The magnetic field in a region in space (where \(x>0\) and \(y>0\) ) is given by \(B=(x-a z) \hat{y}+(x y-b) \hat{z},\) where \(a\) and \(b\) are positive constants. An electron moving with a constant velocity, \(\vec{v}=v_{0} \hat{x},\) enters this region. What are the coordinates of the points at which the net force acting on the electron is zero?

Short Answer

Expert verified
Answer: The coordinates of the point where the net force acting on the electron is zero are \((x, y, z) = \left(a\sqrt{\frac{b}{a}}, \sqrt{\frac{b}{a}}, \sqrt{\frac{b}{a}}\right)\).

Step by step solution

01

Write down the given magnetic field and velocity vectors

The magnetic field is given by \(\vec{B} = (x - az)\hat{y} + (xy - b)\hat{z}\). The electron has a constant velocity vector \(\vec{v} = v_{0}\hat{x}\).
02

Find the magnetic force vector

The magnetic force vector is given by \(\vec{F} = q(\vec{v} \times \vec{B})\). We need to calculate the cross product of \(\vec{v}\) and \(\vec{B}\):\((\vec{v} \times \vec{B}) = |\begin{matrix}\hat{x}&\hat{y}&\hat{z}\\0&v_0&0\\(x-az)&0&(xy-b)\end{matrix}| = v_0((x-az)\hat{z} - (xy-b)\hat{y})\). So, the magnetic force vector is: \(\vec{F} = qv_0((x-az)\hat{z} - (xy-b)\hat{y})\).
03

Set the magnetic force components to zero

For the net force to be zero, the magnetic force should be zero in both y and z directions. Thus, we get two equations: 1. \(qv_0(xy-b) = 0 \Rightarrow xy = b\) 2. \(qv_0(x-az) = 0 \Rightarrow x = az\)
04

Solve equations for coordinates

From equation (2), we get \(x = az\). Now, substitute this value into equation (1):\(a z^2 = b \Rightarrow z^2 = \frac{b}{a}\). Taking the square root, we get \(z = \pm\sqrt{\frac{b}{a}}\). However, we are given that \(x > 0\) and \(y > 0\), so we can take the positive square root for z. Therefore, \(z = \sqrt{\frac{b}{a}}\). Now, we can plug this value for z into equation (2) and obtain \(x = a\sqrt{\frac{b}{a}}\). Finally, we can use equation (1) to get \(y = \frac{b}{a\sqrt{\frac{b}{a}}} = \sqrt{\frac{b}{a}}\).
05

Write down the coordinates of the point

The coordinates of the point where the net force acting on the electron is zero are \((x, y, z) = \left(a\sqrt{\frac{b}{a}}, \sqrt{\frac{b}{a}}, \sqrt{\frac{b}{a}}\right)\).

