Problem

A long solenoid with 1000 turns uniformly distributed over a length of $0.5 \, \mathrm{m}$ produces a magnetic field of $2.5 \times 10^{-3} \, \mathrm{T}$ at its center. Find the current flowing through the solenoid.

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Data

• Number of turns: $N = 1000$,
• Length of solenoid: $L = 0.5 \, \mathrm{m}$,
• Magnetic field: $B = 2.5 \times 10^{-3} \, \mathrm{T}$,
• Permeability of free space: $\mu_0 = 4 \pi \times 10^{-7} \, \mathrm{Wb \cdot A^{-1} \cdot m^{-1}}$, Simplified: $\mu_0 = 12.56 \times 10^{-7} \, \mathrm{Wb \cdot A^{-1} \cdot m^{-1}}$.

To find:

• Current in the solenoid: $I = ?$.

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Prerequisite Concepts

The magnetic field inside a solenoid is given by:

$$B = \frac{N}{L} \cdot \mu_0 \cdot I$$

Rearranging for $I$:

$$I = \frac{B \cdot L}{N \cdot \mu_0}$$

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Solution

1. Substitute the known values:

$$I = \frac{2.5 \times 10^{-3} \cdot 0.5}{1000 \cdot 12.56 \times 10^{-7}}$$

2. Simplify step-by-step:
   • Calculate the numerator:

$$2.5 \times 10^{-3} \cdot 0.5 = 1.25 \times 10^{-3}$$

• Calculate the denominator:

$$1000 \cdot 12.56 \times 10^{-7} = 1.256 \times 10^{-3}$$

• Divide the values:

$$I = \frac{1.25 \times 10^{-3}}{1.256 \times 10^{-3}} \approx 0.99 \, \mathrm{A}$$

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Answer

The current in the solenoid is approximately $0.99 \, \mathrm{A}$.