Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
To find the EMF generated in a solenoid when the magnetic flux changes, we can use Faraday's law of electromagnetic induction. Faraday's law states that the induced electromotive force (EMF) in a coil is directly proportional to the rate of change of magnetic flux through the coil. The formula for EMF is given by:
[tex]\[ \text{EMF} = -N \frac{d\Phi}{dt} \][/tex]
where:
- [tex]\( \text{EMF} \)[/tex] is the electromotive force,
- [tex]\( N \)[/tex] is the number of loops in the solenoid,
- [tex]\( \Phi \)[/tex] (phi) is the magnetic flux,
- [tex]\( \frac{d\Phi}{dt} \)[/tex] is the rate of change of magnetic flux.
Given the data:
- Initial magnetic flux, [tex]\(\Phi_{\text{initial}} = 6.78 \times 10^{-4} \, \text{Wb}\)[/tex]
- Final magnetic flux, [tex]\(\Phi_{\text{final}} = 1.33 \times 10^{-4} \, \text{Wb}\)[/tex]
- Time interval, [tex]\( \Delta t = 0.0333 \, \text{s} \)[/tex]
- Number of loops, [tex]\( N = 605 \)[/tex]
1. Calculate the change in magnetic flux ([tex]\( \Delta \Phi \)[/tex]):
[tex]\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} \][/tex]
Substitute the given values:
[tex]\[ \Delta \Phi = 1.33 \times 10^{-4} \, \text{Wb} - 6.78 \times 10^{-4} \, \text{Wb} \][/tex]
[tex]\[ \Delta \Phi = -5.45 \times 10^{-4} \, \text{Wb} \][/tex]
2. Determine the rate of change of magnetic flux ([tex]\( \frac{d\Phi}{dt} \)[/tex]):
[tex]\[ \frac{\Delta \Phi}{\Delta t} = \frac{-5.45 \times 10^{-4} \, \text{Wb}}{0.0333 \, \text{s}} \][/tex]
[tex]\[ \frac{\Delta \Phi}{\Delta t} = -0.01637 \, \text{Wb/s} \][/tex]
3. Calculate the EMF generated:
[tex]\[ \text{EMF} = -N \frac{\Delta \Phi}{\Delta t} \][/tex]
Substitute [tex]\( N = 605 \)[/tex] and [tex]\( \frac{\Delta \Phi}{\Delta t} = -0.01637 \, \text{Wb/s} \)[/tex]:
[tex]\[ \text{EMF} = -605 \times (-0.01637 \, \text{Wb/s}) \][/tex]
[tex]\[ \text{EMF} = 9.901 \, \text{Volts} \][/tex]
Therefore, the EMF generated in the solenoid is approximately [tex]\(\boxed{9.901 \, \text{Volts}}\)[/tex].
[tex]\[ \text{EMF} = -N \frac{d\Phi}{dt} \][/tex]
where:
- [tex]\( \text{EMF} \)[/tex] is the electromotive force,
- [tex]\( N \)[/tex] is the number of loops in the solenoid,
- [tex]\( \Phi \)[/tex] (phi) is the magnetic flux,
- [tex]\( \frac{d\Phi}{dt} \)[/tex] is the rate of change of magnetic flux.
Given the data:
- Initial magnetic flux, [tex]\(\Phi_{\text{initial}} = 6.78 \times 10^{-4} \, \text{Wb}\)[/tex]
- Final magnetic flux, [tex]\(\Phi_{\text{final}} = 1.33 \times 10^{-4} \, \text{Wb}\)[/tex]
- Time interval, [tex]\( \Delta t = 0.0333 \, \text{s} \)[/tex]
- Number of loops, [tex]\( N = 605 \)[/tex]
1. Calculate the change in magnetic flux ([tex]\( \Delta \Phi \)[/tex]):
[tex]\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} \][/tex]
Substitute the given values:
[tex]\[ \Delta \Phi = 1.33 \times 10^{-4} \, \text{Wb} - 6.78 \times 10^{-4} \, \text{Wb} \][/tex]
[tex]\[ \Delta \Phi = -5.45 \times 10^{-4} \, \text{Wb} \][/tex]
2. Determine the rate of change of magnetic flux ([tex]\( \frac{d\Phi}{dt} \)[/tex]):
[tex]\[ \frac{\Delta \Phi}{\Delta t} = \frac{-5.45 \times 10^{-4} \, \text{Wb}}{0.0333 \, \text{s}} \][/tex]
[tex]\[ \frac{\Delta \Phi}{\Delta t} = -0.01637 \, \text{Wb/s} \][/tex]
3. Calculate the EMF generated:
[tex]\[ \text{EMF} = -N \frac{\Delta \Phi}{\Delta t} \][/tex]
Substitute [tex]\( N = 605 \)[/tex] and [tex]\( \frac{\Delta \Phi}{\Delta t} = -0.01637 \, \text{Wb/s} \)[/tex]:
[tex]\[ \text{EMF} = -605 \times (-0.01637 \, \text{Wb/s}) \][/tex]
[tex]\[ \text{EMF} = 9.901 \, \text{Volts} \][/tex]
Therefore, the EMF generated in the solenoid is approximately [tex]\(\boxed{9.901 \, \text{Volts}}\)[/tex].
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.