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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].
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