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A \(+6.00 \mu\) C point charge is moving at a constant \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in the \(+y\) -direction, relative to a reference frame. At the instant when the point charge is at the origin of this reference frame, what is the magnetic-field vector \(\overrightarrow{\boldsymbol{B}}\) it produces at the following points: (a) \(x=0.500 \mathrm{~m}, y=0, z=0\) (b) \(x=0, y=-0.500 \mathrm{~m}, z=0\) (c) \(x=0, \quad y=0, \quad z=+0.500 \mathrm{~m}\) (d) \(x=0, \quad y=-0.500 \mathrm{~m}\) \(z=+0.500 \mathrm{~m} ?\)

Short Answer

Expert verified
The magnetic field vectors will be found individually after calculating the values and directions for each case using the Biot-Savart law. The answers will be in Tesla unit.

Step by step solution

01

Declaration of Given Parameters

Declare all the given parameters of the problem: the charge Q = +6.00 \(\mu\)C = 6 x 10^-6 C, and its constant speed v = 8.00 x 10^6 m/s which is in the y-direction, hence \(\vec{v}\) = 0i + 8.00 x10^6j + 0k. Different points given at which \(\vec{B}\) is to be calculated are (0.500, 0, 0), (0, -0.500, 0), (0, 0, 0.500), (0, -0.500, 0.500).
02

Use Biot-Savart law

Apply the Biot-Savart law to calculate the magnetic field vector at each point. The vector \(\vec{r}\) will be the position of the point (x, y, z) minus the position of the point charge located at the origin. Then, calculate the cross product of \(\vec{v}\) and \(\vec{r}\) and insert it into the formula for each case. Don’t forget to use the absolute value of the cross product for the numerator and for the denominator the cubic distance from the charge.
03

Solve Individual Cases

Each point represents a different case, so they'll need to be solved individually using previous steps and adding the specific coordinates.
04

Find the Magnetic Field Vector

After all calculations are done, the magnitude and direction of each magnetic field vector at each point will be found, satisfying the problem’s conditions.

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Key Concepts

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

Biot-Savart Law
The Biot-Savart Law is crucial for understanding the magnetic field caused by a moving electric charge. It is a mathematical description articulating how currents produce magnetic fields and could be seen as an equivalent of Coulomb's law in the magnetic realm. According to this law, a moving charge or current element generates a magnetic field that is directly proportional to the magnitude of the electric current and inversely proportional to the square of the distance from the point of interest to the current element.

The law is expressed by the formula:
\[\textbf{B} = \frac{\mu_0}{4\pi} \int \frac{Id\textbf{s} \times \hat{r}}{r^2}\]where \(\textbf{B}\) is the magnetic field, \(\mu_0\) is the permeability of free space, \(I\) is the current, \(d\textbf{s}\) is an infinitesimal element of the current path, \(\hat{r}\) is the unit vector from the current element to the point of observation, and \(r\) is the distance between them. For a point charge moving with a velocity \(\vec{v}\), the current '\(I\)' is replaced by the charge \(q\) times velocity \(\vec{v}\) (analogous to current being charge per unit time). The cross product \(d\textbf{s} \times \hat{r}\) stresses the vector nature of the magnetic field, which is perpendicular both to the current and the direction to the point of interest.
Magnetic Field Calculation
Calculating the magnetic field produced by a moving charge involves determining both the magnitude and direction of the magnetic field vector. The magnetic field \(\textbf{B}\) at a distance from a moving charge can be calculated using a modification of the Biot-Savart Law, particularly by the formula:
\[\textbf{B} = \frac{\mu_0}{4\pi} \frac{q\textbf{v} \times \vec{r}}{r^3}\]where \(q\) is the electric charge, \(\textbf{v}\) is the velocity vector of the charge, and \(\vec{r}\) is the position vector from the charge to the point of interest. The cross product \(\textbf{v} \times \vec{r}\) ensures that the resulting magnetic field is perpendicular to the plane formed by the velocity of the charge and the line connecting the charge to the point at which the field is being calculated.

For a point charge in motion, this formula encapsulates the effects of both the speed and direction of the charge as well as the relative position of the point where the field is being measured.
Electric Charge Motion
The concept of electric charge motion is fundamental in magnetism. When an electric charge moves, it generates a magnetic field, which is a cornerstone of electromagnetic theory. The magnetic field is not static; it varies as the velocity of the charge changes. In the context of our problem, the charge is in motion along the y-axis at a constant velocity, which simplifies the problem as the magnetic field at any given point is solely dependent on this consistent motion.

To fully understand the scenario, one should note that only a moving charge produces a magnetic field—the magnitude of this magnetic field is proportional to the speed of the charge and inversely proportional to the square of the distance from the point of observation to the charge. The consistent velocity vector \(\vec{v}\) interacts with the position vectors of the points in question to give us a dynamic picture of the resulting magnetic field at any specific location.
Cross Product in Magnetism
The cross product in magnetism represents the vector multiplication that is essential in calculating the direction and magnitude of a magnetic field produced by a moving charge. Specifically, when using the Biot-Savart Law to calculate the magnetic field \(\textbf{B}\), the cross product \(\vec{v} \times \vec{r}\) is used, which yields a vector that is perpendicular to both the velocity vector \(\vec{v}\) of the moving charge and the position vector \(\vec{r}\).

