Chapter 28: Problem 37
An electron moves through a uniform magnetic field given by \(\vec{B}=B_{x} \hat{\mathrm{i}}+\left(-3.0 B_{x}\right) \hat{\mathrm{j}}\). At a particular instant, the electron has velocity \(\vec{v}=(2.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\) and the magnetic force acting on it is \(\left(6.4 \times 10^{-19} \mathrm{~N}\right) \hat{\mathrm{k}}\). Find \(B_{x^{-}}\)
Short Answer
Step by step solution
Understanding the Lorentz Force
Setting Up the Cross Product
Calculating the Determinant
Applying the Lorentz Force Equation
Solving for \( B_x \)
Conclusion: Finding \( B_x \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnetic Field
Magnetic fields are vectors, denoted typically by \( \vec{B} \), consisting of both a magnitude and a direction. These fields can be generated by various sources, such as a magnet or a current-carrying wire, following the right-hand rule for direction.
- Uniform magnetic fields have the same magnitude and direction at every point.
- The SI unit for magnetic field strength is the Tesla (T).
Cross Product
For vectors \(\vec{A}\) and \(\vec{B}\), the cross product \(\vec{A}\times\vec{B}\) is found using the determinant of a matrix that includes the unit vectors and the components of the original vectors. In our exercise:
\[\vec{v} \times \vec{B} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \2.0 & 4.0 & 0 \B_x & -3.0B_x & 0 \end{vmatrix}\]
By calculating this determinant, we find that the cross product results in a vector along the \(\hat{\mathrm{k}}\) direction. This result is crucial to understanding the forces involved in the exercise:
- The cross product's magnitude depends not only on the magnitudes of the original vectors but also on the sine of the angle between them.
- The direction follows the right-hand rule: if you point the fingers of your right hand from \(\vec{v}\) to \(\vec{B}\), your thumb points in the direction of \(\vec{v} \times \vec{B}\).
Charged Particle Motion
The Lorentz force \(\vec{F} = q(\vec{v} \times \vec{B})\) is perpendicular to both the velocity \(\vec{v}\) and the magnetic field \(\vec{B}\), creating a centripetal motion. In this exercise:
- The electron, having a negative charge, moves in a manner opposite to that of a positive charge.
- The direction of the force can be determined using the right-hand rule, which must be adjusted for the electron's negative charge.
Elementary Charge
The value of \(e\) is \(1.6 \times 10^{-19}\) Coulombs, and it is fundamental to understanding interactions in electric and magnetic fields.
- This charge unit is used to quantify the charge of particles in electromagnetism equations, like the Lorentz force.
- In the exercise, the Lorentz force equation \(\vec{F} = q(\vec{v} \times \vec{B})\) specifically uses \(-e\) for an electron.