Problem 56 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

The force components are \(0, 2.08 \times 10^{-3}, 3.59 \times 10^{-3}\) N.

Step 1: Understand the Equation

The force on a charged particle moving in a magnetic field is calculated using the formula \( \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \), where \( q \) is the charge, \( \mathbf{v} \) is the velocity vector, and \( \mathbf{B} \) is the magnetic field vector.

Step 2: Define Velocity and Magnetic Field Vectors

Identify the velocity vector \( \mathbf{v} = \langle v_x, v_y, v_z \rangle = \langle -1.68 \times 10^4, -3.11 \times 10^4, 5.85 \times 10^4 \rangle \) m/s and the magnetic field vector \( \mathbf{B} = \langle 0.650, 0, 0 \rangle \) T.

Step 3: Calculate the Cross Product \( \mathbf{v} \times \mathbf{B} \)

Use the formula for a cross product to find \( \mathbf{v} \times \mathbf{B} \):\[\mathbf{v} \times \mathbf{B} =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \-1.68 \times 10^4 & -3.11 \times 10^4 & 5.85 \times 10^4 \0.650 & 0 & 0 \\end{vmatrix}\]Calculating it, the resulting vector components are: \( v_x B_y - v_y B_x = 0 - (-3.11 \times 10^4 \cdot 0.650) \),\( v_z B_x - v_x B_z = 5.85 \times 10^4 \cdot 0.650 - 0 \),\( v_x B_y - v_y B_x = (-1.68 \times 10^4 \cdot 0) - (-3.11 \times 10^4 \cdot 0) = 0 \).This results in \( \mathbf{v} \times \mathbf{B} = \langle 0, 3.11 \times 10^4 \times 0.650, 5.85 \times 10^4 \times 0.650 \rangle \).

Step 4: Substitute Values into Cross Product

Substituting the numerical values for \( v_y \) and \( v_z \) into the expression yields:\[ 0, -(-3.11 \times 10^4 \cdot 0.650) = 20215, \quad(5.85 \times 10^4 \cdot 0.650) = 38025 \]. Hence, the cross product \( \mathbf{v} \times \mathbf{B} = \langle 0, 20215, 38025 \rangle \).

Step 5: Calculate the Magnetic Force \( \mathbf{F} \)

The force vector \( \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \) is calculated by multiplying the charge \( q = 9.45 \times 10^{-8} \) C with the cross product vector \( \langle 0, 20215, 38025 \rangle \):\( F_x = q \cdot 0 \),\( F_y = q \cdot 20215 = 9.45 \times 10^{-8} \cdot 20215 \),\( F_z = q \cdot 38025 = 9.45 \times 10^{-8} \cdot 38025 \).

Step 6: Compute Force Components

Calculate each component:\( F_x = 0 \),\( F_y = 2.08 \times 10^{-3} \) N,\( F_z = 3.59 \times 10^{-3} \) N.Thus, the components of the force on the particle are \( \langle 0, 2.08 \times 10^{-3}, 3.59 \times 10^{-3} \rangle \) N.

Key Concepts

Cross Product

The cross product is an essential mathematical operation used in physics to determine the resulting vector from two vectors. In the context of magnetic forces, it helps us find the direction and magnitude of the force acting on a charged particle moving through a magnetic field.

A cross product between vectors \( \mathbf{a} \) and \( \mathbf{b} \) is calculated using the formula:
\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \a_x & a_y & a_z \b_x & b_y & b_z \\end{vmatrix}\]
This determinant expands to give:\( (a_y b_z - a_z b_y)\hat{i} + (a_z b_x - a_x b_z)\hat{j} + (a_x b_y - a_y b_x)\hat{k} \). The resulting vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \).

In our exercise, we calculated the cross product \( \mathbf{v} \times \mathbf{B} \) by following the above steps, resulting in a vector that describes how the particle's velocity interacts with the magnetic field.

Magnetic Field

A magnetic field is a vector field surrounding magnets and electric currents, representing the magnetic force exerted in the surrounding space. It is characterized by both a direction and magnitude, measured in Tesla (T).

The magnetic field plays a crucial role in determining the motion of charged particles. This magnetic force is perpendicular to both the charged particle's velocity and the magnetic field itself.

In our exercise, the magnetic field \( \mathbf{B} \) is uniform and directed along the \(+ x\)-axis, with a magnitude of 0.650 T. This uniformity simplifies calculations as the field can be represented as a vector \( \mathbf{B} = \langle 0.650, 0, 0 \rangle \). When using the cross product to find the force \( \mathbf{F} \), the direction of the resulting force depends significantly on this magnetic field's orientation.

Charged Particle

A charged particle is any particle, such as an electron or proton, that carries an electric charge. When a charged particle moves through a magnetic field, it experiences a magnetic force.

This force \( \mathbf{F} \) is defined by the Lorentz force equation:
\[\mathbf{F} = q(\mathbf{v} \times \mathbf{B})\]
Where:

• \( q \) is the charge of the particle
• \( \mathbf{v} \) is the velocity vector of the particle
• \( \mathbf{B} \) is the magnetic field vector

This force is perpendicular to both the velocity vector and the magnetic field.

In the given exercise, our particle has a small charge of \( 9.45 \times 10^{-8} \) C and a specified velocity vector. By calculating \( \mathbf{v} \times \mathbf{B} \) and then multiplying it by the particle's charge, we determine the force components acting on the particle at the given moment.