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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 \(\vec{\boldsymbol{v}}_{1}=\left(9.20 \times 10^{5} \mathrm{m} / \mathrm{s}\right) \hat{\imath}\) Charge \(q_{2}=-2.90 \times 10^{-6} \mathrm{C}\) is at the point \((0.150 \mathrm{m}, 0,0)\) and has velocity \(\vec{v}_{2}=\left(-5.30 \times 10^{5} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{J}}\) . 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 magnetic force on \( q_2 \) is approximately \( 3.90 \times 10^{-6} \text{ N} \) in the positive x-direction.

Step by step solution

01

Formula for Magnetic Force

The magnetic force between two moving charges is given by the Lorentz force formula. For a charge \( q_2 \) moving with velocity \( \vec{v}_2 \) in the magnetic field \( \vec{B} \) due to \( q_1 \), the force can be calculated using: \[ \vec{F} = q_2(\vec{v}_2 \times \vec{B}) \] where \( \vec{B} \) is the magnetic field produced by \( q_1 \).
02

Magnetic Field by Moving Charge

The magnetic field \( \vec{B} \) produced by a moving point charge \( q_1 \) is given by Biot-Savart Law: \[ \vec{B} = \frac{\mu_0}{4\pi} \frac{q_1 \vec{v}_1 \times \vec{r}}{r^3} \] where \( \vec{r} \) is the position vector from \( q_1 \) to \( q_2 \), \( r \) is the magnitude of \( \vec{r} \), and \( \mu_0 = 4\pi \times 10^{-7} \text{ T·m/A} \) is the permeability of free space.
03

Calculate Displacement Vector and its Magnitude

The displacement vector \( \vec{r} \) from \( q_1 \) to \( q_2 \) is \( (0.150 \hat{\imath} - 0.250 \hat{\jmath} + 0 \hat{k}) \text{ m} \). Calculate its magnitude: \[ r = \sqrt{(0.150)^2 + (-0.250)^2 + 0^2} \approx 0.291 \text{ m} \].
04

Calculate Cross Product \( \vec{v}_1 \times \vec{r} \)

Perform the cross product \( \vec{v}_1 \times \vec{r} \): \[ (9.20 \times 10^5 \hat{\imath}) \times (0.150 \hat{\imath} - 0.250 \hat{\jmath}) = (0 \hat{\imath} + 0 \hat{\jmath} - 230000 \hat{k}) \text{ m}^2/\text{s} \] since \( \hat{\imath} \times \hat{\imath} = 0 \) and \( \hat{\imath} \times \hat{\jmath} = \hat{k} \).
05

Calculate Magnetic Field \( \vec{B} \)

Substitute the values into the Biot-Savart Law:\[ \vec{B} = \frac{4\pi \times 10^{-7}}{4\pi} \frac{4.80 \times 10^{-6} q_1 \times (-230000 \hat{k})}{(0.291)^3} \ \approx 2.54 \times 10^{-6} \hat{k} \text{ T} \].
06

Calculate Magnetic Force \( \vec{F} \) on \( q_2 \)

Use the magnetic force formula with \( \vec{v}_2 = (-5.30 \times 10^5 \hat{\jmath}) \text{ m/s} \):\[ \vec{F} = -2.90 \times 10^{-6} ((-5.30 \times 10^5 \hat{\jmath}) \times (2.54 \times 10^{-6} \hat{k})) \ = -2.90 \times 10^{-6} (1.3462 \times \hat{\imath}) \text{ N} \ \approx 3.90 \times 10^{-6} \hat{\imath} \text{ N} \] as \( \hat{\jmath} \times \hat{k} = \hat{\imath} \).
07

Determine the Magnitude and Direction

The magnitude of the force \( \vec{F} \) is \(|\vec{F}| = 3.90 \times 10^{-6} \text{ N}\). The direction is along the \( \hat{\imath} \) axis, meaning it points in the positive x-direction.

