A 460-turn solenoid of length 30.0 cm and radius 4.00 cm carries a current of 3.50 A. What is the magnetic field strength inside the coil at its midpoint?
-The units T·m²/s are equivalent to:

Question

Quizplus · Accepted Answer

The magnetic field inside a solenoid is given by B = μ_0nI, where μ_0 is the permeability of free space, n is the number of turns per unit length, and I is the current. At the midpoint of the solenoid, n = N/L, where N is the total number of turns and L is the length of the solenoid. Substituting the given values gives: n = N/L = 460/(0.3 m) = 1533.33 m^-1 B = μ_0nI = (4π×10^-7 T·m/A)(1533.33 m^-1)(3.5 A) = 1.72×10^-3 T The units T·m²/s are equivalent to volts (V), which is not a valid choice. The correct unit for magnetic field strength is tesla (T), which has units of N/(A·m) or N·s/(C·m). The other options (W, N/m, and webers) are not relevant to the question.

A 460-Turn Solenoid of Length 30