Problem 1 - University Physics with Modern Physics

Question

A particle with a charge of \(-1.24 \times 10^{-8} \mathrm{C}\) is moving with instantaneous velocity \(\vec{v}=\left(4.19 \times 10^{4} \mathrm{m} / \mathrm{s}\right) \hat{\imath}+(-3.85 \times\) \(10^{4} \mathrm{m} / \mathrm{s} ) \hat{\boldsymbol{J}}\) . What is the force exerted on this particle by a mag- netic field (a) \(\vec{\boldsymbol{B}}=(1.40 \mathrm{T}) \hat{\boldsymbol{i}}\) and \((\mathrm{b}) \vec{\boldsymbol{B}}=(1.40 \mathrm{T}) \hat{\boldsymbol{k}} ?\)

Step-by-Step Solution

Answer

(a) Force is \(-0.668 \hat{\boldsymbol{k}} \text{ N}\); (b) Force is \(0.668 \hat{\imath} + 0.728 \hat{\boldsymbol{j}} \text{ N}\).

Step 1: Understanding the Formula

Force on a charged particle moving in a magnetic field can be calculated using the formula: \( \vec{F} = q(\vec{v} \times \vec{B}) \), where \(q\) is the charge, and \(\vec{v}\) and \(\vec{B}\) are the velocity and magnetic field vectors.

Step 2: Cross Product Calculation for Part (a)

For \( \vec{B} = (1.40 \text{ T}) \hat{\boldsymbol{i}} \), the cross product \( \vec{v} \times \vec{B} \) is calculated as follows:\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{\imath} & \hat{\boldsymbol{j}} & \hat{\boldsymbol{k}} \ 4.19 \times 10^4 & -3.85 \times 10^4 & 0 \ 1.40 & 0 & 0 \end{vmatrix} = (0) \hat{\boldsymbol{i}} + (0) \hat{\boldsymbol{j}} + (1.40 \times 3.85 \times 10^4) \hat{\boldsymbol{k}} \]Calculating gives \( \vec{v} \times \vec{B} = 5.39 \times 10^4 \hat{\boldsymbol{k}} \).

Step 3: Force Calculation for Part (a)

Now apply the cross product result to the force equation: \[ \vec{F} = q(\vec{v} \times \vec{B}) = (-1.24 \times 10^{-8}) \cdot (5.39 \times 10^4) \hat{\boldsymbol{k}} = -0.668 \hat{\boldsymbol{k}} \text{ N}. \] So, the force is \(-0.668 \hat{\boldsymbol{k}} \; \text{N}\).

Step 4: Cross Product Calculation for Part (b)

For \( \vec{B} = (1.40 \text{ T}) \hat{\boldsymbol{k}} \), the cross product \( \vec{v} \times \vec{B} \) is calculated as follows:\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{\imath} & \hat{\boldsymbol{j}} & \hat{\boldsymbol{k}} \ 4.19 \times 10^4 & -3.85 \times 10^4 & 0 \ 0 & 0 & 1.40 \end{vmatrix} = \left( (-3.85 \times 1.40) \hat{\imath} - (4.19 \times 1.40) \hat{\boldsymbol{j}} \right) \]Calculating gives: \( \vec{v} \times \vec{B} = -5.39 \times 10^4 \hat{\imath} - 5.87 \times 10^4 \hat{\boldsymbol{j}} \).

Step 5: Force Calculation for Part (b)

Now apply the result to the force equation: \[ \vec{F} = q(\vec{v} \times \vec{B}) = (-1.24 \times 10^{-8}) (\begin{bmatrix} -5.39 \times 10^4 \hat{\imath} - 5.87 \times 10^4 \hat{\boldsymbol{j}} \end{bmatrix}) \]\( \vec{F} = 0.668 \hat{\imath} + 0.728 \hat{\boldsymbol{j}} \text{ N} \)

Key Concepts

cross product

When we look at the forces acting on a charged particle moving through a magnetic field, the cross product plays a critical role. The cross product, denoted by the symbol \( \times \), combines two vectors to produce a third vector that is perpendicular to the plane formed by the initial two vectors. This operation is essential in physics, especially for calculating the magnetic force. For a particle with velocity vector \( \vec{v} \) and magnetic field vector \( \vec{B} \), the resulting vector from the cross product \( \vec{v} \times \vec{B} \) will point in a direction that is perpendicular to both \( \vec{v} \) and \( \vec{B} \). This perpendicular nature is what determines the direction of the magnetic force exerted on the particle.

