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. $2.4\times10^{-4}\text{ H}$
• B. $3.9\times10^{-4}\text{ H}$
• C. $1.28\times10^{-3}\text{ H}$
• D. $3.14\times10^{-3}\text{ H}$

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Mutual inductance formula.
For a coil wound on a long solenoid, $M = \dfrac{\mu_{0}N_{1}N_{2}A}{l}$, where $A$ is the cross-section area.

Step 2: Find the area.
Diameter $2$ cm means radius $r = 0.01$ m. \[ A = \pi r^{2} = \pi\times 10^{-4} \text{ m}^{2} \]
Step 3: List the values.
$l = 2$ m, $N_{1}=2000$, $N_{2}=1000$, $\mu_{0}=4\pi\times 10^{-7}$.

Step 4: Put them in.
\[ M = \frac{(4\pi\times 10^{-7})(2000)(1000)(\pi\times 10^{-4})}{2} \]
Step 5: Simplify the powers.
\[ M = \frac{4\pi^{2}\times 10^{-7}\times 2\times 10^{6}\times 10^{-4}}{2} = 4\pi^{2}\times 10^{-5} \]
Step 6: Use $\pi^{2}\approx 9.87$.
\[ M \approx 4\times 9.87\times 10^{-5} \approx 3.9\times 10^{-4} \text{ H} \]This is option 2.
\[ \boxed{M \approx 3.9\times 10^{-4} \text{ H}} \]