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Chapter 27: Problem 76

A wire 25.0\(\mathrm { cm }\) long lies along the \(z\) -axis and carries a current of 7.40 A in the \(+ z\) -direction. The magnetic field is uniform and has components \(B _ { x } = - 0.242\) T, \(B _ { y } = - 0.985 \mathrm { T }\) , and \(B _ { z } = - 0.336 \mathrm { T }\) (a) Find the components of the magnetic force on the wire. (b) What is the magnitude of the magnetic force on the wire?

Short Answer

Expert verified
The force components are \(F_x = 1.82225\) N, \(F_y = 0.6216\) N, and \(F_z = 0\) N. The magnitude is approximately 1.822 N.

Step by step solution

01

Use the Magnetic Force Formula

To find the components of the magnetic force on the wire, we need to use the formula for the magnetic force: \( \vec{F} = I \cdot (\vec{L} \times \vec{B}) \), where \(I\) is the current, \(\vec{L}\) is the length vector of the wire, and \(\vec{B}\) is the magnetic field vector.
02

Define the Length Vector

Since the wire is aligned along the \(z\)-axis and is 25.0 cm long, we convert this to meters: \( \vec{L} = 0.25 \mathbf{k} \). The length vector, therefore, is \(\vec{L} = 0 \mathbf{i} + 0 \mathbf{j} + 0.25 \mathbf{k} \).
03

Write the Magnetic Field Vector

The magnetic field vector is given as \(\vec{B} = -0.242 \mathbf{i} - 0.985 \mathbf{j} - 0.336 \mathbf{k} \).
04

Calculate the Cross Product \(\vec{L} \times \vec{B}\)

Using the determinant method, calculate the cross product \(\vec{L} \times \vec{B}\):\[\vec{L} \times \vec{B} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 0 & 0.25 \ -0.242 & -0.985 & -0.336\end{vmatrix}= \left(0 \cdot -0.336 - 0.25 \cdot -0.985\right) \mathbf{i}- \left(0 \cdot -0.242 - 0.25 \cdot -0.336\right) \mathbf{j} + \left(0 \cdot -0.985 - 0 \cdot -0.242 \right) \mathbf{k}\]This simplifies to \(\vec{L} \times \vec{B} = 0.24625 \mathbf{i} + 0.084 \mathbf{j} + 0 \mathbf{k} \).
05

Calculate the Components of the Magnetic Force

Now, multiply the current \(I\) by the cross product results from Step 4 to find the components of the magnetic force:\(\vec{F} = 7.40 \left( 0.24625 \mathbf{i} + 0.084 \mathbf{j} + 0 \mathbf{k} \right)\).This gives us \(\vec{F} = 1.82225 \mathbf{i} + 0.6216 \mathbf{j} \).
06

Find the Magnitude of the Magnetic Force

The magnitude \(|\vec{F}|\) of the force can be found using the Pythagorean theorem:\[|\vec{F}| = \sqrt{(1.82225)^2 + (0.6216)^2} \approx \sqrt{3.3195}\approx 1.822 \text{ N}\]

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Key Concepts

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

Cross Product
The cross product is a mathematical operation used when working with vectors in three-dimensional space. It helps to find a vector perpendicular to two given vectors. The formula for the cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is given by \( \vec{A} \times \vec{B} \). This operation is crucial when calculating forces such as the magnetic force on a current-carrying wire.
  • The result of a cross product is a vector, which differs from scalar operations that result in a single number.
  • The direction of the resulting vector is perpendicular to both original vectors, determined by the right-hand rule.
  • This concept is essential for understanding how vector magnitudes and directions relate to physical forces.
In the given exercise, the cross product was used to determine the influence of the magnetic field components on the wire's current, which lies along the \(z\)-axis.
Magnetic Field Components
Magnetic fields affect how charged particles move and exert forces on current-carrying wires. These fields each have different components, described by vectors in space with \( B_x, B_y, \) and \( B_z \) representing the magnetic field's strength along the \(x\), \(y\), and \(z\) axes, respectively.
  • The behavior of a particle or wire segment in a magnetic field depends on the individual components of that field.
  • The field components in the exercise were given as \( B_x = -0.242 \) T, \( B_y = -0.985 \) T, and \( B_z = -0.336 \) T, illustrating a three-dimensional representation of the field.
  • The sign and magnitude of each component influence the resulting force's direction and intensity.
These components act together to form the complete magnetic field influencing the wire, with each having a specific effect on the wire's movement and the resulting magnetic force.
Vector Calculations
Vector calculations are fundamental in physics when analyzing forces like magnetic force. Vectors have both magnitude and direction, which makes them ideal for representing physical phenomena.
  • Vectors are often represented in terms of their components along the \(x\), \(y\), and \(z\) directions.
  • When calculating the magnetic force on a current-carrying wire, vectors are used to describe both the length of the wire and the magnetic field.
  • In the exercise, the length vector \( \vec{L} = 0 \mathbf{i} + 0 \mathbf{j} + 0.25 \mathbf{k} \) represents the wire, which lies along the \(z\)-axis.
Understanding how to manipulate and multiply these vectors through operations like the cross product forms the backbone of solving force-related problems in physics.
Determinant Method
The determinant method is a mathematical technique often used in physics to compute the cross products of vectors efficiently. It involves using a specific 3x3 matrix format with the standard unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) as the first row, components of the first vector as the second row, and components of the second vector as the third row.
  • The determinant process involves multiplying diagonally, which helps calculate the cross product quickly.
  • For instance, in the exercise, this method is utilized with the vectors \( \vec{L} \) and \( \vec{B} \) to find \( \vec{L} \times \vec{B} \).
  • This matrix-based technique is key to simplifying calculations, making it an invaluable tool in vector analysis.
By employing the determinant method, you can more easily manage the complexity of vector cross product calculations, leading to precise and accurate results for expressions like the magnetic force.

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Most popular questions from this chapter

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?

A particle with charge \(9.45 \times 10 ^ { - 8 } \mathrm { C }\) is moving in a region where there is a uniform magnetic field of 0.650 T in the \(+ x\) -direction. At a particular instant of time the velocity of the particle has components \(v _ { x } = - 1.68 \times 10 ^ { 4 } \mathrm { m } / \mathrm { s } , v _ { y } = - 3.11 \times\) \(10 ^ { 4 } \mathrm { m } / \mathrm { s } ,\) and \(v _ { z } = 5.85 \times 10 ^ { 4 } \mathrm { m } / \mathrm { s } .\) What are the components of the force on the particle at this time?

A deuteron (the nucleus of an isotope of hydrogen) 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 potential difference would the deuteron have to be accelerated to acquire this speed?

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.

An electron experiences a magnetic force of magnitude \(4.60 \times 10^{-15} \mathrm{N}\) when moving at an angle of \(60.0^{\circ}\) with respect to a magnetic field of magnitude \(3.50 \times 10^{-3} \mathrm{T.}\) Find the speed of the electron.
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