In mathematical form, the cross product of two vectors \( \vec{A} = a_1 \hat{\imath} + a_2 \hat{\boldsymbol{j}} + a_3 \hat{\boldsymbol{k}} \) and \( \vec{B} = b_1 \hat{\imath} + b_2 \hat{\boldsymbol{j}} + b_3 \hat{\boldsymbol{k}} \) is given by:

• \( (a_2b_3 - a_3b_2) \hat{\imath} \)
• \( (a_3b_1 - a_1b_3) \hat{\boldsymbol{j}} \)
• \( (a_1b_2 - a_2b_1) \hat{\boldsymbol{k}} \)

This calculation involves the use of determinants, which can easily be represented by a 3x3 matrix. Understanding how to apply the cross product correctly is vital for solving problems related to the magnetic force.

magnetic field

A magnetic field is an invisible field that exerts a magnetic force on moving electric charges, magnetic dipoles, and other magnetic materials. It is represented by the vector symbol \( \vec{B} \) and is measured in Tesla (T). In physics, magnetic fields are crucial as they describe how magnets exert forces at a distance.

Magnetic fields arise from the motion of electric charges, such as in electric currents, or from intrinsic properties of particles, giving rise to permanent magnetism. A magnetic field can be thought of as a region around a magnet where magnetic forces can be felt. This means that when a charged particle moves through a magnetic field, the field can change the direction of its motion.

 

• A particle moving perpendicular to a magnetic field will experience a force perpendicular to both its velocity and the magnetic field.
• The resulting path is usually circular, with the magnetic force acting as the centripetal force.

In the given exercise, as the particle interacts with different components of the magnetic field, the force it experiences changes direction as dictated by the vector nature of magnetic interactions.

velocity of charged particle

The velocity of a charged particle, denoted by the vector \( \vec{v} \), characterizes both the speed and direction of the particle's motion. It plays a fundamental role in determining how the particle interacts with a magnetic field.

In the given problem, the velocity vector is specified as \( \vec{v} = (4.19 \times 10^4 \text{ m/s}) \hat{\imath} + (-3.85 \times 10^4 \text{ m/s}) \hat{\boldsymbol{J}} \). This notation specifies components of the velocity along the \( x \)-axis and \( y \)-axis:

• \(4.19 \times 10^4 \text{ m/s} \) along the \( \hat{\imath} \) direction
• \(-3.85 \times 10^4 \text{ m/s} \) along the \( \hat{\boldsymbol{J}} \) direction

The velocity magnitude can be calculated using the Pythagorean theorem if needed, but more importantly, it's the direction that dictates how it interacts with a magnetic field.

When analyzing a particle's motion in a magnetic field, knowing the velocity allows us to use the right-hand rule to predict the force's direction; pointing your fingers in the direction of \( \vec{v} \), curling them towards \( \vec{B} \) gives the force direction (thumb direction). Understanding velocity components is essential for applying the force equation and interpreting movements in magnetic fields.

calculation of force in physics

In physics, calculating force, especially for charged particles in a magnetic field, is a crucial concept. The magnetic force \( \vec{F} \) on a charged particle can be determined using the formula:\[ \vec{F} = q(\vec{v} \times \vec{B}) \]
where:

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

This relationship highlights the dependency of the magnetic force on the cross product of velocity and magnetic field vectors, emphasizing the directionality and magnitude of these cross elements. The negative charge or positive charge of the particle also changes the force's direction (affecting which direction the resulting vector points due to negative sign adjustment).

In the original exercise, for both parts (a) and (b), after calculating the cross product, this result is multiplied by the particle's charge to determine \( \vec{F} \). This multiplication considers both magnitude and direction, with care taken for negative charges as they reverse the direction of the product. Understanding this process helps students correctly apply and solve the force equation in various physical contexts involving magnetic fields.