Question:
A 2 m long solenoid with diameter 2 cm and 2000 turns has a secondary coil of 1000 turns wound closely near its midpoint. The mutual inductance between the two coils is
A 2 m long solenoid with diameter 2 cm and 2000 turns has a secondary coil of 1000 turns wound closely near its midpoint. The mutual inductance between the two coils is
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When computing with $\pi^2$, approximate it as $10$ for a quick estimation, or use $9.87$ to get closer to the exact scientific choice.
Updated On: Jun 3, 2026
- $2.4\times10^{-4}\text{ H}$
- $3.9\times10^{-4}\text{ H}$
- $1.28\times10^{-3}\text{ H}$
- $3.14\times10^{-3}\text{ H}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The mutual inductance $M$ between a long primary solenoid and a concentric secondary coil wound closely around its center is given by the formula $M = \frac{\mu_{0} N_{1} N_{2} A}{l}$.
Step 2: Meaning
Here, $N_{1}$ is the primary turns, $N_{2}$ is the secondary turns, $l$ is the length of the primary solenoid, and $A$ is the cross-sectional area of the inner solenoid ($A = \pi r^2 = \pi (\frac{d}{2})^2$).
Step 3: Analysis
Given values: $l = 2\text{ m}$, $d = 2\text{ cm} = 0.02\text{ m}$, $r = 1\text{ cm} = 0.01\text{ m}$, $N_{1} = 2000$, $N_{2} = 1000$, and $\mu_{0} = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$. The area is $A = \pi \times (10^{-2})^2 = \pi \times 10^{-4}\text{ m}^2$. Substituting these values into the formula: $$M = \frac{(4\pi \times 10^{-7}) \times 2000 \times 1000 \times (\pi \times 10^{-4})}{2}$$ $$M = \frac{4\pi^2 \times 10^{-7} \times 2 \times 10^{6} \times 10^{-4}}{2} = 4\pi^2 \times 10^{-5}$$ Since $\pi^2 \approx 9.87$: $$M \approx 4 \times 9.87 \times 10^{-5} = 39.48 \times 10^{-5} \approx 3.9 \times 10^{-4}\text{ H}$$
Step 4: Conclusion
The calculation matches the value $3.9\times10^{-4}\text{ H}$.
Final Answer: (B)
The mutual inductance $M$ between a long primary solenoid and a concentric secondary coil wound closely around its center is given by the formula $M = \frac{\mu_{0} N_{1} N_{2} A}{l}$.
Step 2: Meaning
Here, $N_{1}$ is the primary turns, $N_{2}$ is the secondary turns, $l$ is the length of the primary solenoid, and $A$ is the cross-sectional area of the inner solenoid ($A = \pi r^2 = \pi (\frac{d}{2})^2$).
Step 3: Analysis
Given values: $l = 2\text{ m}$, $d = 2\text{ cm} = 0.02\text{ m}$, $r = 1\text{ cm} = 0.01\text{ m}$, $N_{1} = 2000$, $N_{2} = 1000$, and $\mu_{0} = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$. The area is $A = \pi \times (10^{-2})^2 = \pi \times 10^{-4}\text{ m}^2$. Substituting these values into the formula: $$M = \frac{(4\pi \times 10^{-7}) \times 2000 \times 1000 \times (\pi \times 10^{-4})}{2}$$ $$M = \frac{4\pi^2 \times 10^{-7} \times 2 \times 10^{6} \times 10^{-4}}{2} = 4\pi^2 \times 10^{-5}$$ Since $\pi^2 \approx 9.87$: $$M \approx 4 \times 9.87 \times 10^{-5} = 39.48 \times 10^{-5} \approx 3.9 \times 10^{-4}\text{ H}$$
Step 4: Conclusion
The calculation matches the value $3.9\times10^{-4}\text{ H}$.
Final Answer: (B)
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