An electron is observed executing uniform circular motion with a frequency of 1.00 10^8 rev s ⁻¹ inside a tightly-wound, long solenoid. If the solenoid carries a current of 2.00 A , determine the number of turns per metre in the solenoid.

Step-by-step solution

The frequency of revolution of an electron in a magnetic field is given by f = qB 2 m . The magnetic field inside a long solenoid is B = ₀ n I , where n is the number of turns per unit length. Substituting B into the frequency formula, we get: f = q ₀ n I 2 m Rearranging for n : n = 2 m f q ₀ I Substituting the standard values m = 9.1 10⁻³¹ kg , q = 1.6 10⁻¹⁹ C , ₀ = 4 10⁻⁷ T m A ⁻¹ , and the given values f = 1.00 10^8 Hz , I = 2.00 A : n = 2 9.1 10⁻³¹ 10^8 1.6 10⁻¹⁹ 4 10⁻⁷ 2 n = 9.1 10⁻²³ 6.4 10⁻²⁶ n = 1421.875 14