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An electron experiences a force \(\overrightarrow{\mathbf{F}}=(3.8 \hat{\mathbf{i}}-2.7 \hat{\mathbf{j}}) \times 10^{-13} \mathbf{N}\) when passing through a magnetic field \(\overrightarrow{\mathbf{B}}=(0.85 \mathrm{~T}) \hat{\mathbf{k}}\). Determine the electron's velocity.

Short Answer

Expert verified
The velocity vector is approximately \(v_x = 1.99 \times 10^6 \, \text{m/s}\) and \(v_y = -2.79 \times 10^6 \, \text{m/s}\).

Step by step solution

01

Understanding the formula

The force experienced by an electron moving through a magnetic field is given by the Lorentz force formula: \( \overrightarrow{\mathbf{F}} = q(\overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{B}}) \), where \(q\) is the charge of the electron. Rearrange this to find velocity: \( \overrightarrow{\mathbf{v}} = \frac{\overrightarrow{\mathbf{F}}}{q \times \overrightarrow{\mathbf{B}}} \).
02

Set Up Known Values

We are given \( \overrightarrow{\mathbf{F}} = (3.8 \hat{\mathbf{i}} - 2.7 \hat{\mathbf{j}}) \times 10^{-13} \mathrm{~N} \) and \( \overrightarrow{\mathbf{B}} = (0.85 \mathrm{~T}) \hat{\mathbf{k}} \) while the charge of the electron is \( q = -1.6 \times 10^{-19} \mathrm{C} \).
03

Calculating the velocity components

Using \( \overrightarrow{\mathbf{F}} = q(\overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{B}}) \), the cross product can be zero if velocity is aligned with the force direction orthogonal to the magnetic field direction \(\hat{\mathbf{k}}\). Break \(\overrightarrow{\mathbf{v}} = v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}}\). The force components as \((3.8 \hat{\mathbf{i}} - 2.7 \hat{\mathbf{j}}) \times 10^{-13}\) implies formulating and solving:1. \(3.8 \times 10^{-13} \mathrm{~N} = q \times 0.85 v_y \)2. \(-2.7 \times 10^{-13} \mathrm{~N} = -q \times 0.85 v_x \)
04

Solve for the velocity components

From the formulated equations:1. Solve for \(v_y\): \[3.8 \times 10^{-13} = (-1.6 \times 10^{-19})(0.85) v_y \Rightarrow v_y = \frac{3.8 \times 10^{-13}}{-1.36 \times 10^{-19}}\]And compute the value for \(v_y\).2. Solve for \(v_x\): \[-2.7 \times 10^{-13} = (-1.6 \times 10^{-19})(0.85) v_x \Rightarrow v_x = \frac{-2.7 \times 10^{-13}}{-1.36 \times 10^{-19}}\]And compute the value for \(v_x\).
05

Final Calculations and Solution

Compute the speed of the electron vector by vector \((v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}}) \):\[v_x = \frac{2.7 \times 10^{-13}}{1.36 \times 10^{-19}}, \quad v_y = \frac{-3.8 \times 10^{-13}}{1.36 \times 10^{-19}}.\]Calculate these and write out the complete velocity as components.

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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 force field that exerts a magnetic force on moving electric charges, like electrons. It surrounds a magnet and exerts forces on other magnets or magnetic materials within the field. The strength and direction of a magnetic field are represented by the symbol \(\overrightarrow{\mathbf{B}}\) and it is measured in Teslas (T).
The direction of the magnetic field can be defined using the "right-hand rule," where if you point your thumb in the direction of current or motion of the charge, your fingers wrap in the direction of the field lines. In the given exercise, the magnetic field is moving in the \(\hat{\mathbf{k}}\) direction, which means it is perpendicular to the plane of \(v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}}\).
Electron Velocity
Velocity is a vector quantity that represents both the speed and direction of an object's movement. For an electron experiencing a magnetic force in a magnetic field, its velocity causes the interaction defined by the Lorentz force formula. The exercise focuses on finding the electron's velocity \(\overrightarrow{\mathbf{v}}\) by rearranging the formula:
\[ \overrightarrow{\mathbf{v}} = \frac{\overrightarrow{\mathbf{F}}}{q \times \overrightarrow{\mathbf{B}}} \]
To find velocity, the force and magnetic field's cross product must be resolved into components. In this task, they are measured along the \( \hat{\mathbf{i}} \) and \( \hat{\mathbf{j}} \) directions, influenced by the force values \(3.8 \times 10^{-13} \hat{\mathbf{i}}\) and \(-2.7 \times 10^{-13} \hat{\mathbf{j}}\). The components of velocity can be determined using these equations:
  • \(v_y = \frac{3.8 \times 10^{-13}}{-1.36 \times 10^{-19}}\)
  • \(v_x = \frac{-2.7 \times 10^{-13}}{1.36 \times 10^{-19}}\)
Once calculated, \(v_x\) and \(v_y\) provide the complete velocity vector, indicating both magnitude and direction relative to the field.
Electron Charge
The electron charge \(q\) is a fundamental property of electrons, reflecting the amount of electric charge it carries. This charge is constant and universally accepted as \( -1.6 \times 10^{-19} \) Coulombs (C).
Being negative, the electron's charge influences the direction of force and velocity in electromagnetic interactions according to the Lorentz force law. This means when an electron moves under the influence of a magnetic field, the force direction is opposite to the field vector due to the negative charge, resulting in an opposite direction to positive test charges.
Understanding electron charge is crucial for forming and solving equations to determine other properties, such as velocity components in the task detailed above. The charge allows us to resolve the force equation given, such that:
  • The negative charge needs to be considered when determining component alignment: \(-qv_y \times B\) affects the \(i\)-component.
  • \(qv_x \times B\) helping determine the direction in the \(j\)-component.
Consistently applying this value in calculations leads to accurate results in solving for unknown variables.

