Physics Moving Charges and Magnetism 1 Mark
Appeared in: 2025 (2 sets)
You are required to design an air-filled solenoid of inductance 0.016 H having a length 0.81 m and radius 0.02 m. The number of turns in the solenoid should be (A) 2592 (B) 2866 (C) 2976 (D) 3140
✓ Verified CBSE Marking Scheme Official Evaluation Steps & EL10 Insight

Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]

The self-inductance ($L$) of an air-filled solenoid is given by the formula:

$$L = \frac{\mu_0 N^2 A}{l}$$

Where:

  • $L = 0.016\text{ H}$ (Inductance)
  • $l = 0.81\text{ m}$ (Length)
  • $r = 0.02\text{ m}$ (Radius)
  • $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$ (Permeability of free space)
  • $A = \pi r^2$ (Cross-sectional area)

Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]

First, calculate the cross-sectional area ($A$):

$$A = \pi (0.02)^2 = \pi \times 4 \times 10^{-4}\text{ m}^2$$

Rearranging the formula for $N$:

$$N^2 = \frac{L \cdot l}{\mu_0 \cdot A}$$

Substitute the values:

$$N^2 = \frac{0.016 \times 0.81}{(4\pi \times 10^{-7}) \times (\pi \times 4 \times 10^{-4})}$$ $$N^2 = \frac{0.01296}{16\pi^2 \times 10^{-11}}$$

Using $\pi^2 \approx 9.87$:

$$N^2 = \frac{0.01296}{16 \times 9.87 \times 10^{-11}} = \frac{0.01296}{157.92 \times 10^{-11}} \approx 8206686.9$$ $$N = \sqrt{8206686.9} \approx 2864.7$$

Step 3 (Final Concluded Answer): [0.25 Mark]

Rounding to the nearest integer provided in the options, we get $N \approx 2866$.

Correct Option: (B) 2866

💡 EL10 Concept Insight: Think of a solenoid like a magnetic "storage tank"; the more turns of wire you wrap around the core, the more effectively you can trap magnetic energy, which is why the inductance grows so rapidly as you add more loops.