Inside a long solenoid that carries a current of 5.00 A , the magnetic field B is found to be 3.14 × 10^ -2 T . Determine the number of turns per unit length for this solenoid.

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Correct answer: D. 5000 turns/m

The magnetic field inside a long solenoid is given by the formula:

B = \mu_{0} n I

Rearranging for the number of turns per unit length

n

:

n = \dfrac{B}{\mu_{0} I}

Substituting the given values

B = 3.14 \times 10^{-2} \text{ T}

,

I = 5.00 \text{ A}

, and

\mu_{0} = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}

:

n = \dfrac{3.14 \times 10^{-2}}{4\pi \times 10^{-7} \times 5.00}

Using

\pi \approx 3.14

:

n = \dfrac{3.14 \times 10^{-2}}{4 \times 3.14 \times 10^{-7} \times 5.00}

n = \dfrac{10^{-2}}{20 \times 10^{-7}}

n = \dfrac{10^{-2}}{2 \times 10^{-6}}

n = 0.5 \times 10^{4} = 5000 \text{ turns/m}

Inside a long solenoid that carries a current of 5.00 A , the...

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