Question

The magnetic field strenght of a solenoid is 0.0270T. Its radius is 0.40 m and length is 0.40 m. How many turns are there in the solenoid if the steady current passing through it is 12.0 A?

Answer 1

The magnetic field inside the solenoid is given as follows:

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

where $I = 12A$ is the current, $L = 0.4m$ is the length of solenoid, $\mu_0 = 1.26\times 10^{-6}N/A^2$ is the magnetic constant, $N$ - number of turns. Substituting $B = 0.027T$, get:

$$N = \dfrac{BL}{\mu_0I} = \dfrac{0.027\cdot 0.4}{1.26\times 10^{-6}\cdot 12} \approx 710turns$$

Answer. 710 turns.