Problem 3 - Fundamentals of Physics

Question

An electron that has an instantaneous velocity of $$ \vec{v}=\left(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}+\left(3.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{j}} $$ is moving through the uniform magnetic field \(\vec{B}=(0.030 \mathrm{~T}) \hat{\mathrm{i}}-\) \((0.15 \mathrm{~T}) \mathrm{j} .\) (a) Find the force on the electron due to the magnetic field. (b) Repeat your calculation for a proton having the same velocity.

Step-by-Step Solution

Answer

(a) \( \vec{F_e} = 6.24 \times 10^{-14} \hat{k} \) N, (b) \( \vec{F_p} = -6.24 \times 10^{-14} \hat{k} \) N.

Step 1: Write Down Formula for Magnetic Force

The magnetic force on a charged particle moving in a magnetic field is given by the equation: \( \vec{F} = q \left( \vec{v} \times \vec{B} \right) \). Here, \( q \) is the charge of the particle, \( \vec{v} \) is its velocity, and \( \vec{B} \) is the magnetic field.

Step 2: Determine Charges

The charge of the electron is \( q_e = -1.6 \times 10^{-19} \) C. The charge of the proton is \( q_p = 1.6 \times 10^{-19} \) C.

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

Express both \( \vec{v} \) and \( \vec{B} \) in vector form: \( \vec{v} = \left( 2.0 \times 10^6 \right) \hat{i} + \left( 3.0 \times 10^6 \right) \hat{j} \) and \( \vec{B} = (0.030) \hat{i} - (0.15) \hat{j} \). Using the determinant method for cross products, compute: \[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2.0 \times 10^6 & 3.0 \times 10^6 & 0 \ 0.030 & -0.15 & 0 \end{vmatrix} = \left( 0 \hat{i} + 0 \hat{j} + \left(2.0 \times 10^6 \times (-0.15) - 3.0 \times 10^6 \times 0.030 \right) \hat{k}\right) \].

Step 4: Solve the Cross Product Expression

Simplify the cross product computed in Step 3: \( \vec{v} \times \vec{B} = 0 \hat{i} + 0 \hat{j} + \left( -0.3 \times 10^6 - 0.09 \times 10^6 \right) \hat{k} \). Calculate to get \( \vec{v} \times \vec{B} = 0 \hat{i} + 0 \hat{j} - 0.39 \times 10^6 \hat{k} \) \( \mathrm{m}^2/\mathrm{s}^2 \).

Step 5: Calculate Force on Electron

Use the formula \( \vec{F} = q_e (\vec{v} \times \vec{B}) \) to find the force on the electron. Substituting, \( \vec{F_e} = -1.6 \times 10^{-19} \times 0 \hat{i} + 0 \hat{j} - 0.39 \times 10^6 \hat{k} \), \( \vec{F_e} = - (1.6 \times 0.39 \times 10^{-19} \times 10^6) \hat{k} \). This simplifies to \( \vec{F_e} \approx 6.24 \times 10^{-14} \hat{k} \) N.

Step 6: Calculate Force on Proton

Now find the force on the proton using \( \vec{F} = q_p (\vec{v} \times \vec{B}) \). Substitute \( \vec{F_p} = 1.6 \times 10^{-19} \times 0 \hat{i} + 0 \hat{j} - 0.39 \times 10^6 \hat{k} \) \( \vec{F_p} = 1.6 \times 0.39 \times 10^{-19} \times 10^6 \hat{k} \). This means \( \vec{F_p} \approx -6.24 \times 10^{-14} \hat{k} \) N.

Key Concepts

Cross Product

In physics, the cross product is a powerful tool for calculating the force acting on a charged particle within a magnetic field. It's an essential method when dealing with vectors as it helps compute the force direction and magnitude relative to the vectors.

Given two vectors, the cross product yields a third vector perpendicular to the plane formed by these vectors. This concept is particularly important in magnetism, where the magnetic force on a moving charge is calculated using the formula:

• \( \vec{F} = q \left( \vec{v} \times \vec{B} \right) \)

Here, the cross product \( \vec{v} \times \vec{B} \) indicates both the direction and magnitude of the force.

Notably, if \( \vec{v} \) and \( \vec{B} \) are parallel, the magnitude of their cross product is zero, implying no force. Using the determinant method is very effective, particularly with vectors expressed in component form.

Electron Charge

The electron is a fundamental particle with a well-known charge of \( q_e = -1.6 \times 10^{-19} \) C. This negative charge plays a crucial role in determining the direction of the magnetic force acting on the electron.

When you substitute the electron's charge into the magnetic force formula, it causes the resulting force vector to flip in direction compared to a positively charged particle.

This effect is essential when calculating the magnetic forces on particles in electric and magnetic fields, as seen in particle accelerators or television screens. Understanding the nature of electron charge not only aids in electric field calculations but also impacts magnetic properties, influencing how electrons interact in various physical settings.

Magnetic Field

A magnetic field, denoted by \( \vec{B} \), is an invisible field exerted by magnets or moving charges, which can exert forces on other nearby moving charges. It is measured in Tesla (T).

For instance, in this exercise, the magnetic field is represented by the vector \( \vec{B} = (0.030 \mathrm{~T}) \hat{i} - (0.15 \mathrm{~T}) \hat{j} \).

This field determines the force direction and magnitude acting on charges moving through it depending on its orientation with respect to the charge's velocity. Knowing the magnetic field's strength and direction allows us to accurately predict how charged particles like electrons and protons are deflected, a principle used in MRI machines and compass operation.

Proton Charge

Protons, like electrons but with opposite charge, carry a positive charge of \( q_p = 1.6 \times 10^{-19} \) C. This positive charge influences how protons interact within a magnetic field compared to electrons.

In calculating forces in magnetic fields, substituting the proton's charge into the magnetic force equation results in a force vector with the opposite direction of an electron's force vector given the same velocity and magnetic field.

Understanding proton charge is vital in fields like nuclear physics and chemistry, where it helps explain atomic structure and the behavior of elements. It’s also key in particle physics experiments, where differences in electron and proton charge interactions are fundamental to studying matter's properties.