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Chapter 27: Problem 36

A straight, vertical wire carries a current of 1.20 \(\mathrm{A}\) downward in a region between the poles of a large superconducting electromagnet, 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 south, (b) 0.00706 N east, (c) 0.00612 N north-east.

Step by step solution

01

Understanding the problem

We need to find the magnetic force on a vertical wire carrying a current of 1.20 A for a 1.00-cm section in different orientations of a uniform magnetic field of 0.588 T. The direction of the current is down the wire. We will use the magnetic force formula.
02

Using the magnetic force equation

The magnetic force experienced by a current-carrying wire in a magnetic field is given by \[ F = I \cdot L \cdot B \cdot \sin(\theta) \]where:- \( F \) is the magnetic force,- \( I \) is the current (1.20 A),- \( L \) is the length of the wire (1.00 cm converted to meters: 0.01 m),- \( B \) is the magnetic field strength (0.588 T),- \( \theta \) is the angle between the direction of the current and the magnetic field.
03

Calculating the force for case (a) - Magnetic field direction east

For part (a), the magnetic field is directed east. The current is downward (vertically), so the angle \( \theta \) between the current and magnetic field is 90 degrees. Using \( \sin(90^\circ) = 1 \), the force \( F \) is \[ F = 1.20 \cdot 0.01 \cdot 0.588 \cdot 1 = 0.007056 \, \text{N} \]The direction of the force is given by the right-hand rule, which points south.
04

Calculating the force for case (b) - Magnetic field direction south

In part (b), the magnetic field is directed south. The angle between the current and the magnetic field remains 90 degrees. Thus, \[ F = 1.20 \cdot 0.01 \cdot 0.588 \cdot 1 = 0.007056 \, \text{N} \]The right-hand rule indicates the magnetic force direction is east.
05

Calculating the force for case (c) - Magnetic field direction 30 degrees south of west

For part (c), the magnetic field makes an angle of 30 degrees south of west. The angle \( \theta \) between the current and the magnetic field is 90 + 30 = 120 degrees.Using \( \sin(120^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \),\[ F = 1.20 \cdot 0.01 \cdot 0.588 \cdot \frac{\sqrt{3}}{2} = 0.006116 \, \text{N} \]Using the right-hand rule, the direction of the force is north-east.

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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 used to determine the direction of the magnetic force on a current-carrying wire in a magnetic field. To apply this rule, extend your right hand so that your thumb, index, and middle fingers are perpendicular to each other.
  • Your thumb represents the direction of the conventional current, which is the flow of positive charges.
  • Your index finger indicates the direction of the magnetic field.
  • Your middle finger then points in the direction of the force experienced by the wire.
In this exercise, applying the right-hand rule helps to discern the direction of the force on our vertical current-carrying wire. For example, when the magnetic field points east and the current flows downward, your index finger points east, and your thumb points down, indicating the force's direction as being towards the south.
Magnetic Field Direction
Understanding the direction of the magnetic field is crucial when calculating the magnetic force on a current-carrying wire. The magnetic field (\( B \)) is a vector field, meaning it has both magnitude and direction.It's represented by lines starting from the north pole of a magnet to the south pole. The direction of the magnetic field is tangent to the field lines at any point. In the exercise provided, the magnetic field is described in specific directions: east, south, or at an angle "30 degrees south of west."By knowing these orientations, you can apply the equation for magnetic force, \( F = I \cdot L \cdot B \cdot \sin(\theta) \), and correctly compute the force by considering the angle \( \theta \)between the wire and the field direction. For example, when the field is east and the wire's current direction is vertical, the angle is 90 degrees, giving an optimal interaction for calculating force.
Current-Carrying Wire in a Magnetic Field
A current-carrying wire in a magnetic field experiences a force due to the interaction between the current and the magnetic field. This force is calculated using the equation \( F = I \cdot L \cdot B \cdot \sin(\theta) \), which factors in the current \( I \), length of the wire \( L \), magnetic field \( B \), and the sine of the angle \( \theta \)between the current's and field's directions.In the given exercise, the wire is oriented vertically, carrying a current of 1.20 A. Depending on the magnetic field鈥檚 direction, the angle \( \theta \) changes which affects the force magnitude and direction.
  • If the magnetic field is east, making \( \theta = 90^\circ \), the force direction is perpendicular to both the field and current, pointing south.
  • With a field southward, the orientation again results in \( \theta = 90^\circ\), changing the force direction to east.
  • When the field is 30 degrees south of west, \( \theta = 120^\circ\), creating a force towards the northeast.
Understanding these interactions allows for precise predictions of the force direction and magnitude, demonstrating the crucial aspects of electromagnetism in a practical context.

