Problem 76 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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 \(\vec{L} = 0 \mathbf{i} + 0 \mathbf{j} + 0.25 \mathbf{k} \).

Step 3: Write the Magnetic Field Vector

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

Step 4: 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 \(\vec{L} \times \vec{B} = 0.24625 \mathbf{i} + 0.084 \mathbf{j} + 0 \mathbf{k} \).

Step 5: 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 \(\vec{F} = 1.82225 \mathbf{i} + 0.6216 \mathbf{j} \).

Step 6: Find the Magnitude of the Magnetic Force

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

Key Concepts

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.