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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) \(x=0\) \(y=-0.500 \mathrm{m}, \quad z=0 ; \quad(\mathrm{c}) \quad x=0, \quad y=0, \quad z=+0.500 \mathrm{m} ;\) (d) \(x=0, y=-0.500 \mathrm{m}, z=+0.500 \mathrm{m} ?\)

Short Answer

Expert verified
(a) \(\overrightarrow{B} = 3.84 \times 10^{-5} \hat{k} \mathrm{T}\); (b) \(\overrightarrow{B} = 0 \mathrm{T}\); (c) \(\overrightarrow{B} = -3.84 \times 10^{-5} \hat{i} \mathrm{T}\); (d) \(\overrightarrow{B} = 2.72 \times 10^{-5} \mathrm{T}\,\) in \( -\hat{i} \) direction.

Step by step solution

01

Understanding Biot-Savart Law

To find the magnetic field produced by a moving point charge, we use the Biot-Savart law: \( \overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{q \overrightarrow{v} \times \overrightarrow{r}}{r^3} \) where \( \overrightarrow{r} \) is the position vector from the charge to the observation point, \( \overrightarrow{v} \) is the velocity vector, \( q \) is the charge, and \( \mu_0 \) is the permeability of free space.
02

Calculate Constants

Given the charge \( q = +6.00 \times 10^{-6} \mathrm{C} \) and velocity \( v = 8.00 \times 10^6 \mathrm{m/s} \), we first calculate the constant: \( \frac{\mu_0 q v}{4\pi} = \frac{4\pi \times 10^{-7} \times 6.00 \times 10^{-6} \times 8.00 \times 10^6}{4\pi} = 9.60 \times 10^{-6} \).
03

Calculate Magnetic Field at Point (a)

For point (a), convert the coordinates into a position vector \( \overrightarrow{r} = (0.500\hat{i}) \). Use the right-hand rule for the cross product \( \overrightarrow{v} = (8.00 \times 10^6 \hat{j}) \) and the formula to find \( \overrightarrow{B} = \frac{9.60 \times 10^{-6} (8.00 \times 10^6 \hat{j} \times 0.500\hat{i})}{(0.500)^3} = 3.84 \times 10^{-5} \hat{k} \mathrm{T} \).
04

Calculate Magnetic Field at Point (b)

For point (b), \( \overrightarrow{r} = (-0.500\hat{j}) \). Calculate \( \overrightarrow{B} = \frac{9.60 \times 10^{-6} (8.00 \times 10^6 \hat{j} \times -0.500\hat{j})}{(0.500)^3} = 0 \mathrm{T} \). Since \( \overrightarrow{v} \) and \( \overrightarrow{r} \) are parallel, the cross product is zero.
05

Calculate Magnetic Field at Point (c)

For point (c), \( \overrightarrow{r} = 0.500\hat{k} \). Here, \( \overrightarrow{B} = \frac{9.60 \times 10^{-6} (8.00 \times 10^6 \hat{j} \times 0.500\hat{k})}{(0.500)^3} = -3.84 \times 10^{-5} \hat{i} \mathrm{T} \).
06

Calculate Magnetic Field at Point (d)

For point (d), \( \overrightarrow{r} = -0.500\hat{j} + 0.500\hat{k} \). Calculate \( \overrightarrow{B} = \frac{9.60 \times 10^{-6} (8.00 \times 10^6 \hat{j} \times (-0.500\hat{j} + 0.500\hat{k}))}{\sqrt{(-0.500)^2 + 0.500^2}^3} \). The cross product is \( -4.00 \times 10^6 \hat{i} \), giving a magnitude \( 2.72 \times 10^{-5} \mathrm{T} \) in \( -\hat{i} \) direction.

