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

Short Answer

Expert verified
The speed of the electron is approximately \( 9.46 \times 10^{6} \, \mathrm{m/s} \).

Step by step solution

01

Understanding the Magnetic Force Equation

The magnetic force on a charged particle like an electron moving in a magnetic field can be calculated using the formula \( F = qvB\sin\theta \), where \( F \) is the magnetic force, \( q \) is the charge of the electron, \( v \) is the velocity of the electron, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity vector and the magnetic field vector.
02

Identifying Known Values and Constants

In the exercise, you are given that the magnetic force \( F = 4.60 \times 10^{-15} \, \mathrm{N} \), the magnetic field \( B = 3.50 \times 10^{-3} \, \mathrm{T} \), and the angle \( \theta = 60.0^{\circ} \). The charge of an electron \( q \) is a known constant \( -1.6 \times 10^{-19} \, \mathrm{C} \).
03

Rearranging the Equation to Solve for Velocity

Rearrange the magnetic force equation to solve for the velocity \( v \) of the electron: \[ v = \frac{F}{qB\sin\theta} \]. This manipulation allows you to isolate \( v \) on one side of the equation, making it easier to solve.
04

Substituting Known Values into the Equation

Substitute the given values and constants into the rearranged formula: \[ v = \frac{4.60 \times 10^{-15} \mathrm{N}}{(1.6 \times 10^{-19} \mathrm{C}) (3.50 \times 10^{-3} \mathrm{T})\sin(60.0^{\circ})} \]. Make sure to convert the angle from degrees to radians or use direct sin values from trigonometric tables.
05

Calculating the Velocity

First, determine the value of \( \sin(60.0^{\circ}) \), which is approximately \( 0.866 \). Substitute this into the equation and perform the calculations: \[ v = \frac{4.60 \times 10^{-15}}{1.6 \times 10^{-19} \times 3.50 \times 10^{-3} \times 0.866} \]. Simplifying this will give you the velocity of the electron.
06

Final Answer

After performing the calculations, you will find that the velocity of the electron is approximately \( 9.46 \times 10^{6} \, \mathrm{m/s} \).

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Key Concepts

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

electron velocity
Electron velocity is an essential concept in understanding how charged particles, like electrons, behave in the presence of magnetic fields. When an electron moves, it possesses kinetic energy, which makes it crucial to know its speed or velocity. Velocity is not just the speed of an electron but also describes its direction in space. This is important when considering magnetic forces since the force depends on both the speed and direction of the electron relative to the magnetic field. The magnetic force equation helps to derive velocity from the given magnetic force, charge, and magnetic field strength. Remember that the force equation is given by: \( F = qvB\sin\theta \). Here, \( v \), the electron's velocity, is calculated by rearranging the equation as \( v = \frac{F}{qB\sin\theta} \). By substituting known values, we calculated velocity as approximately \( 9.46 \times 10^{6} \, \mathrm{m/s} \), showcasing the electron's incredible speed.
magnetic field
The magnetic field is a force field created by moving electric charges and magnetic dipoles, and it exerts a force on particles that are moving relative to it, such as electrons. The strength of this field is measured in Teslas (\(\mathrm{T}\)). In our exercise, the magnetic field strength is given as \(3.50 \times 10^{-3} \mathrm{T}\).A magnetic field has both a magnitude and a direction. This vector nature means that the force experienced by a particle, like an electron, depends on its velocity direction relative to the magnetic field. This is why the force is calculated using \(\sin\theta\), where \(\theta\) is the angle between the electron’s velocity and the magnetic field direction. For this exercise, the angle is given as \(60.0^{\circ}\). The trigonometric function \(\sin(60.0^{\circ})\) impacts the magnitude of the force, showing the significance of both angular and field considerations.
charge of electron
The charge of an electron is a fundamental physical constant significant in electromagnetism. It is a negative elementary charge denoted as \( q = -1.6 \times 10^{-19} \, \mathrm{C} \). This uniform charge is one of the building blocks of atoms, determining how electrons interact with electric and magnetic fields.In a magnetic field, the sign and magnitude of an electron's charge are crucially involved in the calculation of the magnetic force using the formula \( F = qvB\sin\theta \). In computations involving magnetic fields, only the magnitude of the charge is typically used, since the direction of force (given by the right-hand rule) accounts for the charge being negative.The constancy of the electron's charge implies that no matter the field it’s in or the speed at which it travels, this value remains unchanged, simplifying many calculations in physics. Understanding the electron's charge helps glean insights into many behavioral properties and interactions within electromagnetic contexts.

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Most popular questions from this chapter

(a) An \(\mathrm{}^{16} \mathrm{O}\) nucleus (charge \(+8 e )\) moving horizontally from west to east with a speed of 500 \(\mathrm{km} / \mathrm{s}\) experiences a magnetic force of 0.00320 \(\mathrm{nN}\) vertically downward. Find the magnitude and direction of the weakest magnetic field required to produce this force. Explain how this same force could be caused by a larger magnetic field. (b) An electron moves in a uniform, horizontal, 2.10 -T magnetic field that is toward the west. What must the magnitude and direction of the minimum velocity of the electron be so that the magnetic force on it will be 4.60 pN, vertically upward? Explain how the velocity could be greater than this minimum value and the force still have this same magnitude and direction.

An electron moves at \(2.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) through a region in which there is a magnetic field of unspecified direction and magnitude \(7.40 \times 10^{-2} \mathrm{T}\) (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?

A wire 25.0\(\mathrm { cm }\) long lies along the \(z\) -axis and carries a current of 7.40 A in the \(+ z\) -direction. The magnetic field is uniform and has components \(B _ { x } = - 0.242\) T, \(B _ { y } = - 0.985 \mathrm { T }\) , and \(B _ { z } = - 0.336 \mathrm { T }\) (a) Find the components of the magnetic force on the wire. (b) What is the magnitude of the magnetic force on the wire?

A coil with magnetic moment 1.45\(\mathrm { A } \cdot \mathrm { m } ^ { 2 }\) is oriented initially with its magnetic moment antiparallel to a uniform \(0.835 - \mathrm { T }\) magnetic field. What is the change in potential energy of the coil when it is rotated \(180 ^ { \circ }\) so that its magnetic moment is parallel to the field?

A particle with charge \(q\) is moving with speed \(v\) in the \(- y\) -direction. It is moving in a uniform magnetic field \(\vec { B } =\) \(B _ { x } \hat { i } + B _ { y } \hat { J } + B _ { z } \hat { k } .\) (a) What are the components of the force \(\vec { \boldsymbol { F } }\) exerted on the particle by the magnetic field? (b) If \(q > 0 ,\) what must the signs of the components of \(\vec { B }\) be if the components of \(\vec { \boldsymbol { F } }\) are all nonnegative? (c) If \(q < 0\) and \(B _ { x } = B _ { y } = B _ { z } > 0 ,\) find the direction of \(\vec { \boldsymbol { F } }\) and find the magnitude of \(\vec { \boldsymbol { F } }\) in terms of \(| q | , v ,\) and \(B _ { x }\)

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