Answer :
Certainly! Let's find the magnitude of the electric field by following this step-by-step approach:
1. Understand the Problem:
- We are given the diameter of a circular loop of wire, [tex]\(0.626 \, \text{m}\)[/tex], and the electric flux through the loop when it is oriented such that the flux is maximum, [tex]\(7.50 \times 10^5 \, \text{N} \cdot \text{m}^2/\text{C}\)[/tex].
- We need to determine the magnitude of the electric field [tex]\(E\)[/tex].
2. Calculate the Radius of the Loop:
- The diameter [tex]\(D\)[/tex] of the loop is [tex]\(0.626 \, \text{m}\)[/tex], hence the radius [tex]\(r\)[/tex] is given by:
[tex]\[ r = \frac{D}{2} = \frac{0.626 \, \text{m}}{2} = 0.313 \, \text{m} \][/tex]
3. Calculate the Area of the Circular Loop:
- The area [tex]\(A\)[/tex] of the circular loop can be found using the formula for the area of a circle, [tex]\(A = \pi r^2\)[/tex]:
[tex]\[ A = \pi r^2 = \pi (0.313 \, \text{m})^2 = \pi \times 0.098169 \, \text{m}^2 = 0.30777869067953845 \, \text{m}^2 \][/tex]
4. Relate Electric Flux and Electric Field:
- The electric flux [tex]\(\Phi_E\)[/tex] through the loop is related to the electric field [tex]\(E\)[/tex] and the area [tex]\(A\)[/tex] by the formula:
[tex]\[ \Phi_E = E \times A \][/tex]
Rearranging for the electric field [tex]\(E\)[/tex], we get:
[tex]\[ E = \frac{\Phi_E}{A} \][/tex]
5. Substitute the Given Values:
- The electric flux is [tex]\(7.50 \times 10^5 \, \text{N} \cdot \text{m}^2/\text{C}\)[/tex], and the area calculated is [tex]\(0.30777869067953845 \, \text{m}^2\)[/tex]. Substituting these values into the rearranged formula, we get:
[tex]\[ E = \frac{7.50 \times 10^5 \, \text{N} \cdot \text{m}^2/\text{C}}{0.30777869067953845 \, \text{m}^2} = 2436815.8768369895 \, \text{N}/\text{C} \][/tex]
6. Match the Calculated Electric Field with the Given Options:
- Based on the options provided:
a. [tex]\(8.88 \times 10^5 \, \text{N}/\text{C}\)[/tex]
b. [tex]\(1.07 \times 10^6 \, \text{N}/\text{C}\)[/tex]
c. [tex]\(2.44 \times 10^6 \, \text{N}/\text{C}\)[/tex]
d. [tex]\(4.24 \times 10^6 \, \text{N}/\text{C}\)[/tex]
e. [tex]\(6.00 \times 10^6 \, \text{N}/\text{C}\)[/tex]
The closest match to our calculated value of [tex]\(2436815.8768369895 \, \text{N}/\text{C}\)[/tex] is option c:
[tex]\[ 2.44 \times 10^6 \, \text{N}/\text{C} \][/tex]
Hence, the magnitude of the electric field [tex]\(E\)[/tex] is:
[tex]\[ \boxed{2.44 \times 10^6 \, \text{N}/\text{C}} \][/tex]
1. Understand the Problem:
- We are given the diameter of a circular loop of wire, [tex]\(0.626 \, \text{m}\)[/tex], and the electric flux through the loop when it is oriented such that the flux is maximum, [tex]\(7.50 \times 10^5 \, \text{N} \cdot \text{m}^2/\text{C}\)[/tex].
- We need to determine the magnitude of the electric field [tex]\(E\)[/tex].
2. Calculate the Radius of the Loop:
- The diameter [tex]\(D\)[/tex] of the loop is [tex]\(0.626 \, \text{m}\)[/tex], hence the radius [tex]\(r\)[/tex] is given by:
[tex]\[ r = \frac{D}{2} = \frac{0.626 \, \text{m}}{2} = 0.313 \, \text{m} \][/tex]
3. Calculate the Area of the Circular Loop:
- The area [tex]\(A\)[/tex] of the circular loop can be found using the formula for the area of a circle, [tex]\(A = \pi r^2\)[/tex]:
[tex]\[ A = \pi r^2 = \pi (0.313 \, \text{m})^2 = \pi \times 0.098169 \, \text{m}^2 = 0.30777869067953845 \, \text{m}^2 \][/tex]
4. Relate Electric Flux and Electric Field:
- The electric flux [tex]\(\Phi_E\)[/tex] through the loop is related to the electric field [tex]\(E\)[/tex] and the area [tex]\(A\)[/tex] by the formula:
[tex]\[ \Phi_E = E \times A \][/tex]
Rearranging for the electric field [tex]\(E\)[/tex], we get:
[tex]\[ E = \frac{\Phi_E}{A} \][/tex]
5. Substitute the Given Values:
- The electric flux is [tex]\(7.50 \times 10^5 \, \text{N} \cdot \text{m}^2/\text{C}\)[/tex], and the area calculated is [tex]\(0.30777869067953845 \, \text{m}^2\)[/tex]. Substituting these values into the rearranged formula, we get:
[tex]\[ E = \frac{7.50 \times 10^5 \, \text{N} \cdot \text{m}^2/\text{C}}{0.30777869067953845 \, \text{m}^2} = 2436815.8768369895 \, \text{N}/\text{C} \][/tex]
6. Match the Calculated Electric Field with the Given Options:
- Based on the options provided:
a. [tex]\(8.88 \times 10^5 \, \text{N}/\text{C}\)[/tex]
b. [tex]\(1.07 \times 10^6 \, \text{N}/\text{C}\)[/tex]
c. [tex]\(2.44 \times 10^6 \, \text{N}/\text{C}\)[/tex]
d. [tex]\(4.24 \times 10^6 \, \text{N}/\text{C}\)[/tex]
e. [tex]\(6.00 \times 10^6 \, \text{N}/\text{C}\)[/tex]
The closest match to our calculated value of [tex]\(2436815.8768369895 \, \text{N}/\text{C}\)[/tex] is option c:
[tex]\[ 2.44 \times 10^6 \, \text{N}/\text{C} \][/tex]
Hence, the magnitude of the electric field [tex]\(E\)[/tex] is:
[tex]\[ \boxed{2.44 \times 10^6 \, \text{N}/\text{C}} \][/tex]