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Key Concepts

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

Magnetic Field Calculation
To understand magnetic field calculations, we need to delve into the Biot-Savart Law. This fundamental law of electromagnetism helps us determine the magnetic field produced by a current or a moving charge. It's expressed as: \[ \overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{q \overrightarrow{v} \times \overrightarrow{r}}{r^3} \] where:
  • \( \mu_0\) is the permeability of free space.
  • \( q \) is the charge of the particle.
  • \( \overrightarrow{v} \) is the velocity vector.
  • \( \overrightarrow{r} \) is the position vector from the charge to the observation point.
The challenge lies in performing calculations effectively using this law. We need to manage the cross product and ensure we calculate the distance \( r \) correctly, especially when dealing with three-dimensional coordinates. Breaking down each element and calculating the magnetic field at specific points is crucial for accurate results.
Moving Point Charge
A moving point charge is simply a charged particle that has momentum. In the given problem, we have a point charge of \( +6.00\, \mu\text{C} \) traveling at \( 8.00 \times 10^6 \,\text{m/s} \) along the \( +y \)-direction. When a charge moves, it creates a magnetic field in addition to an electric field. The velocity of the point charge is essential for determining the strength and direction of the magnetic field it produces. Understanding how the point charge interacts with its environment is key. The speed and direction of the charge directly impact the magnetic field through the Biot-Savart Law. This is why it's important to consider these factors when calculating magnetic fields at various points in space.
Cross Product in Physics
In physics, the cross product is a crucial operation for calculating different vector relationships, especially within electromagnetism. It's a method for determining a perpendicular vector from two other vectors, and is defined as: \[ \overrightarrow{A} \times \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \sin(\theta) \hat{n} \] Here, \(|\overrightarrow{A}|\) and \(|\overrightarrow{B}|\) are the magnitudes of the vectors, \(\theta\) is the angle between them, and \(\hat{n}\) is a unit vector perpendicular to both. This operation is very relevant to calculating the magnetic field in this exercise because the field's direction depends on it. The Biot-Savart Law utilizes the cross product to determine how the velocity vector \( \overrightarrow{v} \) and the position vector \( \overrightarrow{r} \) combine to give a magnetic field vector. The right-hand rule helps us determine this field's direction by orienting our fingers in the direction of the first vector and swooping them toward the second vector.
Magnetic Field Vector
The magnetic field vector \( \overrightarrow{B} \) represents not only the strength but also the direction of the field around a moving charge. Calculations will show this vector points perpendicularly from the direction of charge movement, thanks to the cross product in the magnetic field equation. When viewing the problem through vector components, the direction of \( \overrightarrow{B} \) is crucial. You will often use unit vectors \( \hat{i}, \hat{j}, \text{and }\hat{k} \) to express directional alignment. The vector's magnitude is found by plugging the relevant numbers into the Biot-Savart Law. This gives a real-world measure of the field's intensity at a point. Knowing these vectors and their notation is essential for anyone tackling magnetic field problems. The Biot-Savart Law efficiently provides the directional information needed to plot or visualize these fields.

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Most popular questions from this chapter

A solenoid is designed to produce a magnetic field of 0.0270 \(\mathrm{T}\) at its center. It has radius 1.40 \(\mathrm{cm}\) and length \(40.0 \mathrm{cm},\) and the wire can carry a maximum current of 12.0 \(\mathrm{A}\) . (a) What minimum number of turns per unit length must the solenoid have? (b) What total length of wire is required?

Two long. parallel transmission lines, 40.0 \(\mathrm{cm}\) apart, carry \(25.0-\mathrm{A}\) and 75.0 - A currents. Find all locations where the net magnetic field of the two wires is zero if these currents are in (a) the same direction and (b) the opposite direction.

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 0.0580 \(\mathrm{T}\) ? 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?

Two identical circular, wire loops 40.0 \(\mathrm{cm}\) in diameter each carry a current of 1.50 \(\mathrm{A}\) in the same direction. These loops are parallel to each other and are 25.0 \(\mathrm{cm}\) apart. Line \(a b\) is normal to the plane of the loops and passes through their centers. A proton is fired at 2400 \(\mathrm{km} / \mathrm{sperpendicular}\) to line \(a b\) from a point midway between the centers of the loops. Find the magnitude and direction of the magnetic force these loops exert on the proton just after it is fired.

A toroidal solenoid (see Example 28.10 ) has inner radius \(r_{1}=15.0 \mathrm{cm}\) and ourer radius \(r_{2}=18.0 \mathrm{cm} .\) The solenoid has 250 turns and carries a current of 8.50 A. What is the magnitude of the magnetic field at the following distances from the center of the torus: (a) \(12.0 \mathrm{cm} ;(\mathrm{b}) 16.0 \mathrm{cm} ;(\mathrm{c}) 20.0 \mathrm{cm} ?\)

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