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A particle with charge 7.80 \(\mu\)C is moving with velocity \(\vec{v} =\) - 13.80 \(\times\) 103m/s\(\hat{\jmath}\). The magnetic force on the particle is measured to be \(\overrightarrow{F} =\) (7.60 \(\times\) 10\(^{-3}\) N)\(\hat{\imath}\) - (5.20 \(\times\) 10\(^{-3}\) N)\(\hat{k}\). (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\overrightarrow{B}\) \(\cdot\) \(\overrightarrow{F}\). What is the angle between \(\overrightarrow{B}\) and \(\overrightarrow{F}\)?

Short Answer

Expert verified
\( B_x, B_z \) are calculated; \( B_y \) undetermined; angle = \(90^\circ\).

Step by step solution

01

Write down the Lorentz Force Equation

The magnetic force on a charged particle is given by the Lorentz force equation: \( \overrightarrow{F} = q ( \overrightarrow{v} \times \overrightarrow{B} ) \), where \( q \) is the charge, \( \overrightarrow{v} \) is the velocity vector, and \( \overrightarrow{B} \) is the magnetic field vector.
02

Express the Known Quantities

Given: Charge \( q = 7.80 \, \mu C = 7.80 \times 10^{-6} \, C \), velocity \( \overrightarrow{v} = -13.80 \times 10^{3} \, m/s \, \hat{\jmath} \), and force \( \overrightarrow{F} = (7.60 \times 10^{-3} \, N) \, \hat{\imath} - (5.20 \times 10^{-3} \, N) \, \hat{k} \).
03

Set Up Cross Product for Force

The force is calculated by the cross product, \( \overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B}) \). Let \( \overrightarrow{B} = B_x \hat{\imath} + B_y \hat{\jmath} + B_z \hat{k} \). Compute \( \overrightarrow{v} \times \overrightarrow{B} \) using the determinant formula for the cross product.
04

Solve for Magnetic Field Components

From the cross product \( \overrightarrow{v} \times \overrightarrow{B} \), \( \overrightarrow{F} = q (B_z \hat{\imath} + B_x \hat{k}) = (7.60 \times 10^{-3} \, N) \hat{\imath} - (5.20 \times 10^{-3} \, N) \hat{k} \). Solve these equations: - \( 7.80 \times 10^{-6} \, C \times -13.80 \times 10^3 \, m/s \times B_z = 7.60 \times 10^{-3} \, N \) - \( 7.80 \times 10^{-6} \, C \times -13.80 \times 10^3 \, m/s \times B_x = -5.20 \times 10^{-3} \, N \).Obtain \( B_z \) and \( B_x \), noting that \( B_y \) cannot be determined from the force.
05

Check Undetermined Components

Since the velocity is in the \( \hat{\jmath} \) direction, the component of the magnetic field in the same direction, \( B_y \), produces no force and is thus undetermined by the force measurement.
06

Calculate Scalar Product and Angle

The scalar product of \( \overrightarrow{B} \cdot \overrightarrow{F} \) results in zero since the cross product and force are orthogonal. Therefore, \( \cos \theta = 0 \) implies the angle \( \theta = 90^\circ \).

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Key Concepts

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

Magnetic Field Components
When a charged particle, like the one in this exercise, moves through a magnetic field, it experiences a force. According to the Lorentz Force Equation, this force is a cross product of the velocity and the magnetic field vectors. Given the force and velocity, we can calculate some components of the magnetic field but not all.

With the force (\(\overrightarrow{F}\)) and velocity (\(\overrightarrow{v}\)) provided, we can find the components of the magnetic field (\(\overrightarrow{B}\)) using the cross product. Here, the force is only in the \(\hat{\imath}\) and \(\hat{k}\) directions, indicating that the magnetic field components \(B_x\) and \(B_z\) can be solved.

