Problem 56 - University Physics with Modern Physics

Question

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 \(q_2\)?

Step-by-Step Solution

Answer

The force is \(6.815 \times 10^{-12}\) N in the \(-\hat{i}\) direction.

Step 1: 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·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.

Step 2: 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:\[ \vec{v}_1 \times \vec{r} = (9.20 \times 10^5 \hat{i}) \times ( -0.25 \hat{j}) = -2.30 \times 10^5 \hat{k} \] m²/s.

Step 3: 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.

Step 4: 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 \( \vec{B}_1 = -4.44 \times 10^{-6} \hat{k} \):\[ \vec{v}_2 \times \vec{B}_1 = (-5.30 \times 10^5 \hat{j}) \times (-4.44 \times 10^{-6} \hat{k}) \]\[ \vec{v}_2 \times \vec{B}_1 = 2.35 \times 10^{-6} \hat{i} \] m²/s².

Step 5: 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 (\( -\hat{i} \)).

Step 6: 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 (\( -\hat{i} \)).

Key Concepts

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’s 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.