Problem 5 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

The speed of the electron is approximately 9.52 × 10⁶ m/s.

Step 1: Understanding the Problem

We are given a magnetic force, an angle with respect to a magnetic field, and the magnitude of the magnetic field. Our task is to find the speed of the electron. The formula for the magnetic force is: \( F = qvB\sin(\theta) \), where \( F \) is the force, \( q \) is the charge of the electron, \( v \) is the speed, \( B \) is the magnetic field, and \( \theta \) is the angle.

Step 2: Identify Known Quantities

The given quantities include the magnetic force \( F = 4.60 \times 10^{-15} \) N, the angle \( \theta = 60.0^\circ \), and the magnetic field \( B = 3.50 \times 10^{-3} \) T. The charge of the electron \( q \) is a known constant, \( q = 1.60 \times 10^{-19} \) C.

Step 3: Rearrange Formula for Speed

From the magnetic force formula \( F = qvB\sin(\theta) \), we want to solve for the speed \( v \). Rearrange the formula to get: \[ v = \frac{F}{qB\sin(\theta)} \].

Step 4: Substitute the Known Values

Insert the known values into the formula: \( v = \frac{4.60 \times 10^{-15}}{1.60 \times 10^{-19} \times 3.50 \times 10^{-3} \times \sin(60^\circ)} \). Recognize that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \).

Step 5: Perform the Calculation

Calculate the denominator first: \( 1.60 \times 10^{-19} \times 3.50 \times 10^{-3} \times 0.866 = 4.83 \times 10^{-22} \). Then, calculate the speed: \( v = \frac{4.60 \times 10^{-15}}{4.83 \times 10^{-22}} \approx 9.52 \times 10^{6} \) m/s.

Key Concepts

Magnetic Field

A magnetic field is a crucial concept in physics, particularly in electromagnetism. It refers to the field surrounding a magnetic material or a moving electric charge. This field exerts a force on other nearby magnetic materials and moving charges. Understanding magnetic fields is fundamental in various applications, from industrial magnetics to medical imaging.
Key characteristics of a magnetic field include:

• Magnitude: It signifies the strength of the magnetic field, generally expressed in Tesla (T).
• Direction: Magnetic field lines represent the direction, showing the path a free north pole would take under the influence of the field.

In the given problem, the electron experiences a magnetic force when moving relative to the magnetic field. This interaction highlights how magnetic fields can act on charges, setting them into motion or altering their trajectory.

Electron Speed Calculation

To determine the speed of an electron under the influence of a magnetic field, we employ the concept of magnetic force, which is expressed by the formula: \( F = qvB\sin(\theta) \)
Here, q is the charge of the electron, v is its speed, B is the magnetic field's magnitude, and \(\theta\) is the angle between the electron's velocity vector and the magnetic field.
Steps to find the electron's speed:

• Rearrange the formula: The formula is manipulated to solve for \(v\), resulting in \( v = \frac{F}{qB\sin(\theta)} \). This is a fundamental trick in physics allowing any variable to be isolated.
• Substitution: Once the formula is rearranged, insert the given values: \( F = 4.60 \times 10^{-15} \) N, \( q = 1.60 \times 10^{-19} \) C, etc., ensuring all units are consistent.
• Calculation: Plug the values into the formula, carefully perform arithmetic to ensure accuracy, and solve for the speed of the electron.

This method effectively translates known quantities into the unknown, thus providing the desired result of electron speed:
\( v \approx 9.52 \times 10^{6} \) m/s.

Sin Function in Physics

The sine function \(\sin(\theta)\) is an important trigonometric function employed in physics due to its properties of relating angles to ratios. In scenarios involving vectors, the sine function helps relate the components of the vectors forming an angle. This function is central in calculating the magnetic force since it accounts for the component of velocity perpendicular to the magnetic field.
Crucial points about the sine function in physics:

• Application in forces: It helps calculate components of forces or vectors when they are at an angle, especially crucial for calculations involving non-parallel directions.
• Use in magnetic context: In the formula \( F = qvB\sin(\theta) \), \(\sin\theta\) determines the effective contribution of the velocity in the plane perpendicular to the magnetic field.
• Trigonometric value: Knowing common sine values, like \(\sin(60^{\circ}) \), which is \( \frac{\sqrt{3}}{2} \approx 0.866 \), is vital for simplifying problems and ensuring calculations are accurate.

Through understanding and using these components, one effectively bridges the gap between mathematical theory and practical physics applications, aiding in problems like calculating the resultant force acting on a moving charge.