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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 used to calculate the magnetic field generated by a moving charge or an electric current. This law is crucial for understanding how magnetic fields behave and interact with charges.
According to the Biot-Savart Law, the magnetic field \( \vec{B} \) due to a moving charge \( q_1 \) with velocity \( \vec{v}_1 \) is given by:
  • \( \vec{B} = \frac{\mu_0}{4\pi} \frac{q_1 \vec{v}_1 \times \vec{r}}{r^3} \)
Here, \( \mu_0 \) is the permeability of free space, \( \vec{r} \) is the position vector from the moving charge to the point where the magnetic field is being calculated, and \( r \) is the magnitude of this vector.
This formula tells us that the magnetic field is directly proportional to the charge and its velocity, and inversely proportional to the cube of the distance from the charge.
Lorentz force
The Lorentz force is the force exerted on a moving charge by a magnetic field. It's a cornerstone concept in electromagnetism and explains how charged particles move through magnetic environments.
For a charge \( q_2 \) moving with a velocity \( \vec{v}_2 \) in a magnetic field \( \vec{B} \) created by another charge, the force \( \vec{F} \) can be expressed as:
  • \( \vec{F} = q_2(\vec{v}_2 \times \vec{B}) \)
This force depends on the charge, its speed, and the orientation of its path relative to the magnetic field. Importantly, the direction of the Lorentz force is perpendicular to both the velocity of the charge and the magnetic field, determined by the right-hand rule. This means a charge moving parallel to a magnetic field experiences no Lorentz force.
Cross Product
A cross product is a vector operation essential in calculating certain physical phenomena, like magnetic force and field. It's used to find a vector perpendicular to two given vectors.
The cross product \( \vec{v}_1 \times \vec{r} \), where \( \vec{v}_1 \) is the velocity and \( \vec{r} \) is the position vector, represents a vector that is orthogonal to the plane containing \( \vec{v}_1 \) and \( \vec{r} \). The magnitude of the cross product is given by:
  • \( |\vec{v}_1 \times \vec{r}| = |\vec{v}_1||\vec{r}| \sin\theta \)
where \( \theta \) is the angle between \( \vec{v}_1 \) and \( \vec{r} \). In the magnetic field scenario, the cross product helps determine the direction and magnitude of the field produced by moving charges, which is crucial for understanding interactive forces between charged particles.
Magnetic Field Calculation
Calculating the magnetic field produced by a moving charge involves using the Biot-Savart Law and integrating over the path of the charge. In simpler cases, like a point charge, we can directly apply the Biot-Savart formula.
Using the position vectors and velocities of the charges, we determine the magnetic field \( \vec{B} \) at a specific point. With charges like \( q_1 \) and \( q_2 \), first calculate the position vector \( \vec{r} \) and then perform the necessary cross product to find components of the magnetic field.
As illustrated:
  • \( \vec{B} = \frac{\mu_0}{4\pi} \frac{q_1 \vec{v}_1 \times \vec{r}}{r^3} \)
Once \( \vec{B} \) is found, it can be used in the Lorentz force formula to find the magnetic influence a charge exerts on another. This step-by-step calculation is crucial, as it allows for accurate prediction of magnetic interactions in many scientific and engineering applications.

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Most popular questions from this chapter

A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 \(\mathrm{cm}\) carries a current of 0.25 A. The relative permeability of the core is \(80 .\) (a) What is the magnetic field in the core? (b) What part of the magnetic field is due to atomic currents?

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 1.75 nm from each of them.

As a new electrical technician, you are designing a large solenoid to produce a uniform \(0.150-\) T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 1.40 \(\mathrm{m}\) long and 2.80 \(\mathrm{cm}\) in diameter. What current will you need to produce the necessary field?

Two long, straight conducting wires with linear mass density \(\lambda\) are suspended from cords so that they are each horizontal, parallel to each other, and a distance \(d\) apart. The back ends of the wires are connected to each other by a slack, low-resistance connecting wire. A charged capacitor (capacitance \(C )\) is now added to the system; the positive plate of the capacitor (initial charge \(+Q_{0}\) ) is connected to the front end of one of the wires, and the negative plate of the capacitor (initial charge \(-Q_{0} )\) is connected to the front end of the other wire (Fig. P28.87). Both of these connections are also made by slack, low-resistance wires. When the connection is made, the wires are pushed aside by the repulsive force between the wires, and each wire has an initial horizontal velocity of magnitude \(v_{0} .\) Assume that the time constant for the capacitor to discharge is negligible compared to the time it takes for any appreciable displacement in the position of the wires to occur. (a) Show that the initial speed \(v_{0}\) of either wire is given by $$ v_{0}=\frac{\mu_{0} Q_{0}^{2}}{4 \pi \lambda R C d} $$ where \(R\) is the total resistance of the circuit. (b) To what height \(h\) will each wire rise as a result of the circuit connection?

(a) How large a current would a very long, straight wire have to carry so that the magnetic field 2.00 \(\mathrm{cm}\) from the wire is equal to 1.00 G (comparable to the earth's northward-pointing magnetic field)? (b) If the wire is horizontal with the current running from east to west, at what locations would the magnetic field of the wire point in the same direction as the horizontal component of the earth's magnetic field? (c) Repeat part (b) except the wire is vertical with the current going upward.

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