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Sagot :
Certainly! Let's solve the problem by breaking it down into its two parts.
### Part (a)
#### Calculating the Average Induced EMF
1. Given Values:
- Radius of the circular coil, [tex]\( r = 0.2 \)[/tex] meters (20 cm)
- Magnetic field strength, [tex]\( B = 0.2 \)[/tex] Teslas (T)
- Time interval, [tex]\( \Delta t = 0.5 \)[/tex] seconds (s)
2. Calculate the Area of the Coil:
The area [tex]\( A \)[/tex] of a circle is given by:
[tex]\[ A = \pi \times r^2 \][/tex]
Substituting the given radius:
[tex]\[ A = \pi \times (0.2)^2 = 0.04\pi \, \text{m}^2 \][/tex]
3. Initial Magnetic Flux:
Magnetic flux [tex]\( \Phi \)[/tex] through the coil is given by:
[tex]\[ \Phi = B \times A \][/tex]
[tex]\[ \Phi_{\text{initial}} = 0.2 \times 0.04\pi = 0.008\pi \, \text{Wb} \, (\text{Weber}) \][/tex]
4. Final Magnetic Flux:
Since the coil is pulled out of the magnetic field, the final magnetic flux [tex]\( \Phi_{\text{final}} \)[/tex] is zero:
[tex]\[ \Phi_{\text{final}} = 0 \, \text{Wb} \][/tex]
5. Change in Magnetic Flux:
The change in magnetic flux [tex]\( \Delta \Phi \)[/tex] is:
[tex]\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 0 - 0.008\pi = -0.008\pi \, \text{Wb} \][/tex]
6. Average Induced EMF:
According to Faraday's Law of Induction, the induced EMF [tex]\( \mathcal{E} \)[/tex] is given by:
[tex]\[ \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} \][/tex]
Substituting the values:
[tex]\[ \mathcal{E} = -\left(-\frac{0.008\pi}{0.5}\right) = \frac{0.008\pi}{0.5} = 0.016\pi \, \text{V} \approx 0.0503 \, \text{V} \][/tex]
Hence, the average induced EMF during this interval is approximately [tex]\( 0.0503 \, \text{V} \)[/tex].
### Part (b)
#### Calculating the Motional EMF
1. Given Values:
- Length of the bar, [tex]\( l = 2 \)[/tex] meters (m)
- Speed of the bar, [tex]\( v = 5 \)[/tex] meters per second (m/s)
- Magnetic field strength, [tex]\( B = 0.2 \)[/tex] Teslas (T)
2. Motional EMF:
The motional EMF [tex]\( \mathcal{E}_{\text{motional}} \)[/tex] is given by:
[tex]\[ \mathcal{E}_{\text{motional}} = B \times l \times v \][/tex]
Substituting the values:
[tex]\[ \mathcal{E}_{\text{motional}} = 0.2 \times 2 \times 5 = 2 \, \text{V} \][/tex]
Therefore, the motional EMF induced in the bar and rails is [tex]\( 2 \, \text{V} \)[/tex].
### Summary:
a) The average induced EMF during the interval is approximately [tex]\( 0.0503 \, \text{V} \)[/tex] or [tex]\( 0.016\pi \, \text{V} \)[/tex].
b) The motional EMF induced in the bar and rails is [tex]\( 2 \, \text{V} \)[/tex].
### Part (a)
#### Calculating the Average Induced EMF
1. Given Values:
- Radius of the circular coil, [tex]\( r = 0.2 \)[/tex] meters (20 cm)
- Magnetic field strength, [tex]\( B = 0.2 \)[/tex] Teslas (T)
- Time interval, [tex]\( \Delta t = 0.5 \)[/tex] seconds (s)
2. Calculate the Area of the Coil:
The area [tex]\( A \)[/tex] of a circle is given by:
[tex]\[ A = \pi \times r^2 \][/tex]
Substituting the given radius:
[tex]\[ A = \pi \times (0.2)^2 = 0.04\pi \, \text{m}^2 \][/tex]
3. Initial Magnetic Flux:
Magnetic flux [tex]\( \Phi \)[/tex] through the coil is given by:
[tex]\[ \Phi = B \times A \][/tex]
[tex]\[ \Phi_{\text{initial}} = 0.2 \times 0.04\pi = 0.008\pi \, \text{Wb} \, (\text{Weber}) \][/tex]
4. Final Magnetic Flux:
Since the coil is pulled out of the magnetic field, the final magnetic flux [tex]\( \Phi_{\text{final}} \)[/tex] is zero:
[tex]\[ \Phi_{\text{final}} = 0 \, \text{Wb} \][/tex]
5. Change in Magnetic Flux:
The change in magnetic flux [tex]\( \Delta \Phi \)[/tex] is:
[tex]\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 0 - 0.008\pi = -0.008\pi \, \text{Wb} \][/tex]
6. Average Induced EMF:
According to Faraday's Law of Induction, the induced EMF [tex]\( \mathcal{E} \)[/tex] is given by:
[tex]\[ \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} \][/tex]
Substituting the values:
[tex]\[ \mathcal{E} = -\left(-\frac{0.008\pi}{0.5}\right) = \frac{0.008\pi}{0.5} = 0.016\pi \, \text{V} \approx 0.0503 \, \text{V} \][/tex]
Hence, the average induced EMF during this interval is approximately [tex]\( 0.0503 \, \text{V} \)[/tex].
### Part (b)
#### Calculating the Motional EMF
1. Given Values:
- Length of the bar, [tex]\( l = 2 \)[/tex] meters (m)
- Speed of the bar, [tex]\( v = 5 \)[/tex] meters per second (m/s)
- Magnetic field strength, [tex]\( B = 0.2 \)[/tex] Teslas (T)
2. Motional EMF:
The motional EMF [tex]\( \mathcal{E}_{\text{motional}} \)[/tex] is given by:
[tex]\[ \mathcal{E}_{\text{motional}} = B \times l \times v \][/tex]
Substituting the values:
[tex]\[ \mathcal{E}_{\text{motional}} = 0.2 \times 2 \times 5 = 2 \, \text{V} \][/tex]
Therefore, the motional EMF induced in the bar and rails is [tex]\( 2 \, \text{V} \)[/tex].
### Summary:
a) The average induced EMF during the interval is approximately [tex]\( 0.0503 \, \text{V} \)[/tex] or [tex]\( 0.016\pi \, \text{V} \)[/tex].
b) The motional EMF induced in the bar and rails is [tex]\( 2 \, \text{V} \)[/tex].
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