about the work it can do on charges confined in a conductor. b) Since only an electric field can do work on charges, it is actually the electric fields doing the work here. c) This apparent work is due to another type of force.

What is a good rule of thumb for designing a simple magnetic coil? Specifically, given a circular coil of radius \(\sim 1 \mathrm{~cm},\) what is the approximate magnitude of the magnetic field, in gauss per amp per turn? (Note: \(1 \mathrm{G}=0.0001 \mathrm{~T}\).) a) \(0.0001 \mathrm{G} /(\mathrm{A}\) -turn \()\) b) \(0.01 \mathrm{G} /(\) A-turn \()\) c) \(1 \mathrm{G} /(\mathrm{A}\) -turn \()\) d) \(100 \mathrm{G} /(\mathrm{A}\) -turn \()\)

A long, straight wire carries a current of 2.5 A. a) What is the strength of the magnetic field at a distance of \(3.9 \mathrm{~cm}\) from the wire? b) If the wire still carries \(2.5 \mathrm{~A}\), but is used to form a long solenoid with 32 turns per centimeter and a radius of \(3.9 \mathrm{~cm}\) what is the strength of the magnetic field at the center of the solenoid?
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