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Chapter 28: Problem 56

At a particular instant, charge \(q_1 = +\)4.80 \(\times\) 10\(^{-6}\) C is at the point (0, 0.250 m, 0) and has velocity \(\vec{v_1}\) = (9.20 \(\times\) 10\(^5\) m/s)\(\hat{\imath}\). Charge \(q_2 = -\)2.90 \(\times\) 10\(^{-6}\) C is at the point (0.150 m, 0, 0) and has velocity \(\vec{v_2} =\) (-5.30 \(\times\) 10\(^5\) m/s)\(\hat{\jmath}\). 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 force is \(6.815 \times 10^{-12}\) N in the \(-\hat{i}\) direction.

Step by step solution

01

Determine the Magnetic Field Created by q1

The magnetic field created by a moving charge is given by Biot-Savart Law. In this exercise, the magnetic field due to charge \( q_1 \) at the location of \( q_2 \) is:\[ \vec{B}_1 = \frac{\mu_0}{4\pi} \cdot \frac{q_1 \vec{v}_1 \times \vec{r}}{r^3} \]Where \( \mu_0 = 4\pi \times 10^{-7} \) T路m/A, \( \vec{v}_1 \) is the velocity of \( q_1 \), and \( \vec{r} \) is the position vector from \( q_1 \) to \( q_2 \). First, calculate the position vector \( \vec{r} = (0.15 \hat{i} - 0.25 \hat{j}) \) m. The magnitude \( r = \sqrt{0.15^2 + 0.25^2} \approx 0.2915 \) m.
02

Calculate the Cross Product

The cross product \( \vec{v}_1 \times \vec{r} \) is calculated as follows:\[ \vec{v}_1 \times \vec{r} = (9.20 \times 10^5 \hat{i}) \times (0.15 \hat{i} - 0.25 \hat{j}) \]Using the properties of cross products, \( \hat{i} \times \hat{j} = \hat{k} \) and \( \hat{i} \times \hat{i} = 0 \), we have:\[ \vec{v}_1 \times \vec{r} = (9.20 \times 10^5 \hat{i}) \times ( -0.25 \hat{j}) = -2.30 \times 10^5 \hat{k} \] m虏/s.
03

Compute the Magnetic Field B1

Substitute the values into the Biot-Savart law to find \( \vec{B}_1 \):\[ \vec{B}_1 = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{4.80 \times 10^{-6} \cdot (-2.30 \times 10^5 \hat{k})}{(0.2915)^3} \]\[ \vec{B}_1 = 10^{-7} \cdot \frac{4.80 \times 10^{-6} \cdot (-2.30 \times 10^5)}{0.02472} \hat{k} = -4.44 \times 10^{-6} \hat{k} \] T.
04

Determine the Magnetic Force on q2

The force on a charge due to a magnetic field is found using the equation:\[ \vec{F} = q_2 (\vec{v}_2 \times \vec{B}_1) \]Substitute \( \vec{v}_2 = -5.30 \times 10^5 \hat{j} \) and \( \vec{B}_1 = -4.44 \times 10^{-6} \hat{k} \):\[ \vec{v}_2 \times \vec{B}_1 = (-5.30 \times 10^5 \hat{j}) \times (-4.44 \times 10^{-6} \hat{k}) \]\[ \vec{v}_2 \times \vec{B}_1 = 2.35 \times 10^{-6} \hat{i} \] m虏/s虏.
05

Compute the Final Magnetic Force

Now, substitute back into the magnetic force formula:\[ \vec{F} = (-2.90 \times 10^{-6} C) \cdot (2.35 \times 10^{-6} \hat{i}) \]\[ \vec{F} = -6.815 \times 10^{-12} \hat{i} \] N.The direction of the force is in the negative \( x \)-direction (\( -\hat{i} \)).
06

Conclude with Magnitude and Direction

The magnitude of the magnetic force is \( 6.815 \times 10^{-12} \) N, and its direction is along the negative \( x \)-axis (\( -\hat{i} \)).

