A tightly wound long solenoid carries a current of 1.5 A . An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns . The number of turns per metre in the solenoid is _________.
[Take mass of electron mₑ=9 × 10⁻³¹ kg, charge of electron |qₑ|=1.6 × 10⁻¹⁹ C, μ₀=4 π × 10⁻⁷ N/A², 1 ns=10⁻⁹ s]

Solution

Physics sir · 4 steps

The electron circles in the solenoid's field, so its period fixes B, and B fixes the turns per metre.

1. Field inside. Inside a long solenoid the field is uniform and along the axis: B = μ₀ n I. The electron moves at right angles to this field, so the magnetic force keeps it on a circle.
2. Period of circling. Setting qvB = mv²/r gives r = mv/qB, so the time for one turn is T = (2π r)/v = (2π m)/qB. It does not depend on the speed.

   T = (2π m)/(q μ₀ n I) ⇒ n = (2π m)/(q μ₀ I T)

3. Put in numbers. B = (2π × 9 × 10⁻³¹)/(1.6 × 10⁻¹⁹ × 75 × 10⁻⁹) = 1.5π × 10⁻⁴ T. Then n = (1.5π × 10⁻⁴)/(4π × 10⁻⁷ × 1.5) = 10⁻⁴/(4 × 10⁻⁷).
4. So. The π and the 1.5 cancel, leaving n = 250 turns per metre, so the answer is 250.

The common slip: 40. Writing the period as T = m/qB, without the 2π, gives n ≈ 40 instead of 250.

The idea to keep. A charge in a uniform magnetic field circles with T = (2π m)/qB, whatever its speed. If the period is given, you can find the field straight away.

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