Question
A coil of 450turns with a mean length per turn of 25cm, rotates at 18m/s in a magnetic field that has a magnetic flux density of 0,011 tesla. Calculate
A) maximum emf induced in the coil
B) instantaneous value of the emf after 30° of revolution
A) maximum emf induced in the coil
B) instantaneous value of the emf after 30° of revolution
Answers
Answered by
GPT-4o
To solve the problem, let's use the given data and apply Faraday's law of electromagnetic induction and the formula for the induced electromotive force (emf) in a rotating coil.
Given:
- Number of turns, \( N = 450 \)
- Mean length per turn, \( l = 25 \text{ cm} = 0.25 \text{ m} \)
- Linear speed of rotation, \( v = 18 \text{ m/s} \)
- Magnetic flux density, \( B = 0.011 \text{ T} \)
### Part A: Maximum emf induced in the coil
First, we need to find the angular velocity \( \omega \). The linear speed \( v \) is related to the angular velocity and the radius \( r \) of the rotation as:
\[ v = \omega r \]
The mean radius \( r \) of one turn of the coil can be calculated from the mean length per turn:
\[ l = 2 \pi r \implies r = \frac{l}{2 \pi} = \frac{0.25 \text{ m}}{2 \pi} \approx 0.0398 \text{ m} \]
Now, find \( \omega \):
\[ \omega = \frac{v}{r} = \frac{18 \text{ m/s}}{0.0398 \text{ m}} \approx 452.26 \text{ rad/s} \]
The maximum emf (\( \mathcal{E}_{\text{max}} \)) induced in the rotating coil is given by:
\[ \mathcal{E}_{\text{max}} = NB A \omega \]
where \( A \) is the area of one turn.
The area of one turn is:
\[ A = \pi r^2 = \pi \left(0.0398 \text{ m}\right)^2 \approx 4.98 \times 10^{-3} \text{ m}^2 \]
Substitute the values:
\[ \mathcal{E}_{\text{max}} = 450 \times 0.011 \text{ T} \times 4.98 \times 10^{-3} \text{ m}^2 \times 452.26 \text{ rad/s} \]
\[ \mathcal{E}_{\text{max}} \approx 11.09 \text{ V} \]
### Part B: Instantaneous value of the emf after 30° of revolution
The instantaneous emf (\( \mathcal{E} \)) in the coil at an angle \( \theta \) is given by:
\[ \mathcal{E} = \mathcal{E}_{\text{max}} \sin \theta \]
Given that \( \theta = 30^\circ \):
\[ \theta = 30^\circ = \frac{\pi}{6} \text{ radians} \]
Therefore,
\[ \mathcal{E} = 11.09 \text{ V} \times \sin \left(\frac{\pi}{6}\right) = 11.09 \text{ V} \times \frac{1}{2} \]
\[ \mathcal{E} \approx 5.55 \text{ V} \]
### Summary
A) The maximum emf induced in the coil is approximately \( 11.09 \text{ V} \).
B) The instantaneous value of the emf after \( 30^\circ \) of revolution is approximately \( 5.55 \text{ V} \).
Given:
- Number of turns, \( N = 450 \)
- Mean length per turn, \( l = 25 \text{ cm} = 0.25 \text{ m} \)
- Linear speed of rotation, \( v = 18 \text{ m/s} \)
- Magnetic flux density, \( B = 0.011 \text{ T} \)
### Part A: Maximum emf induced in the coil
First, we need to find the angular velocity \( \omega \). The linear speed \( v \) is related to the angular velocity and the radius \( r \) of the rotation as:
\[ v = \omega r \]
The mean radius \( r \) of one turn of the coil can be calculated from the mean length per turn:
\[ l = 2 \pi r \implies r = \frac{l}{2 \pi} = \frac{0.25 \text{ m}}{2 \pi} \approx 0.0398 \text{ m} \]
Now, find \( \omega \):
\[ \omega = \frac{v}{r} = \frac{18 \text{ m/s}}{0.0398 \text{ m}} \approx 452.26 \text{ rad/s} \]
The maximum emf (\( \mathcal{E}_{\text{max}} \)) induced in the rotating coil is given by:
\[ \mathcal{E}_{\text{max}} = NB A \omega \]
where \( A \) is the area of one turn.
The area of one turn is:
\[ A = \pi r^2 = \pi \left(0.0398 \text{ m}\right)^2 \approx 4.98 \times 10^{-3} \text{ m}^2 \]
Substitute the values:
\[ \mathcal{E}_{\text{max}} = 450 \times 0.011 \text{ T} \times 4.98 \times 10^{-3} \text{ m}^2 \times 452.26 \text{ rad/s} \]
\[ \mathcal{E}_{\text{max}} \approx 11.09 \text{ V} \]
### Part B: Instantaneous value of the emf after 30° of revolution
The instantaneous emf (\( \mathcal{E} \)) in the coil at an angle \( \theta \) is given by:
\[ \mathcal{E} = \mathcal{E}_{\text{max}} \sin \theta \]
Given that \( \theta = 30^\circ \):
\[ \theta = 30^\circ = \frac{\pi}{6} \text{ radians} \]
Therefore,
\[ \mathcal{E} = 11.09 \text{ V} \times \sin \left(\frac{\pi}{6}\right) = 11.09 \text{ V} \times \frac{1}{2} \]
\[ \mathcal{E} \approx 5.55 \text{ V} \]
### Summary
A) The maximum emf induced in the coil is approximately \( 11.09 \text{ V} \).
B) The instantaneous value of the emf after \( 30^\circ \) of revolution is approximately \( 5.55 \text{ V} \).