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Key Concepts

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

Magnetic Force on a Moving Charge
Understanding how a magnetic field influences a moving charge is fundamental to the study of electromagnetism. The magnetic force exerted on a moving charge can be calculated using the simple formula:
\( \textbf{F}_\text{mag} = q(\textbf{v} \times \textbf{B}) \),
where \( q \) is the charge, \( \textbf{v} \) is the velocity of the charge, and \( \textbf{B} \) is the magnetic field. This cross product indicates that the direction of the force is perpendicular to both the velocity of the charge and the magnetic field. Additionally, the magnitude of this force is dependent on the angle between the velocity and the magnetic field vectors.
For a charge moving perpendicular to the magnetic field, the force is at its maximum, while for a charge moving parallel to the field, the force is zero, as there is no component of velocity across the field lines. When dealing with problems where the net force is zero, we are often finding the points at which the velocity and magnetic field are parallel, thus nullifying the magnetic force.
  • The direction of the force follows the right-hand rule.
  • The magnitude is zero when the charge moves parallel to the magnetic field lines.
  • If the charge moves at an angle to the field, the force is proportional to the sine of the angle between \( \textbf{v} \) and \( \textbf{B} \).
Cross Product in Physics
The cross product, or vector product, is a binary operation on two vectors in three-dimensional space and is denoted by the symbol '\( \times \)'. The cross product between two vectors \( \textbf{a} \) and \( \textbf{b} \) is given by:
\( \textbf{a} \times \textbf{b} = |a||b|\text{sin}(\theta)\textbf{n} \),
where \( |a| \) and \( |b| \) are magnitudes of vectors \( \textbf{a} \) and \( \textbf{b} \), \( \theta \) is the angle between them, and \( \textbf{n} \) is a unit vector perpendicular to the plane containing \( \textbf{a} \) and \( \textbf{b} \), with direction given by the right-hand rule.
In physics, the cross product is vital in calculating forces which are directional and dependent on the orientation of vectors, such as the magnetic force on a moving charge. Finding the cross product can be done using the determinant of a matrix constructed from the two vectors, when expressed in a coordinate system:
\[ ( \textbf{a} \times \textbf{b} ) = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ a_x & a_y & a_z \ b_x & b_y & b_z \ \end{vmatrix} \]
This process yields a new vector whose components determine the force's direction in each of the coordinate system's axes.
  • The cross product results in a vector perpendicular to the original plane.
  • The magnitude depends on the sine of the angle between the vectors, meaning it is maximum when they are perpendicular.
Lorentz Force
The Lorentz force is the combination of electric and magnetic forces on a point charge due to electromagnetic fields. It is given by:
\( \textbf{F} = q(\textbf{E} + \textbf{v} \times \textbf{B}) \),
where \( q \) is the charge of the particle, \( \textbf{E} \) is the electric field, \( \textbf{v} \) is the velocity of the particle, and \( \textbf{B} \) is the magnetic field. In the absence of an electric field, the Lorentz force equation simplifies to just the magnetic force component, as discussed earlier.
The significance of the Lorentz force lies in its central role in explaining how charged particles behave in electromagnetic fields, influencing the design of devices from electric motors to particle accelerators. In analyzing the motion of charged particles, it's crucial to consider both components of the Lorentz force, if applicable, to accurately predict their trajectories.
  • It governs the motion of charged particles in electromagnetic fields.
  • The magnetic component of the force is perpendicular to the velocity and the magnetic field.
Coordinate System in Physics
A coordinate system in physics allows for the description of the position and movement of objects in space. Most commonly, we use the three-dimensional Cartesian coordinate system, which consists of three mutually perpendicular axes: x, y, and z. By knowing the coordinates on these axes, we can describe the exact location of a point in space.
When applying forces in physics, especially those that result from vector calculations like the cross product, understanding the orientation of these axes and how movements along them correlate to real-world directions is vital. In the exercise provided, the magnetic field and the motion of the electron were described using unit vectors along these axes. This enables us to perform vector operations like the cross product to find the magnetic force on the electron.
By solving for when this force is zero, we effectively seek the coordinates in this system where the particle experiences no net force, which means it can continue moving unaffected by the magnetic field.
  • It provides a systematic way to represent physical quantities like displacement, velocity, and acceleration.
  • Applying vector operations correctly depends on identifying the correct axes and their orientations.

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Most popular questions from this chapter

A conducting rod of length \(L\) slides freely down an inclined plane, as shown in the figure. The plane is inclined at an angle \(\theta\) from the horizontal. A uniform magnetic field of strength \(B\) acts in the positive \(y\) -direction. Determine the magnitude and the direction of the current that would have to be passed through the rod to hold it in position on the inclined plane.

Initially at rest, a small copper sphere with a mass of \(3.00 \cdot 10^{-6} \mathrm{~kg}\) and a charge of \(5.00 \cdot 10^{-4} \mathrm{C}\) is accelerated through a \(7000 .-\mathrm{V}\) potential difference before entering a magnetic field of magnitude \(4.00 \mathrm{~T}\), directed perpendicular to its velocity. What is the radius of curvature of the sphere's motion in the magnetic field?

An electron with a speed of \(4.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) enters a uniform magnetic field of magnitude \(0.040 \mathrm{~T}\) at an angle of \(35^{\circ}\) to the magnetic field lines. The electron will follow a helical path. a) Determine the radius of the helical path. b) How far forward will the electron have moved after completing one circle?

It would be mathematically possible, for a region with zero current density, to define a scalar magnetic potential analogous to the electrostatic potential: \(V_{B}(\vec{r})=-\int_{\vec{r}_{0}}^{\vec{r}} \vec{B} \cdot d \vec{s},\) or \(\vec{B}(\vec{r})=-\nabla V_{B}(\vec{r}) .\) However, this has not been done. Explain why not.

A proton is accelerated from rest by a potential difference of \(400 .\) V. The proton enters a uniform magnetic field and follows a circular path of radius \(20.0 \mathrm{~cm} .\) Determine the magnitude of the magnetic field.

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