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A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 \(\mathrm{km} / \mathrm{s}\) im the \(+x\) -direction experiences a force of \(2.25 \times 10^{-16} \mathrm{N}\) in the \(+y\) -direction, and an electron moving at 4.75 \(\mathrm{km} / \mathrm{s}\) in the \(-\mathrm{z}\) -direction experiences a force of \(8.50 \times 10^{-16} \mathrm{N}\) . (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the \(-y\) -direction at 3.2 \(\mathrm{km} / \mathrm{s} ?\)

Short Answer

Expert verified
The magnetic field is 0.09375 T in the +z direction; force on the electron is 4.8 x 10^-16 N in the -x direction.

Step by step solution

01

Identify the Relevant Equation

To solve the problem, we use the formula for the magnetic force on a charged particle: \[ F = q(v \times B) \]where \( F \) is the force, \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field.
02

Calculate the Magnetic Field Using the Proton

For the proton:- Charge of a proton \( q_p = 1.6 \times 10^{-19} \mathrm{C} \).- Velocity \( v_p = 1.50 \times 10^3 \mathrm{m/s} \) in the \(+x\) direction.- Force \( F_p = 2.25 \times 10^{-16} \mathrm{N} \) in the \(+y\) direction.Using the right-hand rule, the magnetic field \( B \) is directed along the \(+z\) axis.Thus, \[ F_p = q_p v_p B \]\[ B = \frac{F_p}{q_p v_p} = \frac{2.25 \times 10^{-16}}{1.6 \times 10^{-19} \times 1.5 \times 10^3} = 9.375 \times 10^{-2} \mathrm{T} \]
03

Verify Using the Electron's Motion

For the electron:- Charge of an electron \( q_e = -1.6 \times 10^{-19} \mathrm{C} \).- Velocity \( v_e = 4.75 \times 10^3 \mathrm{m/s} \) in the \(-z\) direction.- Force \( F_e = 8.50 \times 10^{-16} \mathrm{N} \).Using the force direction (same direction implies opposite charge) and right-hand rule, the calculated field should hold true:\[ F_e = |q_e| v_e B \]Replacing the values, \( B \) should agree with the previously calculated value.
04

Calculate Force on Electron in -y Direction

Now that we have \( B = 9.375 \times 10^{-2} \mathrm{T} \), let's calculate the force on an electron moving in the \(-y\) direction:- Velocity \( v'_e = 3.2 \times 10^3 \mathrm{m/s} \) in the \(-y\) direction.The magnetic force \( F'_e = q_e v'_e B \) will act in the \(-x\) direction by the right-hand rule.\[ F'_e = 1.6 \times 10^{-19} \times 3.2 \times 10^3 \times 9.375 \times 10^{-2} \]\[ F'_e = 4.8 \times 10^{-16} \mathrm{N} \]
05

Summarize the Results

- The magnitude of the magnetic field is \( 9.375 \times 10^{-2} \mathrm{T} \) and is directed in the \(+z\)-direction.- The force on a second electron moving at 3.2 \( \mathrm{km/s} \) in the \(-y\)-direction is \( 4.8 \times 10^{-16} \mathrm{N} \) in the \(-x\)-direction.

