Question

A 200-turn solenoid having a length of 25 cm and diameter of 10 cm carries a current of 0.30 A. What is the magnitude of the magnetic field B near the center of the solenoid?

Answer 1

The magnetic fields is given as follows:

$$B = \mu_0\dfrac{NI}{L}$$

where $\mu_0 = 4\pi\times 10^{-7}H/m$ is the magnetic constant, $N = 200$ number of turns, $I = 0.30A$ is the current, $L = 25cm = 0.25m$ is the length of the solenoid. Thus, obtain:

$$B = 4\pi\times 10^{-7}\dfrac{H}{m}\cdot \dfrac{200\cdot 0.30A}{0.25m} \approx 3.0\times 10^{-4}T$$

Answer. $3.0\times 10^{-4}T$.