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Most popular questions from this chapter

A 3.40-g bullet moves with a speed of \(155 \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field of \(5.00 \times 10^{-5} \mathrm{~T}\). If the bullet possesses a net charge of \(18.5 \times 10^{-9} \mathrm{C},\) by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled \(1.00 \mathrm{~km} ?\)

A proton (mass \(m_{\mathrm{p}}\) ), a deuteron \(\left(m=2 m_{\mathrm{p}}, Q=e\right)\), and an alpha particle \(\left(m=4 m_{\mathrm{p}}, Q=2 e\right)\) are accelerated by the same potential difference \(V\) and then enter a uniform magnetic field \(\overrightarrow{\mathbf{B}},\) where they move in circular paths perpendicular to \(\overrightarrow{\mathbf{B}}\). Determine the radius of the paths for the deuteron and alpha particle in terms of that for the proton.

(1I) In a probe that uses the Hall effect to measure magnetic fields, a 12.0 -A current passes through a 1.50 -cm-wide 1.30 -mm-thick strip of sodium metal. If the Hall emf is 1.86\(\mu V\) , what is the magnitude of the magnetic field (take it perpendic- ular to the flat face of the strip)? Assume one free electron per atom of Na, and take its specific gravity to be \(0.971 .\)

A uniform conducting rod of length \(d\) and mass \(m\) sits atop a fulcrum, which is placed a distance \(d / 4\) from the rod's left-hand end and is immersed in a uniform magnetic field of magnitude \(B\) directed into the page (Fig. \(27-57\) ). An object whose mass \(M\) is 8.0 times greater than the rod's mass is hung from the rod's left-hand end. What current (direction and magnitude) should flow through the rod in order for it to be "balanced" (i.e., be at rest horizontally) on the fulcrum? (Flexible connecting wires which \(d\) exert negligible force on the rod are not shown.)

(II) A Hall probe used to measure magnetic field strengths consists of a rectangular slab of material (free-electron density \(n )\) with width \(d\) and thickness \(t,\) carrying a current \(I\) along its length \(\ell\) . The slab is immersed in a magnetic field of magnitude \(B\) oriented perpendicular to its rectangular face (of area \(\ell d ),\) so that a Hall emf \(\mathscr{E}_{\mathrm{H}}\) is produced across its width \(d .\) The probe's magnetic sensitivity, defined as \(K_{\mathrm{H}}=8_{\mathrm{H}} / I B,\) indicates the magnitude of the Hall emf achieved for a given applied magnetic field and current. A slab with a large \(K_{H}\) is a good candidate for use as a Hall probe. (a) Show that \(K_{H}=1 /\) ent. Thus, a good Hall probe has small values for both \(n\) and \(t\) . \((b)\) As possible candidates for the material used in a Hall probe, consider \(\left(\) i) a typical metal \(\left(n \approx 1 \times 10^{29} / \mathrm{m}^{3}\right)\) and \right. (ii) a (doped) semiconductor \(\left(n \approx 3 \times 10^{22} / \mathrm{m}^{3}\right) .\) Given that a semiconductor slab can be manufactured with a thickness of \(0.15 \mathrm{mm},\) how thin \((\mathrm{nm})\) should a metal slab be to yield a \(K_{\mathrm{H}}\) value equal to that of the semiconductor slab? Compare this metal slab thickness with the 0.3 -nm size of a typical metal atom. (c) For the typical semiconductor slab described in part \((b),\) what is the expected value for \(\mathscr{E}_{\mathrm{H}}\) when \(I=100 \mathrm{mA}\) and \(B=0.1 \mathrm{T}\) ?

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