An ideal solenoid that is 34.0 cm long is carrying a current of 2.00 A.If the magnitude of the magnetic field generated at the center of the solenoid is 9.00 mT,how many turns of wire does this solenoid contain? (μ₀ = 4π × 10^-7 T ∙ m/A) A)860 B)1590 C)2320 D)3180 E)1220

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Quizplus · Accepted Answer

The magnetic field inside a solenoid is given by B = μ₀ * n * I, where n is the number of turns per unit length. Rearranging for n gives n = B / (μ₀ * I). Substituting the given values: n = 9.00 x 10⁻³ T / (4π x 10⁻⁷ T∙m/A * 2.00 A) = 3581 turns/m. The solenoid is 0.34 m long, so the total number of turns is 3581 turns/m * 0.34 m = 1217.54, which rounds to 1220 turns.

An Ideal Solenoid That Is 34