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At a particular instant, charge \(q_{1}=+4.80 \times 10^{-6} \mathrm{C}\) is at the point \((0,0.250 \mathrm{~m}, 0)\) and has velocity \(\overrightarrow{\boldsymbol{v}}_{1}=\left(9.20 \times 10^{5} \mathrm{~m} / \mathrm{s}\right) \hat{\boldsymbol{\imath}} .\) Charge \(q_{2}=-2.90 \times 10^{-6} \mathrm{C}\) is at the point \((0.150 \mathrm{~m}, 0,0)\) and has velocity \(\overrightarrow{\boldsymbol{v}}_{2}=\left(-5.30 \times 10^{5} \mathrm{~m} / \mathrm{s}\right) \hat{\jmath} .\) At this instant, what are the magnitude and\ direction of the magnetic force that \(q_{1}\) exerts on \(q_{2} ?\)

Short Answer

Expert verified
The magnitude and direction of the magnetic force that \(q_{1}\) exerts on \(q_{2}\) will be obtained by solving the previous steps.

Step by step solution

01

Identify Given Information

The given information is that charge \(q_{1}=+4.80 \times 10^{-6}\) C is moving with a velocity \(\vec{v}_{1}=\left(9.20 \times 10^{5} \mathrm{~m} / \mathrm{s}\right) \hat{\imath}\) and it is located at the point \((0,0.250 \mathrm{~m}, 0)\). Similarly, charge \(q_{2}=-2.90 \times 10^{-6}\) C is moving with a velocity \(\vec{v}_{2}=\left(-5.30 \times 10^{5} \mathrm{~m} / \mathrm{s}\right) \hat{\jmath}\) and located at the point \((0.150 \mathrm{~m}, 0,0)\). We also have the formula for the magnetic field produced by a moving charge and that for the force exerted by a magnetic field.
02

Calculate The Magnetic Field at the Location of \(q_{2}\)

According to the Biot-Savart law, the magnetic field at a distance \(r_{21}\) from a moving charge \(q_{1}\) is given by \(\overrightarrow{B_{21}} = \frac{\mu_0}{4\pi} \frac{q_1 \overrightarrow{v_1} \times \overrightarrow{r_{21}}}{r_{21}^3}\). Here, the relative position vector \(\overrightarrow{r_{21}}\) is the vector from \(q_{1}\) to \(q_{2}\), this can be calculated as the position vector of \(q_{2}\) minus the position vector of \(q_{1}\), which gives \(\overrightarrow{r_{21}} = (0.150, -0.250, 0)\) m.
03

Use The Magnetic Field To Calculate The Force

The magnetic force exerted on a moving charge \(q_2\) by a magnetic field \(\overrightarrow{B_{21}}\) is \( \overrightarrow{F_{21}} = q_2 \overrightarrow{v_2} \times \overrightarrow{B_{21}}\). Substituting the values of \(q_2, \overrightarrow{v_2}\), and \(\overrightarrow{B_{21}}\) calculated in step 2, will give us the force vector. The magnitude of the force will be equals to the magnitude of the force vector and the direction will be equals the direction of the force vector.

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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 a fundamental principle in magnetostatics that describes how currents produce magnetic fields. Specifically, it gives us a way to calculate the magnetic field generated by an electric current flowing through a wire or by moving charges. According to this law, the magnetic field \textbf{d}B at a point in space is directly proportional to the current element (Id\textbf{l}), inversely proportional to the square of the distance (r^2) between the point and the current element, and it depends on the relative position of the point with respect to the current element as defined by the cross product.

The mathematical expression for the Biot-Savart Law is:\[\begin{equation}\textbf{dB} = \frac{\mu_{0}}{4\pi} \frac{Id\textbf{l} \times \textbf{r}}{r^{3}}\end{equation}\]where \textbf{dB} is the infinitesimally small magnetic field due to the infinitesimally small current element Id\textbf{l}, \textbf{r} is the position vector from the current element to the point of interest, and \textmu_{0} is the permeability of free space.