The direction of this cross product vector determines the orientation of the magnetic field lines around the moving charge. The right-hand rule is often used to visualize the direction of the magnetic field: if you point your right hand's thumb in the direction of the velocity and curl your fingers towards the direction of \(\vec{r}\), your palm faces the direction of the magnetic field.

The magnitude of the cross product is equal to the product of the magnitudes of the two vectors multiplied by the sine of the angle between them, which is ultimately utilized in calculating the magnitude of the magnetic field.

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Most popular questions from this chapter

A wire of length \(20.0 \mathrm{~cm}\) lies along the \(x\) -axis with the center of the wire at the origin. The wire carries current \(I=8.00 \mathrm{~A}\) in the \(-x\) -direction. (a) What is the magnitude \(B\) of the magnetic field of the wire at the point \(y=5.00 \mathrm{~cm}\) on the \(y\) -axis? (b) What is the percent difference between the answer in (a) and the value you obtain if you assume the wire is infinitely long and use Eq. ( 28.9 ) to calculate \(B\) ?

An alpha particle (charge \(+2 e\) ) and an electron move in opposite directions from the same point, each with the speed of \(2.50 \times 10^{5} \mathrm{~m} / \mathrm{s}\) (Fig. E28.4). Find the magnitude and direction of the total magnetic field these charges produce at point \(P,\) which is \(8.65 \mathrm{nm}\) from each charge.

A long, straight, solid cylinder, oriented with its axis in the \(z\) -direction, carries a current whose current density is \(\overrightarrow{\boldsymbol{J}}\). The current density, although symmetric about the cylinder axis, is not constant but varies according to the relationship $$ \begin{array}{rlr} \overrightarrow{\boldsymbol{J}} & =\frac{2 I_{0}}{\pi a^{2}}\left[1-\left(\frac{r}{a}\right)^{2}\right] \hat{k} & \text { for } r \leq a \\ & =\mathbf{0} \quad & \text { for } r \geq a \end{array} $$ where \(a\) is the radius of the cylinder, \(r\) is the radial distance from the cylinder axis, and \(I_{0}\) is a constant having units of amperes. (a) Show that \(I_{0}\) is the total current passing through the entire cross section of the wire. (b) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \geq a\). (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. (d) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \leq a\). How do your results in parts (b) and (d) compare for \(r=a ?\)

The law of Biot and Savart in Eq. ( 28.7 ) generalizes to the case of surface currents as $$ \overrightarrow{\boldsymbol{B}}=\frac{\mu_{0}}{4 \pi} \int \frac{\sigma \overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}}}{r^{2}} d a $$ where \(\sigma\) is the local charge density, \(\overrightarrow{\boldsymbol{v}}\) is the local velocity, and \(d a\) is a differential area element. Re-visit Challenge Problem 28.76 and use the above equation as an alternative means to derive the magnetic field at the center of the cylinder. Use the following steps: (a) Write the charge density \(\sigma\). (b) The origin is at the center of the cylinder. What is the vector \(\vec{v}\) that points from the element with coordinates \((x, y, z)=(x, R \cos \phi, R \sin \phi)\) to the origin? (c) What is the velocity \(\overrightarrow{\boldsymbol{v}}\) of the element? (d) What is the vector product \(\overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}} ?\) (e) An area element on the cylinder may be written as \(d a=R d x d \phi .\) Use this and the previously established information to write the generalized law of Biot and Savart as a double integral. Evaluate the integral to determine the magnetic field \(\vec{B}\) at the center of the cylinder. (f) Is your result consistent with your result in Challenge Problem \(28.76 ?\)

A cylindrical shell with radius \(R_{1}\) and height \(H\) has charge \(Q_{1}\) and rotates around its axis with angular speed \(\omega_{1},\) as shown in Fig. \(\mathbf{P 2 8 . 7 1 .}\). Inside the cylinder, far from its edges, sits a very small disk with radius \(R_{2},\) mass \(M,\) and charge \(Q_{2}\) mounted on a pivot, spinning with a large angular velocity \(\vec{\omega}_{2}\) and oriented at angle \(\theta\) with respect to the axis of the cylinder. The center of the disk is on the axis of the cylinder. The magnetic interaction between the cylinder and the disk causes a precession of the axis of the disk. (a) What is the magnitude of the enclosed current \(I_{\text {encl }}\) surrounded by a loop that has one vertical side that is along the axis of the cylinder and extends beyond the top and bottom of the cylinder? The other vertical side of the loop is very far outside the cylinder. (b) Assume the field is uniform within the cylinder and use Ampere's law to find the magnetic field at the center of the disk. (c) The magnetic moment of the disk has magnitude \(\mu=\frac{1}{4} Q_{2} \omega_{2} R_{2}^{2} .\) What is the magnitude of the torque exerted on the disk? (d) What is the magnitude of the angular momentum of the disk?

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