Answer:
To determine the length of the wire from which the coil is made, we can use the formula for the induced electromotive force (emf) in a generator coil:
\[ \text{emf} = N \cdot B \cdot A \cdot \omega \cdot \sin(\omega t) \]
Where:
- \( \text{emf} \) is the root mean square (rms) value of the induced emf (110 V in this case).
- \( N \) is the number of turns in the coil (80 turns).
- \( B \) is the magnetic field strength (0.500 T).
- \( A \) is the area of the coil.
- \( \omega \) is the angular frequency ( \(2\pi \times \text{frequency}\), where the frequency is 60 Hz).
We are looking to find the length of wire used in the coil, which can be related to the area of the coil. Since each turn of the coil is circular, the area of each turn can be approximated as a circle. The formula for the area of a circle is:
\[ A = \pi r^2 \]
Where \( r \) is the radius of the circle.
The length of wire in the coil is related to the circumference of the circle:
\[ \text{Length of wire} = \text{circumference of circle} = 2\pi r \]
So, our goal is to find the radius of each turn (\( r \)), which will help us determine the length of wire.
First, let's rearrange the formula for emf to solve for \( A \):
\[ A = \frac{\text{emf}}{N \cdot B \cdot \omega} \]
Now, we'll use this formula to find the area of each turn, and then use that to find the length of wire. Let's calculate:
\[ A = \frac{110\, \text{V}}{80 \cdot 0.500\, \text{T} \cdot 2\pi \cdot 60\, \text{Hz}} \]
\[ A ≈ \frac{110}{80 \cdot 0.500 \cdot 2\pi \cdot 60} \, \text{m}^2 \]
\[ A ≈ \frac{110}{240\pi} \, \text{m}^2 \]
Now, we find the radius of each turn using the formula \( A = \pi r^2 \):
\[ \frac{110}{240\pi} = \pi r^2 \]
\[ r^2 = \frac{110}{240} \]
\[ r ≈ \sqrt{\frac{110}{240}} \]
\[ r ≈ \sqrt{\frac{11}{24}} \]
\[ r ≈ \frac{\sqrt{11}}{\sqrt{24}} \]
\[ r ≈ \frac{\sqrt{11}}{4\sqrt{6}} \]
Now, we find the length of wire in the coil using \( \text{Length of wire} = 2\pi r \):
\[ \text{Length of wire} ≈ 2\pi \times \frac{\sqrt{11}}{4\sqrt{6}} \]
\[ \text{Length of wire} ≈ \frac{\sqrt{11}\pi}{2\sqrt{6}} \]
\[ \text{Length of wire} ≈ \frac{\sqrt{66}\pi}{12} \]
So, the length of wire from which the coil is made is approximately \( \frac{\sqrt{66}\pi}{12} \) meters.
Explanation:
