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Chapter 27: Problem 8

A particle with charge \(-5.60 \mathrm{nC}\) is moving in a uniform magnetic field \(\vec{\boldsymbol{B}}=-(1.25 \mathrm{T}) \hat{\boldsymbol{k}}\) . The magnetic force on the particle is measured to be \(\vec{\boldsymbol{F}}=-\left(3.40 \times 10^{-7} \mathrm{N}\right) \hat{\imath}+(7.40 \times\) \(10^{-7} \mathrm{N} ) \hat{J}\) . (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{v} \cdot\) What is the angle between \(\vec{v}\) and \(\vec{F} ?\)

Short Answer

Expert verified
(a) \( v_x = -105.71 \) m/s, \( v_y = 48.57 \) m/s. \( v_z \) is undetermined. (b) \( v_z \) component is not determined. (c) The angle is 90 degrees.

Step by step solution

01

Understand the Force Equation

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

Set Up the Equation for the Cross Product

Given \( \vec{F} = -\left(3.40 \times 10^{-7} \mathrm{N}\right) \hat{\imath} +(7.40 \times 10^{-7} \mathrm{N}) \hat{\jmath} \) and \( \vec{B} = -(1.25 \mathrm{T}) \hat{\boldsymbol{k}} \), the cross-product \( \vec{v} \times \vec{B} \) in vector form is: \[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \ v_x & v_y & v_z \ 0 & 0 & -1.25 \end{vmatrix} = 1.25v_y \hat{\imath} - 1.25v_x \hat{\jmath} \].
03

Relate Cross Product to Given Force

Substitute the cross product into the Lorentz force equation to get: \( q(1.25v_y \hat{\imath} - 1.25v_x \hat{\jmath}) = -3.40 \times 10^{-7} \hat{\imath} + 7.40 \times 10^{-7} \hat{\jmath} \). Set components equal to solve for \( v_x \) and \( v_y \): \(-q \cdot 1.25v_y = -3.40 \times 10^{-7} \) and \( q \cdot 1.25v_x = 7.40 \times 10^{-7} \).
04

Solve for Velocity Components

We know \( q = -5.60 \times 10^{-9} \mathrm{C} \). Solve for \( v_y \) and \( v_x \):\[ v_y = \frac{-3.40 \times 10^{-7}}{1.25q} = \frac{-3.40 \times 10^{-7}}{1.25 \times (-5.60 \times 10^{-9})} = 48.57 \text{ m/s} \] \[ v_x = \frac{7.40 \times 10^{-7}}{1.25q} = \frac{7.40 \times 10^{-7}}{1.25 \times (-5.60 \times 10^{-9})} = -105.71 \text{ m/s} \].
05

Identify Undetermined Velocity Component

The Lorentz force equation does not include \( v_z \), which indicates that the velocity component \( v_z \) is not determined by the given measurements because the magnetic force depends only on the components perpendicular to \( \vec{B} \).
06

Calculate the Scalar Product and Angle

The scalar product \( \vec{v} \cdot \vec{F} \) between perpendicular vectors is zero. Use that \( \vec{v} \) (velocity) and \( \vec{F} \) (force) are perpendicular according to \( \vec{v} \times \vec{B} = \vec{F} \). Therefore, the angle \( \theta \) between \( \vec{v} \) and \( \vec{F} \) is 90 degrees.

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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 a region where a magnetic force can be experienced by charged particles, current-carrying wires, or magnetic materials. It is often represented by the symbol \( \vec{B} \). The field itself is measured in units of Tesla (T).
The direction of this field is conventionally taken from North to South outside of a magnet.
  • Magnetic fields are invisible, but their presence is often felt through the forces they exert.
  • The magnitude of the magnetic field indicates the strength of the force a charged particle will experience.
  • Uniform magnetic fields have the same magnitude and direction everywhere within a defined space.
In the given exercise, the magnetic field is uniform and pointing along the negative z-axis, as noted by \( \vec{B} = -(1.25 \mathrm{T}) \hat{\boldsymbol{k}} \). This tells us about both the direction and the strength of the field where the charged particle is moving.
Cross Product
The cross product is a mathematical operation used to find a vector perpendicular to two given vectors in three-dimensional space. It's essential in calculating the magnetic force on a charged particle.
For vectors \( \vec{A} \) and \( \vec{B} \), the cross product \( \vec{A} \times \vec{B} \) is given by the determinant:
  • \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \end{vmatrix} \]
  • This yields a vector whose direction follows the right-hand rule.
  • The resultant vector is perpendicular to the plane formed by \( \vec{A} \) and \( \vec{B} \).
In the exercise, we find the magnetic force \( \vec{F} \) on the charged particle using the Lorentz force law: \( \vec{F} = q (\vec{v} \times \vec{B}) \). This force depends only on the velocity components that are perpendicular to the magnetic field, emphasizing how the cross product works.
Velocity Components
A particle moving in a magnetic field has velocity components along the coordinate axes represented by \( v_x, v_y, \) and \( v_z \). These components are crucial for determining the forces acting on the particle in the magnetic field's presence. Each component refers to the speed and direction along its respective axis.
When solving for these components in the given scenario:
  • \( v_x \) and \( v_y \) were determined using the Lorentz force equation, revealing how these components contribute to the force.
  • The component \( v_z \) cannot be determined from the force equation because the magnetic force applies only to components perpendicular to the magnetic field.
Thus, understanding these components is essential to calculating the actual movement of charged particles in a magnetic field and predicting how they will be deflected.
Charged Particles
Charged particles are objects with an electrical charge, either positive or negative. This charge causes them to experience forces in electric and magnetic fields. The movement of these particles in a magnetic field is governed by the Lorentz force, which is the focus of this exercise.
Key concepts related to charged particles in a magnetic field include:
  • The charge determines the direction of the force; positive charges deflect in one direction while negative charges in the opposite.
  • The magnitude of a charge affects the strength of the force exerted; larger charges experience stronger forces.
  • The velocity of a charged particle influences its path and the magnitude of the force due to the cross product in the Lorentz equation.
With the particle in our example having a charge of \(-5.60 \mathrm{nC}\), the force experienced alters its motion according to the rules discussed, illuminating crucial aspects of charged particle dynamics in magnetic contexts.

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Most popular questions from this chapter

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 \(B _ { x }\)

A straight, \(2.5-\mathrm{m}\) wire carries a typical household current of 1.5 \(\mathrm{A}\) (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects undernormal household conditions?

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 magnetic field exerts a torque \(\tau\) on a round current carrying loop of wire. What will be the torque on this loop (in terms of \(\tau\) if its diameter is tripled?

A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconducting electromagnet of the cyclotron produces a 2.9 -T magnetic field perpendicular to the proton orbits. (a) When the protons have achieved a kinetic energy of 2.7\(\mathrm { MeV }\) what is the radius of their circular orbit and what is their angular speed? (b) Repeat part (a) when the protons have achieved their final kinetic energy of 5.4\(\mathrm { MeV }\) .
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