Question
A solenoid is wound with 2000 turns per meter. When the current is 5.2 A, what is the magnetic field within the solenoid?
A solenoid is wound with 2000 turns per meter. When the current is 5.2 A, what is the magnetic field within the solenoid?
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Koala Tutor
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The magnetic field inside a solenoid can be calculated using the formula:
B = μ0 * n * I
where B is the magnetic field, μ0 is the permeability of free space (approximately 4π * 10^-7 T·m/A), n is the number of turns per meter, and I is the current.
In this case, n = 2000 turns/m and I = 5.2 A. Substituting these values into the formula:
B = (4π * 10^-7 T·m/A) * 2000 turns/m * 5.2 A
B = 4π * 10^-7 T·m/A * 2000 turns/m * 5.2 A
B = (4π * 10^-7 T·m/A) * (2000 turns/m) * (5.2 A)
B ≈ 5.23 × 10^-3 T
Therefore, the magnetic field within the solenoid is approximately 5.23 millitesla (mT).
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