  • To find \(B_x\) and \(B_z\), use the Lorentz Force Equation and solve for the components by isolating them from the given force equations.
  • Since there is no force in the \(\hat{\jmath}\) direction, \(B_y\) remains undetermined. This is because the magnetic field component parallel to the velocity does not affect the force.
By knowing the fixed components and understanding the relations, we can partially determine the magnetic field.
Scalar Product
The scalar product, also known as the dot product, is a way to multiply two vectors to get a scalar. It is represented by the equation \( \overrightarrow{A} \cdot \overrightarrow{B} = AB \cos \theta \), where \(\theta\) is the angle between them. This process differs greatly from vector multiplication via a cross product.

  • In this exercise, we're asked to calculate the scalar product of the magnetic field and the force: \( \overrightarrow{B} \cdot \overrightarrow{F} \).
  • The important part for this exercise is understanding that the force due to a magnetic field is perpendicular to the velocity of the charge. In this case, the scalar product is zero because \( \cos 90^\circ = 0 \).
This zero result simplifies the problem and proves that \(\overrightarrow{B}\) and \(\overrightarrow{F}\) are orthogonal.
Cross Product
The cross product, or vector product, is a way of multiplying two vectors to get a third vector, which is perpendicular to the plane containing the initial two vectors. For vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\), \( \overrightarrow{A} \times \overrightarrow{B} = |\overrightarrow{A}||\overrightarrow{B}|\sin\theta\overrightarrow{n} \), where \(\overrightarrow{n}\) is a unit vector normal to the plane of \(\overrightarrow{A}\) and \(\overrightarrow{B}\).

The Lorentz Force is an application of the cross product, impacting how forces act directionally on moving charges in a magnetic field.

  • By applying the cross product to the velocity and magnetic field vectors, we can understand why the particle experiences a force in the specified direction.
  • The result shows which components of the velocity vector and magnetic field vector influence the force.
Remember, the direction of the force vector can be found using the right-hand rule, scrolling your finger from \(\overrightarrow{v}\) to \(\overrightarrow{B}\) produces \(\overrightarrow{F}\).
Angle Between Vectors
Understanding the angle between vectors is crucial in vector mathematics. The angle provides insight into the relationship and interaction of vectors. Mathematically, the angle \(\theta\) between two vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) can be found with the formula \( \cos\theta = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|} \).

  • In the case of the magnetic force problem, calculating this angle is simplified due to the perpendicular nature of the force and velocity vectors.
  • Since the dot product of the magnetic field and force vectors is zero, it confirms that their angle is \(90^\circ\).
This 90-degree angle signifies that the vectors are perpendicular. This cross interaction explains why force is maximal when the field and velocity have no parallel components.

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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?

A particle of charge \(q\) > 0 is moving at speed v in the \(+z\)-direction through a region of uniform magnetic field \(\overrightarrow{B}\). The magnetic force on the particle is \(\overrightarrow{F} =\) \(F_0\)(3\(\hat{\imath}\) + 4 \(\hat{\jmath}\)), where \(F_0\) is a positive constant. (a) Determine the components \(B_x\), \(B_y\), and \(B_z\), or at least as many of the three components as is possible from the information given. (b) If it is given in addition that the magnetic field has magnitude 6\(F_0/qv\), determine as much as you can about the remaining components of \(\overrightarrow{B}\).

A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses (see Section 27.5). In one design for such an instrument, ions with mass \(m\) and charge \(q\) are accelerated through a potential difference \(V\). They then enter a uniform magnetic field that is perpendicular to their velocity, and they are deflected in a semicircular path of radius \(R\). A detector measures where the ions complete the semicircle and from this it is easy to calculate \(R\). (a) Derive the equation for calculating the mass of the ion from measurements of \(B\), \(V\), \(R\), and \(q\). (b) What potential difference \(V\) is needed so that singly ionized \(^{12}\)C atoms will have \(R =\) 50.0 cm in a 0.150-T magnetic field? (c) Suppose the beam consists of a mixture of \(^{12}\)C and \(^{14}\)C ions. If \(v\) and \(B\) have the same values as in part (b), calculate the separation of these two isotopes at the detector. Do you think that this beam separation is sufficient for the two ions to be distinguished? (Make the assumption described in Problem 27.59 for the masses of the ions.)

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?

A 150-g ball containing 4.00 \(\times\) 10\(^8\) excess electrons is dropped into a 125-m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 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 enters the field.

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