Question

To manufacture a solenoid of length ' \( l \) ' and inductance ' \( L \) ', the length of the thin wire required is (cross - sectional diameter of a solenoid is considerably less than length, \( \mu_0 = \) permeability of free space)

• A. \( \left[ \frac{4\pi l L}{\mu_0} \right]^{1/2} \)
• B. \( \left[ \frac{2\pi l L}{\mu_0} \right]^{1/2} \)
• C. \( \left[ \frac{8\pi l L^2}{\mu_0} \right]^{\frac{1}{2}} \)
• D. \( \left[ \frac{\pi l^2 L}{\mu_0} \right]^{\frac{1}{2}} \)

Hint:

For solenoids: - $L \propto N^2$ - Relate turns with wire length using circumference

Answer 1

The Correct Option is A

Concept: Inductance of a long solenoid: \[ L = \mu_0 \frac{N^2 A}{l} \]

Step 1: Relate number of turns with wire length.
Let total length of wire = $x$ For tightly wound solenoid: \[ \text{Each turn has circumference} = 2\pi r \] \[ N = \frac{x}{2\pi r} \]

Step 2: Substitute in inductance formula.
\[ L = \mu_0 \frac{\left(\frac{x}{2\pi r}\right)^2 \cdot \pi r^2}{l} \]

Step 3: Simplify.
\[ L = \mu_0 \frac{x^2}{4\pi^2 r^2} \cdot \frac{\pi r^2}{l} = \mu_0 \frac{x^2}{4\pi l} \]

Step 4: Solve for $x$.
\[ x^2 = \frac{4\pi l L}{\mu_0} \] \[ x = \left( \frac{4\pi l L}{\mu_0} \right)^{1/2} \] Answer: \( \left[ \frac{4\pi l L}{\mu_0} \right]^{1/2} \)