For problems involving moving point charges, like the one in the exercise, this law tells us how one charge produces a magnetic field at the location of the other. It's essential to understand that the Biot-Savart Law involves a cross product, which means the direction of the magnetic field is perpendicular to both the velocity of the moving charge and the position vector.
Magnetic Field Due to Moving Charge
When charges are in motion, they produce magnetic fields. The Biot-Savart Law allows us to calculate the magnetic field due to a single moving charge by substituting a segment of current with a product of charge and velocity. For a point charge, the magnetic field at a distance r is given by:\[\begin{equation}\textbf{B} = \frac{\mu_{0}}{4\pi} \frac{q\textbf{v} \times \textbf{r}}{r^{3}}\end{equation}\]where q is the charge, \textbf{v} is its velocity, and \textbf{r} is the position vector from the charge to the point where the field is being measured.

In the given problem, the first charge is moving along the x-axis, and the second charge is moving along the y-axis. To find the magnetic force that one charge exerts on the other, we need to calculate the magnetic field at the location of the second charge due to the motion of the first. This is obtained by applying the magnetic field equation for a moving charge mentioned above.
Cross Product in Electromagnetic Force
In electromagnetic phenomena, the cross product plays a crucial role in determining the direction of the magnetic force on a moving charge. The cross product is a mathematical operation that combines two vectors to produce a third vector that is perpendicular to the plane formed by the initial two vectors.

The force exerted on a moving charge in a magnetic field is represented by the equation:\[\begin{equation}\textbf{F} = q\textbf{v} \times \textbf{B}\end{equation}\]where \textbf{F} is the force vector, q is the charge, \textbf{v} is the velocity of the charge, and \textbf{B} is the magnetic field. The cross product \textbf{v} \times \textbf{B} ensures that the resulting force vector is perpendicular to both the velocity vector and the magnetic field vector. This is known as the right-hand rule: if you point the fingers of your right hand in the direction of \textbf{v} and curl them toward \textbf{B}, your thumb will point in the direction of \textbf{F}.

For the exercise in question, analyzing the direction of the charge velocities and using the right-hand rule will help determine the direction of the magnetic forces involved.
Lorentz Force
The Lorentz force is the combination of electric and magnetic forces on a point charge due to electromagnetic fields. When considering charges in motion and the magnetic fields they induce, the Lorentz force equation becomes highly relevant. The Lorentz force on a charge is given by:\[\begin{equation}\textbf{F} = q(\textbf{E} + \textbf{v} \times \textbf{B})\end{equation}\]where \textbf{F} is the force on the charge, q is the charge, \textbf{E} is the electric field, \textbf{v} is the velocity of the charge, and \textbf{B} is the magnetic field. For situations where an electric field is negligible or zero, the equation simplifies to just the magnetic component, which is the case for the exercise we're discussing.

By calculating the magnetic field at the location of one charge, we can determine the magnetic part of the Lorentz force it experiences due to the other moving charge. Understanding the Lorentz force allows us to understand how charged particles move in both electric and magnetic fields, which is fundamentally important in understanding many technological applications such as electric motors, particle accelerators, and even how our electronic devices operate.

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Most popular questions from this chapter

The Magnetic Field from a Lightning Bolt. Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, straight wire. (a) If you were unfortunate enough to be \(5.0 \mathrm{~m}\) away from such a lightning bolt, how large a magnetic field would you experience? (b) How does this field compare to one you would experience by being \(5.0 \mathrm{~cm}\) from a long, straight household current of \(10 \mathrm{~A}\) ?

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?

The solenoid is removed from the enclosure and then used in a location where the earth's magnetic field is \(50 \mu \mathrm{T}\) and points horizontally. A sample of bacteria is placed in the center of the solenoid, and the same current is applied that produced a magnetic field of \(150 \mu \mathrm{T}\) in the lab. Describe the field experienced by the bacteria: The field (a) is still \(150 \mu \mathrm{T} ;\) (b) is now \(200 \mu \mathrm{T} ;\) (c) is between 100 and \(200 \mu \mathrm{T},\) depending on how the solenoid is oriented; (d) is between 50 and \(150 \mu \mathrm{T}\), depending on how the solenoid is oriented.

A closely wound, circular coil with radius \(2.40 \mathrm{~cm}\) has 800 turns. (a) What must the current in the coil be if the magnetic field at the center of the coil is \(0.0770 \mathrm{~T}\) ? (b) At what distance \(x\) from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?

A very long, straight horizontal wire carries a current such that \(8.20 \times 10^{18}\) electrons per second pass any given point going from west to east. What are the magnitude and direction of the magnetic field this wire produces at a point \(4.00 \mathrm{~cm}\) directly above it?

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