**Question 8** A 2.0 m conductor is formed into a square and placed in the horizontal \(xy\)-plane. A magnetic field is oriented \(30.0^\circ\) above the horizontal with a strength of 1.0 T. What is the magnetic flux through the conductor? - \( \text{A. } 2.0 \, \text{T} \cdot \text{m}^2 \) - \( \text{B. } 0.25 \, \text{T} \cdot \text{m}^2 \) - \( \text{C. } 0.22 \, \text{T} \cdot \text{m}^2 \) - \( \text{D. } 0.12 \, \text{T} \cdot \text{m}^2 \) **Question 9** **Figure 3.1** Click Save and Submit to save and submit. Click Save All Answers to save all answers. --- *Explanation for Educational Context:* - A square loop with a side of 2.0 meters is subjected to a magnetic field. - The magnetic flux \((\Phi_B)\) through the conductor can be calculated using the formula: \[ \Phi_B = B \cdot A \cdot \cos(\theta) \] Where: - \( B \) is the magnetic field strength (1.0 T in this case), - \( A \) is the area of the square conductor which is \(2.0 \, \text{m} \times 2.0 \, \text{m} = 4.0 \, \text{m}^2\), - \(\theta\) is the angle between the magnetic field and the normal to the surface of the conductor, which is \(30.0^\circ\). Substituting the values, we get: \[ \Phi_B = 1.0 \, \text{T} \times 4.0 \, \text{m}^2 \times \cos(30.0^\circ) \] \[ \Phi_B = 4.0 \, \text{T} \cdot \text{m}^2 \times \frac{\sqrt{3}}{2} \approx 2.0 \, \text{T} \cd
**Question 8** A 2.0 m conductor is formed into a square and placed in the horizontal \(xy\)-plane. A magnetic field is oriented \(30.0^\circ\) above the horizontal with a strength of 1.0 T. What is the magnetic flux through the conductor? - \( \text{A. } 2.0 \, \text{T} \cdot \text{m}^2 \) - \( \text{B. } 0.25 \, \text{T} \cdot \text{m}^2 \) - \( \text{C. } 0.22 \, \text{T} \cdot \text{m}^2 \) - \( \text{D. } 0.12 \, \text{T} \cdot \text{m}^2 \) **Question 9** **Figure 3.1** Click Save and Submit to save and submit. Click Save All Answers to save all answers. --- *Explanation for Educational Context:* - A square loop with a side of 2.0 meters is subjected to a magnetic field. - The magnetic flux \((\Phi_B)\) through the conductor can be calculated using the formula: \[ \Phi_B = B \cdot A \cdot \cos(\theta) \] Where: - \( B \) is the magnetic field strength (1.0 T in this case), - \( A \) is the area of the square conductor which is \(2.0 \, \text{m} \times 2.0 \, \text{m} = 4.0 \, \text{m}^2\), - \(\theta\) is the angle between the magnetic field and the normal to the surface of the conductor, which is \(30.0^\circ\). Substituting the values, we get: \[ \Phi_B = 1.0 \, \text{T} \times 4.0 \, \text{m}^2 \times \cos(30.0^\circ) \] \[ \Phi_B = 4.0 \, \text{T} \cdot \text{m}^2 \times \frac{\sqrt{3}}{2} \approx 2.0 \, \text{T} \cd
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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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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![**Question 8**
A 2.0 m conductor is formed into a square and placed in the horizontal \(xy\)-plane. A magnetic field is oriented \(30.0^\circ\) above the horizontal with a strength of 1.0 T. What is the magnetic flux through the conductor?
- \( \text{A. } 2.0 \, \text{T} \cdot \text{m}^2 \)
- \( \text{B. } 0.25 \, \text{T} \cdot \text{m}^2 \)
- \( \text{C. } 0.22 \, \text{T} \cdot \text{m}^2 \)
- \( \text{D. } 0.12 \, \text{T} \cdot \text{m}^2 \)
**Question 9**
**Figure 3.1**
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
---
*Explanation for Educational Context:*
- A square loop with a side of 2.0 meters is subjected to a magnetic field.
- The magnetic flux \((\Phi_B)\) through the conductor can be calculated using the formula:
\[ \Phi_B = B \cdot A \cdot \cos(\theta) \]
Where:
- \( B \) is the magnetic field strength (1.0 T in this case),
- \( A \) is the area of the square conductor which is \(2.0 \, \text{m} \times 2.0 \, \text{m} = 4.0 \, \text{m}^2\),
- \(\theta\) is the angle between the magnetic field and the normal to the surface of the conductor, which is \(30.0^\circ\).
Substituting the values, we get:
\[ \Phi_B = 1.0 \, \text{T} \times 4.0 \, \text{m}^2 \times \cos(30.0^\circ) \]
\[ \Phi_B = 4.0 \, \text{T} \cdot \text{m}^2 \times \frac{\sqrt{3}}{2} \approx 2.0 \, \text{T} \cd](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8689a897-6f19-4f8c-bce2-bc1e854b9491%2F29e16344-9c77-436a-991e-ff107fa0e788%2Fz8nhnsr.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 8**
A 2.0 m conductor is formed into a square and placed in the horizontal \(xy\)-plane. A magnetic field is oriented \(30.0^\circ\) above the horizontal with a strength of 1.0 T. What is the magnetic flux through the conductor?
- \( \text{A. } 2.0 \, \text{T} \cdot \text{m}^2 \)
- \( \text{B. } 0.25 \, \text{T} \cdot \text{m}^2 \)
- \( \text{C. } 0.22 \, \text{T} \cdot \text{m}^2 \)
- \( \text{D. } 0.12 \, \text{T} \cdot \text{m}^2 \)
**Question 9**
**Figure 3.1**
Click Save and Submit to save and submit. Click Save All Answers to save all answers.
---
*Explanation for Educational Context:*
- A square loop with a side of 2.0 meters is subjected to a magnetic field.
- The magnetic flux \((\Phi_B)\) through the conductor can be calculated using the formula:
\[ \Phi_B = B \cdot A \cdot \cos(\theta) \]
Where:
- \( B \) is the magnetic field strength (1.0 T in this case),
- \( A \) is the area of the square conductor which is \(2.0 \, \text{m} \times 2.0 \, \text{m} = 4.0 \, \text{m}^2\),
- \(\theta\) is the angle between the magnetic field and the normal to the surface of the conductor, which is \(30.0^\circ\).
Substituting the values, we get:
\[ \Phi_B = 1.0 \, \text{T} \times 4.0 \, \text{m}^2 \times \cos(30.0^\circ) \]
\[ \Phi_B = 4.0 \, \text{T} \cdot \text{m}^2 \times \frac{\sqrt{3}}{2} \approx 2.0 \, \text{T} \cd
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