An ideal solenoid has a field of magnitude 0.10 T when carrying a current of 200A. The solenoid is 5.0 m long.What is the number of turns for the coils of the solenoid? A)1.0 * 10³ B)2.0 * 10³ C)3.0 * 10³ D)4.0 * 10³

Question

Quizplus · Accepted Answer

The magnetic field inside an ideal solenoid is given by B = μ₀ * (N/L) * I. Solving for N, we get N = B * L / (μ₀ * I). Substituting the given values, N = 0.10 T * 5.0 m / (4π * 10⁻⁷ T*m/A * 200 A) = 2.0 * 10³ turns.

An Ideal Solenoid Has a Field of Magnitude 0