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Most popular questions from this chapter

A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2\(\Omega\) . When the motor is running at full load on a \(120-\mathrm{V}\) line, the emf in the rotor is 105 \(\mathrm{V}\) . (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

A proton \(\left(q=1.60 \times 10^{-19} \mathrm{C}, m=1.67 \times 10^{-27} \mathrm{kg}\right)\) moves in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=(0.500 \mathrm{T}) \hat{\boldsymbol{i}} .\) At \(t=0\) the proton has velocity components \(v_{x}=1.50 \times 10^{5} \mathrm{m} / \mathrm{s}, v_{y}=0,\) and \(v_{z}=2.00 \times 10^{5} \mathrm{m} / \mathrm{s}(\text { see Example } 27.4) .\) (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the \(+x\) -direction, \(\vec{E}=\left(+2.00 \times 10^{4} \mathrm{V} / \mathrm{m}\right) \hat{\imath}\) (b) Will the proton have a component of acceleration in the direction of the electric field?(c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At \(t=T / 2\) , where \(T\) is the period of the circular motion of the proton, what is the \(x\) -component of the displacement of the proton from its position at \(t=0 ?\)

An insulated wire with mass \(m=5.40 \times 10^{-5} \mathrm{kg}\) is bent into the shape of an inverted U such that the horizontal part has a length \(l=15.0 \mathrm{cm} .\) The bent ends of the wire are partially immersed in two pools of mercury, with 2.5 \(\mathrm{cm}\) of each end below the mercury's surface. The entire structure is in a region containing a uniform \(0.00650-\mathrm{T}\) magnetic field directed into the page (Fig. 27.71\()\) . An electrical connection from the mercury pools is made through the ends of the wires. The mercury pools are connected to a \(1.50-\mathrm{V}\) battery and a switch \(\mathrm{S}\) . When switch \(\mathrm{S}\) is closed, the wire jumps 35.0 \(\mathrm{cm}\) into the air, measured from its initial position. (a) Determine the speed \(v\) of the wire as it leaves the mercury. (b) Assuming that the current \(I\) through the wire was constant from the time the switch was closed until the wire left the mercury, determine \(I\) (c) Ignoring the resistance of the mercury and the circuit wires, determine the resistance of the moving wire.

A deuteron (the nucleus of an isotope of bydrogen) has a mass of \(3.34 \times 10^{-27} \mathrm{kg}\) and a charge of \(+e .\) The deuteron travels in a circular path with a radius of 6.96 \(\mathrm{mm}\) in a magnetic field with magnitude 2.50 \(\mathrm{T}\) . (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what to potential difference would the deuteron have to be accelerated to acquire this speed?

The force on a charged particle moving in a magnetic field can be computed as the vector sunt of the forces due to each separate component of the magnetic field. As an example, a particle with charge \(q\) is moving with speed \(v\) in the \(-y\) -direction. It is moving in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=\overrightarrow{\boldsymbol{B}}_{\boldsymbol{x}} \hat{\boldsymbol{i}}+\boldsymbol{B}_{\boldsymbol{y}} \hat{\boldsymbol{j}}+\boldsymbol{B}_{\boldsymbol{z}} \hat{\boldsymbol{k}}\) (a) What are the components of the force \(\overrightarrow{\boldsymbol{F}}\) exerted on the particle by the magnetic field? (b) If \(q>0,\) what must the signs of the components of \(\overrightarrow{\boldsymbol{B}}\) be if the components of \(\overrightarrow{\boldsymbol{F}}\) are all nonnegative? (c) If \(q<0\) and \(B_{x}=B_{y}=B_{z}>0,\) find the direction of \(\vec{F}\) and find the magnitude of \(\vec{F}\) in terms of \(|q|, v,\) and \(B_{x}\) .
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