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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 fundamental in calculating the magnetic field produced by a moving charge. It expresses how moving electric charges generate magnetic fields around them. The law is mathematically represented as:
  • \[ \vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{q \vec{v} \times \vec{r}}{r^3} \]
In this equation, \( \vec{B} \) is the magnetic field vector, \( \mu_0 \) (the permeability of free space) is a constant, \( q \) stands for the charge, \( \vec{v} \) 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 \( \vec{v} \times \vec{r} \) computes a vector perpendicular to both \( \vec{v} \) and \( \vec{r} \).
  • The magnitude \( r \) in the denominator ensures that the field decreases with the square of the distance, emphasizing the inversely squared relationship.
This law particularly aids in understanding how different elements of a current distribution contribute to the resultant magnetic field.
Cross Product
The cross product is essential in physics when describing how two vectors interact, especially in the context of magnetic forces. It's a mathematical operation resulting in a vector that is perpendicular to both original vectors. The expression \( \vec{A} \times \vec{B} \) results in a vector \( \vec{C} \), where:
  • The direction is determined by the right-hand rule. Point your index finger in the direction of \( \vec{A} \) and your middle finger in the direction of \( \vec{B} \); your thumb then points in the direction of \( \vec{C} \).
  • The magnitude is given by \( |\vec{A}| |\vec{B}| \sin(\theta) \), where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \).
In magnetic calculations, consider the cross product of velocity and position vectors where:
  • A vector like \( \hat{i} \times \hat{j} = \hat{k} \) illustrates how unit vectors interact to form a perpendicular vector, crucial in determining the direction of magnetic fields and forces.
  • This property is what allows magnetic fields and forces to have directionality, which is a unique feature of magnetic interactions.
Magnetic Field Calculation
When calculating a magnetic field due to a moving charge, like in the given exercise, it鈥檚 important to follow a step-by-step approach:
  • First, identify the source charge's velocity, \( \vec{v}_1 \), and the charge itself, \( q_1 \).
  • Determine the position vector \( \vec{r} \) from the source charge to the point where the field is being calculated.
  • Confirm the magnitude of \( \vec{r} \) as \( r = \sqrt{x^2 + y^2} \), which helps normalize the influence of distance on the magnetic field.
  • The next step is using the Biot-Savart Law to calculate the magnetic field \( \vec{B} \).
  • The use of the cross product ensures the right directional relationship between \( \vec{v} \) and \( \vec{r} \), standardizing its calculation.
Each of these steps utilizes algebra and vector mathematics to resolve the magnitude and direction of the magnetic field, ensuring accuracy in determining how moving charges affect one another.

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Most popular questions from this chapter

Certain bacteria (such as \(Aquaspirillum\) \(magnetotacticum\)) tend to swim toward the earth's geographic north pole because they contain tiny particles, called magnetosomes, that are sensitive to a magnetic field. If a transmission line carrying 100 A is laid underwater, at what range of distances would the magnetic field from this line be great enough to interfere with the migration of these bacteria? (Assume that a field less than 5\(\%\) of the earth's field would have little effect on the bacteria. Take the earth's field to be 5.0 \(\times\) 10\(^{-5}\) T, and ignore the effects of the seawater.)

A solenoid that is 35 cm long and contains 450 circular coils 2.0 cm in diameter carries a 1.75-A current. (a) What is the magnetic field at the center of the solenoid, 1.0 cm from the coils? (b) Suppose we now stretch out the coils to make a very long wire carrying the same current as before. What is the magnetic field 1.0 cm from the wire's center? Is it the same as that in part (a)? Why or why not?

Two long, parallel transmission lines, 40.0 cm apart, carry 25.0-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 wide, long, insulating belt has a uniform positive charge per unit area \(\sigma\) on its upper surface. Rollers at each end move the belt to the right at a constant speed \(v\). Calculate the magnitude and direction of the magnetic field produced by the moving belt at a point just above its surface. (\(Hint:\) At points near the surface and far from its edges or ends, the moving belt can be considered to be an infinite current sheet like that in Problem 28.73.)

A short current element \(\overrightarrow{dl}\) S = (0.500 mm)\(\hat{\imath}\) carries a current of 5.40 A in the same direction as \(\overrightarrow{dl}\). Point P is located at \(\overrightarrow{r} =\) (-0.730 m)\(\hat{\imath}\) + (0.390 m)\(\hat{k}\) . Use unit vectors to express the magnetic field at \(P\) produced by this current element.
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