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A straight, vertical wire carries a current of 2.60 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude \(B =\) 0.588 T and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00-cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) 30.0\(^\circ\) south of west?

Short Answer

Expert verified
(a) 0.0153 N north, (b) 0.0153 N west, (c) 0.0132 N northeast.

Step by step solution

01

Understand the Problem

The problem involves calculating the magnetic force on a vertical wire in a magnetic field. The wire carries a current of 2.60 A, and we need to find the force on a 1.00 cm (0.01 m) segment of the wire. The magnetic force is affected by the direction of the magnetic field.
02

Use the Force Formula

The force on a current-carrying wire in a magnetic field is given by the formula: \[ F = I L B \sin(\theta) \]where:- \( F \) is the force in Newtons,- \( I = 2.60 \) A is the current,- \( L = 0.01 \) m is the length of the wire segment,- \( B = 0.588 \) T is the magnetic field,- \( \theta \) is the angle between the wire and the magnetic field.
03

Solve for Magnetic Field East

If the magnetic field is directed east, and given that the current is downward, the angle \( \theta \) between the wire and the magnetic field is 90 degrees (since they are perpendicular). Thus, \( \sin(90^\circ) = 1 \). Substitute into the formula:\[ F = 2.60 \times 0.01 \times 0.588 \times 1 = 0.015288 \] N.Using the right-hand rule, the direction of the force will be north.
04

Solve for Magnetic Field South

For a magnetic field directed south, the direction makes the angle \( \theta = 90^\circ \) with the vertical wire. Thus, \( \sin(90^\circ) = 1 \). Substitute into the formula:\[ F = 2.60 \times 0.01 \times 0.588 \times 1 = 0.015288 \] N.Using the right-hand rule, the force direction will be west.
05

Solve for Magnetic Field 30 Degrees South of West

For a magnetic field directed 30 degrees south of west, the direction makes an angle \( \theta = 90^\circ - 30^\circ = 60^\circ \) with the vertical wire. Thus, \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \). Substitute into the formula:\[ F = 2.60 \times 0.01 \times 0.588 \times \frac{\sqrt{3}}{2} = 0.013242 \] N.Using the right-hand rule, the force is directed toward the north-east direction, given the 30° offset.

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Key Concepts

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

Superconducting Electromagnet
Superconducting electromagnets are magnets made from coils of superconducting wire. These magnets are unique because they can carry electricity without losing any energy to resistance. Unlike regular electromagnets, they can generate significantly stronger magnetic fields.
In our exercise, the wire is placed between the poles of a superconducting electromagnet. This means that the magnetic field is both strong and steady. As a result, the interaction between the wire's current and the magnet's magnetic field creates a noticeable force.
Superconducting electromagnets are often used in scientific research and various industrial applications. These magnets are particularly beneficial when a high and consistent magnetic field strength is necessary for precision work or large-scale applications.
Right-Hand Rule
The right-hand rule is a helpful guideline in physics for determining the direction of a magnetic force on a current-carrying wire. It's easy to use, just follow these steps:
  • Point your thumb in the direction of the current's flow. In our case, that's downward.
  • Extend your fingers in the direction of the magnetic field. Depending on the scenario, this could be east, south, or 30 degrees south of west.
  • Your palm will indicate the direction of the magnetic force.

So, in our solutions:
  • For a magnetic field pointing east, the force points north.
  • If the field is south, the force is directed west.
  • At 30 degrees south of west, the force is toward the north-east.
The right-hand rule is a reliable tool for remembering how forces act in electromagnetic contexts.
Angle Between Wire and Magnetic Field
In calculating magnetic force, the angle between the wire and the magnetic field is crucial because it affects the intensity of the force. This is represented mathematically by \( \sin(\theta) \), where \( \theta \) is the angle.
Consider these scenarios:
  • If the wire and the field are perpendicular, like in parts (a) and (b) of the solution, \( \theta = 90^\circ \) leading to \( \sin(90^\circ) = 1 \). This means the force is at its maximum possible strength.
  • When the field moves to 30 degrees south of west, \( \theta = 60^\circ \), as the wire is 90 degrees minus the 30-degree deviation. This alignment gives \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \).
To find the magnetic force, use \[ F = I L B \sin(\theta) \]. This shows how the orientation of the magnetic field directly influences the result by modulating the effective strength of interaction.

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Most popular questions from this chapter

If two deuterium nuclei (charge \(+e\), mass 3.34 \(\times\) 10\(^{-27}\) kg) get close enough together, the attraction of the strong nuclear force will fuse them to make an isotope of helium, releasing vast amounts of energy. The range of this force is about 10\(^{-15}\) m. This is the principle behind the fusion reactor. The deuterium nuclei are moving much too fast to be contained by physical walls, so they are confined magnetically. (a) How fast would two nuclei have to move so that in a head-on collision they would get close enough to fuse? (Assume their speeds are equal. Treat the nuclei as point charges, and assume that a separation of 1.0 \(\times\) 10\(^{-15}\) is required for fusion.) (b) What strength magnetic field is needed to make deuterium nuclei with this speed travel in a circle of diameter 2.50 m?

A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses (see Section 27.5). In one design for such an instrument, ions with mass \(m\) and charge \(q\) are accelerated through a potential difference \(V\). They then enter a uniform magnetic field that is perpendicular to their velocity, and they are deflected in a semicircular path of radius \(R\). A detector measures where the ions complete the semicircle and from this it is easy to calculate \(R\). (a) Derive the equation for calculating the mass of the ion from measurements of \(B\), \(V\), \(R\), and \(q\). (b) What potential difference \(V\) is needed so that singly ionized \(^{12}\)C atoms will have \(R =\) 50.0 cm in a 0.150-T magnetic field? (c) Suppose the beam consists of a mixture of \(^{12}\)C and \(^{14}\)C ions. If \(v\) and \(B\) have the same values as in part (b), calculate the separation of these two isotopes at the detector. Do you think that this beam separation is sufficient for the two ions to be distinguished? (Make the assumption described in Problem 27.59 for the masses of the ions.)

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an angle of 30.0\(^\circ\) above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10 \(\times\) 10\(^{-4}\) Wb through the surface?

A circular loop of wire with area \(A\) lies in the \(xy\)-plane. As viewed along the \(z\)-axis looking in the -\(z\)-direction toward the origin, a current \(I\) is circulating clockwise around the loop. The torque produced by an external magnetic field \(\overrightarrow{B}\) is given by \(\vec{\tau}\) = D(4\(\hat{\imath}\) - 3\(\hat{\jmath}\)), where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U = -\vec{\mu}\) \(\cdot\) \(\overrightarrow{B}\) is negative. The magnitude of the magnetic field is \(B_0 = 13D/IA\). (a) Determine the vector magnetic moment of the current loop. (b) Determine the components \(B_x\), \(B_y\), and \(B_z\) of \(\overrightarrow{B}\).

A particle with charge 7.80 \(\mu\)C is moving with velocity \(\vec{v} =\) - 13.80 \(\times\) 103m/s\(\hat{\jmath}\). The magnetic force on the particle is measured to be \(\overrightarrow{F} =\) (7.60 \(\times\) 10\(^{-3}\) N)\(\hat{\imath}\) - (5.20 \(\times\) 10\(^{-3}\) N)\(\hat{k}\). (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\overrightarrow{B}\) \(\cdot\) \(\overrightarrow{F}\). What is the angle between \(\overrightarrow{B}\) and \(\overrightarrow{F}\)?

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