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Key Concepts

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

Magnetic Field
A magnetic field is an invisible but very real force field around magnetic materials or moving electric charges. It exerts forces on other nearby moving charges and magnetic dipoles. You can visualize it like the wind: invisible, but you can feel its effects.
The direction and magnitude of a magnetic field are important characteristics. They determine how strong the force will be on charged particles moving within the field. For example, in the given exercise, the magnetic field causes a proton and an electron to experience forces. It's important to understand how these forces manifest, which is largely governed by the direction and magnitude of the magnetic field.
The magnetic field (denoted by the symbol \( B \)) is measured in teslas (T) and has both direction and magnitude, making it a vector quantity.
Right-Hand Rule
The right-hand rule is a simple way to find the direction of the magnetic force acting on a moving charged particle.
Using the right-hand rule, you point your fingers in the direction of the particle's velocity, align your palm such that it faces towards the magnetic field direction, and your thumb then points in the direction of the acting force. For a positive charge like a proton, this rule applies directly.
For a negative charge, like an electron, the force direction is exactly opposite to the one predicted by the right-hand rule for positive charges. In the provided exercise, this rule helps us determine the direction of force on the proton in the \(+y\) direction and the magnetic field in the \(+z\) direction.
Charged Particles
Charged particles, like electrons and protons, interact with magnetic fields in interesting ways. Depending on the charge and direction of motion, they will experience different magnitudes and directions of force.
Protons have a positive charge, while electrons are negatively charged. The charge of a proton and electron is exactly equal in magnitude but opposite in sign, approximately \(1.6 \times 10^{-19} \mathrm{C}\). This equal charge magnitude causes them to experience similar but opposite forces when moving in the same magnetic field.
Within a magnetic field, the force on a charged particle can be calculated using the formula \( F = q(v \times B) \), where \( F \) is the force experienced, \( q \) is the charge, \( v \) is the velocity vector, and \( B \) is the magnetic field vector.
Vector Calculations
Vector calculations are essential for solving problems involving magnetic forces, as both the velocity of the particle and the magnetic field are vector quantities.
Vectors have both magnitude and direction, and calculating the net result often involves vector cross products. The cross product in magnetic force calculations is used to determine the direction of the resulting force. The resultant of the vector cross product of the velocity \( v \) and the magnetic field \( B \) vector is also a vector force \( F \).
The formula \( F = q(v \times B) \) signifies that the force is perpendicular to both the velocity and the magnetic field. This concept is crucial when dealing with forces on charged particles, as demonstrated in the solution to the problem. Understanding how to compute vector cross products is key to predicting the correct direction and magnitude of these forces.

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Most popular questions from this chapter

The magnetic flux through one face of a cube is \(+0.120 \mathrm{Wb} .\) (a) What must the total magnetic flux through the other five faces of the cube be? (b) Why didn't you need to know the dimensions of the cube in order to answer part (a) 2 (c) Suppose the magnetic flux is due to a permanent magnet like that shown in Fig. \(27.11 .\) In a sketch, show where the cube in part (a) might be located relative to the magnet.

A \(150-\) g ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a \(125-\mathrm{m}\) vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 \(\mathrm{T}\) and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enrers the field.

An open plastic soda bottle with an opening diameter of 2.5 \(\mathrm{cm}\) is placed on a table. A uniform \(1.75-\mathrm{T}\) magnetic field directed upward and oriented \(25^{\circ}\) from vertical encompasses the bottle. What is the total magnetic flux through the plastic of the soda bottle?

The force on a charged particle moving in a magnetic field can be computed as the vector sunt of the forces due to each separate component of the magnetic field. As an example, a particle with charge \(q\) is moving with speed \(v\) in the \(-y\) -direction. It is moving in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=\overrightarrow{\boldsymbol{B}}_{\boldsymbol{x}} \hat{\boldsymbol{i}}+\boldsymbol{B}_{\boldsymbol{y}} \hat{\boldsymbol{j}}+\boldsymbol{B}_{\boldsymbol{z}} \hat{\boldsymbol{k}}\) (a) What are the components of the force \(\overrightarrow{\boldsymbol{F}}\) exerted on the particle by the magnetic field? (b) If \(q>0,\) what must the signs of the components of \(\overrightarrow{\boldsymbol{B}}\) be if the components of \(\overrightarrow{\boldsymbol{F}}\) are all nonnegative? (c) If \(q<0\) and \(B_{x}=B_{y}=B_{z}>0,\) find the direction of \(\vec{F}\) and find the magnitude of \(\vec{F}\) in terms of \(|q|, v,\) and \(B_{x}\) .

A beam of protons traveling at 1.20 \(\mathrm{km} / \mathrm{s}\) enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular- to its original direction (Fig. 27.48\()\) . The beam travels a distance of 1.18 \(\mathrm{cm}\) while in the field. What is the magnitude of the magnetic field?

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