How much current must pass through a 400-turn ideal solenoid that is 4.0 cm long to generate a 1.0-T magnetic field at the center? (μ0 = 4π × 10-7 T ∙ m/A)
We can use the formula for the magnetic field inside a solenoid:
B = μ0 * n * I
where B is the magnetic field, μ0 is the permeability of free space (4π × 10^-7 T·m/A), n is the number of turns per unit length, and I is the current through the solenoid.
To find n, we need to divide the total number of turns (N) by the length of the solenoid (L):
n = N / L
The total number of turns is given as 400 (N = 400), and the length of the solenoid is given as 4.0 cm, which is equal to 0.04 m (L = 0.04 m). Plugging these values into the equation:
n = 400 / 0.04
n = 10000 turns/m
Now we can plug n, μ0, and B into the formula for the magnetic field:
1.0 T = (4π × 10^-7 T·m/A) * (10000 turns/m) * I
Solving for I:
I = 1.0 T / (4π × 10^-7 T·m/A * 10000 turns/m)
I = 1.0 / (4π × 10^-3 A)
I ≈ 0.0796 A
Therefore, a current of approximately 0.0796 A (or 79.6 mA) is required to generate a 1.0-T magnetic field at the center of the solenoid.