Question

At the centre of an air core solenoid, the value of the magnetic field B is 0.40 mT. If
the current flowing in the solenoid is 0.4 A, calculate the number of turns per cm.

Answer 1

Answer on Question #74264, Physics / Electric Circuits

Question. At the centre of an air core solenoid, the value of the magnetic field $B$ is $0.40 \, mT$. If the current flowing in the solenoid is $0.4 \, A$, calculate the number of turns per cm.

Given. $B = 0.40 \, mT = 0.40 \cdot 10^{-3} \, T; I = 0.4 \, A$.

Find. $n - ?$

Solution.

For an air core solenoid

$$B = \mu_0 n I,$$

where $\mu_0 = 4\pi \cdot 10^{-7} \, \text{H/m. So}$

$$n = \frac{B}{\mu_0 I} = \frac{0.4 \cdot 10^{-3}}{4\pi \cdot 10^{-7} \cdot 0.4} \approx 800 \text{ turns per meter} = 8 \text{ turns per cm}.$$

Answer. $n = 8$ turns per cm.

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