The image contains a problem involving a long, straight wire and a single-turn rectangular loop, both lying in the plane of the page. The wire is parallel to the long sides of the loop and is 0.5 m from the closer side. The problem asks to determine the rate of change of current in the wire when the voltage induced in the loop is 2.5 V. ### Problem Parts **a. Expression for Magnetic Field \( B \):** - Find the magnetic field \( B \) due to the current \( I \) in the long, straight wire at a distance \( r \) away using: \[ B = \frac{\mu_0 I}{2 \pi r} \] **b. Magnetic Flux Calculation:** - Since the magnetic field is not uniform, integration is needed. Find the magnetic flux through the loop as a function of \( I \): \[ \Phi = \int B \, dA \] - **Hint**: Break up the area integral into horizontal and vertical directions. Use appropriate expressions for magnetic field \( B \) to set up the integral. **c. Rate of Change of Current:** - Set the change in magnetic flux \( \frac{d\Phi}{dt} \) equal to the induced voltage and solve for \( \frac{dI}{dt} \). - **Hint**: The solution in part (b) includes only one variable \( I \), simplifying the expression of \( \frac{d\Phi}{dt} \). **Final Result:** - The rate of change of the current in the wire is calculated to be approximately 601120 A/s. ### Diagram The top of the image includes a diagram showing: - A long, straight wire alongside a rectangular loop. - The dimensions of the loop are 0.50 m in width and 3.0 m in length, with a separation of 0.50 m from the wire. This problem is a classic example of electromagnetic induction, illustrating the integration of magnetic fields over areas and the relationship between induced voltage and current changes.
The image contains a problem involving a long, straight wire and a single-turn rectangular loop, both lying in the plane of the page. The wire is parallel to the long sides of the loop and is 0.5 m from the closer side. The problem asks to determine the rate of change of current in the wire when the voltage induced in the loop is 2.5 V. ### Problem Parts **a. Expression for Magnetic Field \( B \):** - Find the magnetic field \( B \) due to the current \( I \) in the long, straight wire at a distance \( r \) away using: \[ B = \frac{\mu_0 I}{2 \pi r} \] **b. Magnetic Flux Calculation:** - Since the magnetic field is not uniform, integration is needed. Find the magnetic flux through the loop as a function of \( I \): \[ \Phi = \int B \, dA \] - **Hint**: Break up the area integral into horizontal and vertical directions. Use appropriate expressions for magnetic field \( B \) to set up the integral. **c. Rate of Change of Current:** - Set the change in magnetic flux \( \frac{d\Phi}{dt} \) equal to the induced voltage and solve for \( \frac{dI}{dt} \). - **Hint**: The solution in part (b) includes only one variable \( I \), simplifying the expression of \( \frac{d\Phi}{dt} \). **Final Result:** - The rate of change of the current in the wire is calculated to be approximately 601120 A/s. ### Diagram The top of the image includes a diagram showing: - A long, straight wire alongside a rectangular loop. - The dimensions of the loop are 0.50 m in width and 3.0 m in length, with a separation of 0.50 m from the wire. This problem is a classic example of electromagnetic induction, illustrating the integration of magnetic fields over areas and the relationship between induced voltage and current changes.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Part b and c incorrect can you show all the steps
![The image contains a problem involving a long, straight wire and a single-turn rectangular loop, both lying in the plane of the page. The wire is parallel to the long sides of the loop and is 0.5 m from the closer side. The problem asks to determine the rate of change of current in the wire when the voltage induced in the loop is 2.5 V.
### Problem Parts
**a. Expression for Magnetic Field \( B \):**
- Find the magnetic field \( B \) due to the current \( I \) in the long, straight wire at a distance \( r \) away using:
\[
B = \frac{\mu_0 I}{2 \pi r}
\]
**b. Magnetic Flux Calculation:**
- Since the magnetic field is not uniform, integration is needed. Find the magnetic flux through the loop as a function of \( I \):
\[
\Phi = \int B \, dA
\]
- **Hint**: Break up the area integral into horizontal and vertical directions. Use appropriate expressions for magnetic field \( B \) to set up the integral.
**c. Rate of Change of Current:**
- Set the change in magnetic flux \( \frac{d\Phi}{dt} \) equal to the induced voltage and solve for \( \frac{dI}{dt} \).
- **Hint**: The solution in part (b) includes only one variable \( I \), simplifying the expression of \( \frac{d\Phi}{dt} \).
**Final Result:**
- The rate of change of the current in the wire is calculated to be approximately 601120 A/s.
### Diagram
The top of the image includes a diagram showing:
- A long, straight wire alongside a rectangular loop.
- The dimensions of the loop are 0.50 m in width and 3.0 m in length, with a separation of 0.50 m from the wire.
This problem is a classic example of electromagnetic induction, illustrating the integration of magnetic fields over areas and the relationship between induced voltage and current changes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee22cf2f-b974-4b00-a3cf-09b388e7d65d%2F068eafda-293a-45dd-80bd-c33fd57f8756%2Fco0hegn.jpeg&w=3840&q=75)
Transcribed Image Text:The image contains a problem involving a long, straight wire and a single-turn rectangular loop, both lying in the plane of the page. The wire is parallel to the long sides of the loop and is 0.5 m from the closer side. The problem asks to determine the rate of change of current in the wire when the voltage induced in the loop is 2.5 V.
### Problem Parts
**a. Expression for Magnetic Field \( B \):**
- Find the magnetic field \( B \) due to the current \( I \) in the long, straight wire at a distance \( r \) away using:
\[
B = \frac{\mu_0 I}{2 \pi r}
\]
**b. Magnetic Flux Calculation:**
- Since the magnetic field is not uniform, integration is needed. Find the magnetic flux through the loop as a function of \( I \):
\[
\Phi = \int B \, dA
\]
- **Hint**: Break up the area integral into horizontal and vertical directions. Use appropriate expressions for magnetic field \( B \) to set up the integral.
**c. Rate of Change of Current:**
- Set the change in magnetic flux \( \frac{d\Phi}{dt} \) equal to the induced voltage and solve for \( \frac{dI}{dt} \).
- **Hint**: The solution in part (b) includes only one variable \( I \), simplifying the expression of \( \frac{d\Phi}{dt} \).
**Final Result:**
- The rate of change of the current in the wire is calculated to be approximately 601120 A/s.
### Diagram
The top of the image includes a diagram showing:
- A long, straight wire alongside a rectangular loop.
- The dimensions of the loop are 0.50 m in width and 3.0 m in length, with a separation of 0.50 m from the wire.
This problem is a classic example of electromagnetic induction, illustrating the integration of magnetic fields over areas and the relationship between induced voltage and current changes.
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