A long tightly wound solenoid is used in a laboratory to generate a uniform magnetic field of 25.12 mT at its center. If the solenoid has 500 turns per 10 cm of its length, what is the magnitude of the steady current

Question

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Correct answer: C. 4 A

The magnetic field at the center of a long tightly wound solenoid is given by the formula:

B = \mu_0 n I

where

n

is the number of turns per unit length and

I

is the current.

Given:

B = 25.12 \text{ mT} = 25.12 \times 10^{-3} \text{ T}

n = \dfrac{500}{10 \text{ cm}} = \dfrac{500}{10 \times 10^{-2} \text{ m}} = 5000 \text{ turns/m}

\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}

Substituting the values into the formula:

25.12 \times 10^{-3} = (4 \times 3.14 \times 10^{-7}) \times 5000 \times I

25.12 \times 10^{-3} = 12.56 \times 10^{-7} \times 5000 \times I

25.12 \times 10^{-3} = 6.28 \times 10^{-3} \times I

I = \dfrac{25.12 \times 10^{-3}}{6.28 \times 10^{-3}}

I = 4 \text{ A}

Answer:

4\text{ A}

A long tightly wound solenoid is used in a laboratory to generate a...

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