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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?

Short Answer

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

Step by step solution

01

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.
02

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, \( a = \frac{1.6512 \times 10^{-14}}{9.11 \times 10^{-31}} \approx 1.812 \times 10^{16} \, m/s^2 \).
03

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, \( a = 0 \, m/s^2 \).
04

Compute Actual Acceleration

Given that the actual acceleration is one-fourth of the largest possible magnitude, the acceleration \( a_{actual} = \frac{1}{4} \times 1.812 \times 10^{16} = 4.53 \times 10^{15} \, m/s^2 \).
05

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 \( \theta \approx 38.8^\circ \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A singly charged ion of \(^7\)Li (an isotope of lithium) has a mass of 1.16 \(\times\) 10\(^{-26}\) kg. It is accelerated through a potential difference of 220 V and then enters a magnetic field with magnitude 0.874 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm. You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV. An alpha particle has charge \(q =\) +2\(e\) and mass \(m =\) 6.64 \(\times\) 10\(^{-27}\) kg. If the magnetic field isn't changed, what will be the orbital radius of the alpha particles?

A straight, vertical wire carries a current of 2.60 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude \(B =\) 0.588 T and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00-cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) 30.0\(^\circ\) south of west?

A proton (\(q\) = 1.60 \(\times\) 10\(^{-19}\) C, \(m =\) 1.67 \(\times\) 10\(^{-27}\) kg) moves in a uniform magnetic field \(\overrightarrow{B} =\) (0.500 T)\(\hat{\imath}\). At \(t =\) 0 the proton has velocity components \(\upsilon_x =\) 1.50 \(\times\) 10\(^5\) m/s, \(\upsilon_y =\) 0, and \(\upsilon_z =\) 2.00 \(\times\) 10\(^5\) m/s (see Example 27.4). (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the +\(x\)-direction, \(\overrightarrow{E} =\) (+2.00 \(\times\) 10\(^4\) V/m)\(\hat{\imath}\). (b) Will the proton have a component of acceleration in the direction of the electric field? (c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At \(t =\) \(T\)/2, where T is the period of the circular motion of the proton, what is the \(x\)-component of the displacement of the proton from its position at \(t =\) 0?

A conducting bar with mass m and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{B}\) fills the region between the rails (\(\textbf{Fig. P27.65}\)). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\), find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\). (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth (11.2 km/s). Let \(B =\) 0.80 T, \(I =\) 2.0 \(\times\) 10\(^3\) A, \(m =\) 25 kg, and \(L =\) 50 cm. For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

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