The magnetic field inside a 200 turns solenoid of radius 10 cm is 2.9 × 10⁻⁴ Tesla. If the solenoid carries a current of 0.29 A , then the length of the solenoid is _______ π cm.

Solution

Physics sir · 3 steps

Only turns per unit length set the field inside a solenoid, so ignore the radius and solve for L.

1. Field inside. Inside a long solenoid the field is uniform and equals μ₀ n I, where n = N/L is the number of turns per unit length. The radius does not appear, so the 10 cm is extra information.

   B = (μ₀ N I)/L ⇒ L = (μ₀ N I)/B

2. Put in numbers. L = (4π × 10⁻⁷ × 200 × 0.29)/(2.9 × 10⁻⁴). Since 0.29/(2.9 × 10⁻⁴) = 1000, this becomes L = 4π × 10⁻⁷ × 2 × 10⁵ = 8π × 10⁻² m.
3. So. L = 8π × 10⁻² m = 8π cm, so the blank is 8.

The common slip: 0.08. Leaving the length in metres gives 0.08π m, so 0.08 instead of 8. Convert to centimetres before reading off the number in front of π.

The idea to keep. For a long solenoid, B = μ₀ n I depends only on turns per unit length and current. The radius never enters, so do not let it distract you.

Answer from the NTA final answer key. Solution by PrepWiser.