Problem 23 - Fundamentals of Physics

Question

\(.\) ILW Figure 29-51 shows a snapshot of a proton moving at velocity \(\vec{v}=(-200 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) toward a long straight wire with current \(i=350 \mathrm{~mA} .\) At the instant shown, the proton's distance from the wire is \(d=2.89 \mathrm{~cm} .\) In unitvector notation, what is the magnetic force on the proton due to the current?

Step-by-Step Solution

Answer

The magnetic force on the proton is approximately \(-7.74 \times 10^{-23} \hat{\mathrm{i}} \; \text{N}\).

Step 1: Use the Biot-Savart Law Formula

To find the magnetic field (\B) at the distance of the proton from the wire, use the formula for the magnetic field around a long straight wire: \[ B = \frac{\mu_0 i}{2 \pi d} \]where \(\mu_0 = 4\pi \times 10^{-7} \; \text{T m/A}\), \(i = 0.35 \; \text{A}\), and \(d = 0.0289 \; \text{m}\).

Step 2: Calculate the Magnetic Field

Substitute the given values into the formula:\[B = \frac{4\pi \times 10^{-7} \times 0.35}{2\pi \times 0.0289}\]Simplify the expression to find the value of \(B\).

Step 3: Simplification Step

Calculate the above expression to get:\[B = \frac{0.35 \times 2 \times 10^{-7}}{0.0289} \approx 2.42 \times 10^{-6} \; \text{T}\]The magnetic field \(\B\) is approximately 2.42 µT.

Step 4: Determine the Magnetic Force Formula

The magnetic force \(\vec{F}_m\) on a moving charge in a magnetic field is given by:\[\vec{F}_m = q \cdot \vec{v} \times \vec{B}\]where \(q = 1.6 \times 10^{-19} \; \text{C}\) is the charge of a proton, \(\vec{v} = (-200 \; \text{m/s}) \hat{\mathrm{j}}\), and \(\vec{B} = B \hat{\mathrm{k}}\).

Step 5: Apply the Cross Product

The force is calculated using the cross product:\[\vec{F}_m = q \cdot (-200 \hat{\mathrm{j}}) \times (2.42 \times 10^{-6} \hat{\mathrm{k}})\]Remember that \(\hat{\mathrm{j}} \times \hat{\mathrm{k}} = \hat{\mathrm{i}}\), which yields:\[\vec{F}_m = q \cdot (-200)(2.42 \times 10^{-6}) \hat{\mathrm{i}}\]

Step 6: Complete the Calculation

Substitute the charge of the proton and simplify: \[\vec{F}_m = (1.6 \times 10^{-19}) \cdot (-200)(2.42 \times 10^{-6}) \hat{\mathrm{i}}\]Calculate the magnitude:\[\vec{F}_m \approx -7.74 \times 10^{-23} \hat{\mathrm{i}} \; \text{N}\]Thus, the magnetic force is approximately \(-7.74 \times 10^{-23} \hat{\mathrm{i}} \; \text{N}\).

Key Concepts

Biot-Savart Law

The Biot-Savart Law is essential in understanding how currents produce magnetic fields. It gives us a mathematical representation to calculate the magnetic field produced by a current-carrying wire at a certain point in space. The law states that every small segment of current contributes to the magnetic field, and it is expressed as:\[ B = \frac{\mu_0 I}{2 \pi r} \]Here, \(B\) is the magnetic field, \(\mu_0\) is the magnetic constant, \(I\) is the current, and \(r\) is the distance from the wire. By integrating over the entire wire length, we account for the contributions of each small segment, allowing precise calculation of the field. This law is particularly useful for straight wires, loops, or solenoids.

In practice, you can use the Biot-Savart Law to determine the magnitude of a magnetic field at any point around a current-carrying wire. It's a foundational principle that assists in analyzing scenarios involving complex magnetic fields and is fundamental to electromagnetism.

Magnetic Field Calculation

Calculating the magnetic field is a critical skill in physics, particularly for understanding how forces act in electromagnetic scenarios. For a straight wire, the Biot-Savart Law simplifies this calculation, as we saw earlier. Plugging the values into the formula, we find:\[ B = \frac{4\pi \times 10^{-7} \times 0.35}{2\pi \times 0.0289} \]This type of substitution and computation is a typical process to find the magnitude of the magnetic field, resulting in units of Tesla (T).
A calculated magnetic field tells us about the influence a current exerts on its surroundings. It allows us to predict how charged particles will move in the field and understand the effects in practical applications like electric motors and generators.

Understanding how to calculate the magnetic field helps connect theoretical concepts to real-world applications, granting insights into how electromagnetic interactions shape technology and nature.

Cross Product in Physics

The cross product in physics is an operation performed on two vectors resulting in another vector. It is crucial in determining forces on moving charges in magnetic fields. Given two vectors \(\vec{A}\) and \(\vec{B}\), their cross product \(\vec{C}\) is:\[ \vec{C} = \vec{A} \times \vec{B} \]This product is perpendicular to both original vectors and has a magnitude equal to the area of the parallelogram that the vectors span. In our problem, the cross product is used to find the magnetic force \(\vec{F}_m\) on a charge, with:\[ \vec{F}_m = q \cdot \vec{v} \times \vec{B} \] Here, \(q\) is the charge, \(\vec{v}\) is the velocity vector, and \(\vec{B}\) is the magnetic field. We see that the force is perpendicular to the velocity and magnetic field, determined using the right-hand rule.
The right-hand rule states that if the fingers of your right hand point in the direction of \(\vec{v}\) and bend towards \(\vec{B}\), your thumb points in the direction of \(\vec{F}_m\). This operation helps in visualizing and computing the directional effects of fields and forces in electromagnetism, making it an invaluable tool in physics.