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A solenoid of length 20 cm area of cross-section 4 cm2 and having 4000 turns is placed inside another solenoid of 2000 turns having a cross-section area 8 cm2 and length 10 cm. The mutual inductance of the system is
- a)0.02 H
- b)0.04 H
- c)0.06 H
- d)0.06 H
Correct answer is option 'A'. Can you explain this answer?
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A solenoid of length 20 cm area of cross-section 4 cm2 and having 4000...
Understanding Mutual Inductance
Mutual inductance (M) between two solenoids can be calculated using the formula:
\[ M = \frac{N_1 \cdot N_2 \cdot \mu_0 \cdot A}{L} \]
Where:
- \( N_1 \) = Number of turns in the first solenoid
- \( N_2 \) = Number of turns in the second solenoid
- \( \mu_0 \) = Permeability of free space (\( 4\pi \times 10^{-7} \, \text{H/m} \))
- \( A \) = Area of cross-section of the inner solenoid
- \( L \) = Length of the outer solenoid
Given Values
- \( N_1 = 4000 \) turns (inner solenoid)
- \( N_2 = 2000 \) turns (outer solenoid)
- \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)
- \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \)
Calculation Steps
1. Convert Area:
\( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)
2. Insert Values into Formula:
\[
M = \frac{4000 \cdot 2000 \cdot (4\pi \times 10^{-7}) \cdot (4 \times 10^{-4})}{0.1}
\]
3. Simplify:
\[
M = \frac{32000000 \cdot 4\pi \times 10^{-11}}{0.1} = 1280000\pi \times 10^{-11}
\]
4. Calculate M:
\( M \approx 0.04 \, \text{H} \)
Conclusion
The mutual inductance of the system is approximately \( 0.04 \, \text{H} \). However, it appears that the correct answer provided in your source indicates \( 0.02 \, \text{H} \). Please verify the parameters and conditions described, as calculations here suggest \( 0.04 \, \text{H} \).
Mutual inductance (M) between two solenoids can be calculated using the formula:
\[ M = \frac{N_1 \cdot N_2 \cdot \mu_0 \cdot A}{L} \]
Where:
- \( N_1 \) = Number of turns in the first solenoid
- \( N_2 \) = Number of turns in the second solenoid
- \( \mu_0 \) = Permeability of free space (\( 4\pi \times 10^{-7} \, \text{H/m} \))
- \( A \) = Area of cross-section of the inner solenoid
- \( L \) = Length of the outer solenoid
Given Values
- \( N_1 = 4000 \) turns (inner solenoid)
- \( N_2 = 2000 \) turns (outer solenoid)
- \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)
- \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \)
Calculation Steps
1. Convert Area:
\( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)
2. Insert Values into Formula:
\[
M = \frac{4000 \cdot 2000 \cdot (4\pi \times 10^{-7}) \cdot (4 \times 10^{-4})}{0.1}
\]
3. Simplify:
\[
M = \frac{32000000 \cdot 4\pi \times 10^{-11}}{0.1} = 1280000\pi \times 10^{-11}
\]
4. Calculate M:
\( M \approx 0.04 \, \text{H} \)
Conclusion
The mutual inductance of the system is approximately \( 0.04 \, \text{H} \). However, it appears that the correct answer provided in your source indicates \( 0.02 \, \text{H} \). Please verify the parameters and conditions described, as calculations here suggest \( 0.04 \, \text{H} \).
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A solenoid of length 20 cm area of cross-section 4 cm2 and having 4000 turns is placed inside another solenoid of 2000 turns having a cross-section area 8 cm2 and length 10 cm. The mutual inductance of the system isa)0.02 Hb)0.04 Hc)0.06 Hd)0.06 HCorrect answer is option 'A'. Can you explain this answer?
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