Problem 6 - University Physics with Modern Physics

Question

An electron moves at 1.40 \(\times\) 10\(^6\) m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.40 \(\times\) 10\(^{-2}\) T. (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

Step-by-Step Solution

Answer

The largest acceleration is approximately 1.812 × 10¹⁶ m/s², the smallest is 0 m/s². The angle \( \theta \) is approximately 38.8°.

Step 1: Use the Formula for Magnetic Force

The force on a charged particle moving through a magnetic field is given by the equation: \( F = qvB \sin \theta \), where \( F \) is the magnetic force, \( q \) is the charge of the particle, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field.

Step 2: Determine Largest Force and Acceleration

The largest force occurs when \( \sin \theta = 1 \) (i.e., \( \theta = 90^\circ \)). For an electron, \( q = 1.6 \times 10^{-19} \, C \). So the force is \( F = (1.6 \times 10^{-19})(1.4 \times 10^6)(7.4 \times 10^{-2}) \). Calculate this to find \( F = 1.6512 \times 10^{-14} \, N \). Acceleration \( a = \frac{F}{m} \), where the mass of an electron \( m = 9.11 \times 10^{-31} \, kg \). Therefore, \( a = \frac{1.6512 \times 10^{-14}}{9.11 \times 10^{-31}} \approx 1.812 \times 10^{16} \, m/s^2 \).

Step 3: Determine Smallest Force and Acceleration

The smallest force occurs when \( \sin \theta = 0 \) (i.e., \( \theta = 0^\circ \) or \( \theta = 180^\circ \)), which implies \( F = 0 \) and therefore, \( a = 0 \, m/s^2 \).

Step 4: Compute Actual Acceleration

Given that the actual acceleration is one-fourth of the largest possible magnitude, the acceleration \( a_{actual} = \frac{1}{4} \times 1.812 \times 10^{16} = 4.53 \times 10^{15} \, m/s^2 \).

Step 5: Determine the Angle Using the Actual Acceleration

The actual acceleration formula based on force is \( a_{actual} = \frac{qvB \sin \theta}{m} \). Since \( a_{actual} = 4.53 \times 10^{15} \, m/s^2 \), you solve for \( \sin \theta \) by rearranging: \( \sin \theta = \frac{a_{actual} \times m}{qvB} \). Substitute the known values to calculate: \( \sin \theta = \frac{4.53 \times 10^{15} \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 1.4 \times 10^6 \times 7.4 \times 10^{-2}} = \frac{4.53}{7.208} \approx 0.628 \). Solve for \( \theta \) to get \( \theta \approx 38.8^\circ \).

Key Concepts

Electron Acceleration

When an electron moves through a magnetic field, it experiences a magnetic force that causes it to accelerate. This force is perpendicular to both the velocity of the electron and the magnetic field.
If you want to calculate this acceleration, you'd use the magnetic force formula:

• Magnetic Force, \( F = qvB \sin \theta \)
• Where \( q \) is the charge of the electron, \( v \) is its velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field.

The acceleration \( a \) can be determined by dividing this force by the mass \( m \) of the electron, so \( a = \frac{F}{m} \).
The maximum possible acceleration occurs when the angle \( \theta \) is \( 90^\circ \), since \( \sin 90^\circ = 1 \).
This translates to the largest force, and thus the largest acceleration. Conversely, if \( \theta \) is \( 0^\circ \) or \( 180^\circ \), \( \sin \theta = 0 \), so there is no force and hence zero acceleration.

Magnetic Field Strength

The strength of the magnetic field, denoted as \( B \), plays a crucial role in determining the force exerted on a charged particle such as an electron.
Magnetic field strength is measured in teslas (T), and in this exercise, it is given as \( 7.40 \times 10^{-2} \) T.
This value directly influences the magnetic force via the formula \( F = qvB \sin \theta \), affecting how strongly the particle is accelerated.

• Increased magnetic field strength \( B \) leads to a proportionally higher force and therefore greater acceleration of the electron, assuming velocity and the angle are constant.
• If either \( v \), \( \theta \), or \( q \) changes, then the effect of any given \( B \) might shift accordingly.

Understanding the impact of magnetic field strength helps predict how charged particles will behave when subjected to a magnetic environment in practical applications.

Angle Between Velocity and Field

The angle \( \theta \) between an electron's velocity \( v \) and the magnetic field \( B \) has a significant impact on the magnitude of the acceleration.
The force experienced by the electron is related to this angle by the sine function in the magnetic force formula \( F = qvB \sin \theta \).

• When the angle \( \theta \) is \( 90^\circ \), \( \sin \theta \) is maximized at 1, resulting in the largest possible force and acceleration.
• If \( \theta \) is \( 0^\circ \) or \( 180^\circ \), the force and, consequently, the acceleration, drop to zero since \( \sin 0^\circ = \sin 180^\circ = 0 \).

In this scenario, when the actual acceleration is one-fourth of the maximum, the value of \( \theta \) must be calculated to achieve the given acceleration. The formula \( \sin \theta = \frac{a}{\left( \frac{qvB}{m} \right)} \) rearranges to solve for \( \theta \), giving us insights into the relative orientation of the velocity to the field and how it affects motion.