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Chapter 27: Problem 40

A straight, vertical wire carries a current of 1.20 A downward in a region between the poles of a large superconductingelectromagnet, where the magnetic field has magnitude \(B=\) 0.588 \(\mathrm{T}\) and is horizontal. What are the magnitude and direction of the magnetic force on a \(1.00-\mathrm{cm}\) 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) 0.00706 N, north; (b) 0.00706 N, west; (c) 0.00611 N, northeast.

Step by step solution

01

Understanding the Problem

To solve the problem, we need to find the magnetic force on a current-carrying wire in a magnetic field. The problem gives us the current, the length of the wire, and the magnetic field in different orientations. We will use the formula for magnetic force on a straight conductor, which is given by: \[ F = I \cdot L \cdot B \cdot \sin(\theta) \] where \( I \) is the current, \( L \) is the length of the wire, \( B \) is the magnetic field, and \( \theta \) is the angle between the direction of the current and the magnetic field. Each part of the problem gives us a different angle \( \theta \) to consider.
02

Calculating Force for East Direction

In this case, the magnetic field is directed to the east, and the current is vertical (downward). The angle between the direction of the current (down) and the magnetic field (east) is \( 90^\circ \). Therefore, \( \sin(90^\circ) = 1 \).\[ F = 1.20 \times 0.01 \times 0.588 \times 1 = 0.007056 \; \text{N} \]The direction of the force will be determined using the right-hand rule. Using the right-hand rule, the force will be directed into the north.
03

Calculating Force for South Direction

Here, the magnetic field is to the south, and the current is downward. Again, the angle between the current and the magnetic field is \( 90^\circ \), so \( \sin(90^\circ) = 1 \).\[ F = 1.20 \times 0.01 \times 0.588 \times 1 = 0.007056 \; \text{N} \]Applying the right-hand rule, the force direction will be toward the west.
04

Calculating Force for 30 degrees South of West

For this scenario, the magnetic field is at an angle \( 30.0^{\circ} \) south of west. To find the correct angle \( \theta \), consider the direction perpendicular to both the wire and field directions. The effective angle with vertical is \( 90^\circ - 30^\circ = 60^\circ \). Therefore, \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \).\[ F = 1.20 \times 0.01 \times 0.588 \times \frac{\sqrt{3}}{2} \approx 0.006114 \; \text{N} \]Using the right-hand rule, the force is directed towards northeast.

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Key Concepts

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

right-hand rule
The right-hand rule is a simple yet powerful tool to determine the direction of the magnetic force on a current-carrying wire. Here鈥檚 how you can use it:
  • Imagine holding the wire with your right hand, aligning your thumb with the direction of the electric current.
  • Your fingers should curl naturally in the direction of the magnetic field lines.
  • The direction your palm faces indicates the direction of the magnetic force acting on the wire, perpendicular to both the current and magnetic field.
This method helps visualize the interaction between the current and magnetic field, ensuring you can easily determine the force's direction without confusion. The right-hand rule is especially useful when considering various angles of interaction, as shown in the exercise.
magnetic field
A magnetic field is an invisible field that exerts force on charged particles, such as electrons, when they move through it. The strength of this field is measured in teslas (T). In the given exercise:
  • The magnetic field has a magnitude of 0.588 T.
  • The direction can vary, such as east, south, or at an angle of 30 degrees south of west.
The magnetic field's direction significantly influences the magnetic force's direction on the wire. When considering the effect of the magnetic field on the wire, you must always note its orientation relative to the electric current, as this determines the angle of interaction and the resulting force.
current-carrying wire
A current-carrying wire is a conductor through which electric current flows. It generates a magnetic field around it and interacts with external magnetic fields. Here's what's essential to know about current-carrying wires:
  • The current in the wire for this problem is 1.20 A downward.
  • The wire segment in question is 1.00 cm long.
Current in the wire creates its own magnetic field, interacting with an external one, causing a force as described by the formula \[ F = I \cdot L \cdot B \cdot \sin(\theta) \]where each symbol represents a physical property. Understanding this interaction is key to solving problems related to electromagnetic force and designing applications like electric motors.
angle of interaction
The angle of interaction (\( \theta \)) is crucial when determining the resulting magnetic force. It is the angle between the direction of the current in the wire and the magnetic field. The force depends on the sine of this angle (\( \sin(\theta) \)), influencing its magnitude:
  • A right angle (\(90^\circ\)) results in maximum force since \(\sin(90^\circ) = 1\).
  • For angles other than \(90^\circ\), the force reduces, proportional to \(\sin(\theta)\).
When the exercise describes different directions of the magnetic field, like east, south, or 30 degrees south of west, it asks us to consider different angles of interaction, affecting the magnitude of the force calculated. Understanding the angle of interaction is key to solving these problems correctly.

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Most popular questions from this chapter

A straight, 2.00 -m, \(150-\mathrm{g}\) wire carries a current in a region where the earth's magnetic field is horizontal with a magnitude of 0.55 gauss. (a) What is the minimum value of the current in this wire so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it? Does it seem likely that such a wire could support this size of current? (b) Show how the wire would have to be oriented relative to the earth's magnetic field to be supported in this way.

An Electromagnetic Rail Gun. A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\vec { B }\) fills the region between the rails (Fig. \(P 27.74 )\) (a) Find the magnitude and direction of the net force on the con- ducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m ,\) find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\) . (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \(( 11.2 \mathrm { km } / \mathrm { s } ) .\) Let \(B = 0.80 \mathrm { T } , \quad I = 2.0 \times 10 ^ { 3 } \mathrm { A } , \quad m = 25 \mathrm { kg }\) and \(L = 50 \mathrm { cm } .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

A plastic circular loop has radius \(R ,\) and a positive charge \(q\) is distributed uniformly around the circumference of the loop. The loop is then rotated around its central axis, perpendicular to the plane of the loop, with angular speed \(\omega .\) If the loop is in a region where there is a uniform magnetic field \(\vec { \boldsymbol { B } }\) directed parallel to the plane of the loop, calculate the magnitude of the magnetic torque on the loop.

An electron in the beam of a TV picture 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?

Crossed \(\vec{E}\) and \(\vec{B}\) Fields. A particle with initial velocity \(\vec{\boldsymbol{v}}_{0}=\left(5.85 \times 10^{3} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{J}}\) enters a region of uniform electric and magnetic fields. The magnetic field in the region is \(\vec{\boldsymbol{B}}=\) \(-(1.35 \mathrm{T}) \hat{k}\) . Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) \(+0.640 \mathrm{nC}\) and \((\mathrm{b})-0.320 \mathrm{nC.}\) You can ignore the weight of the particle.
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