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You have a solenoid that is carrying a current of 1.69 A. If the solenoid has 4,464 turns and is 0.24 m long, what is the magnetic field inside the solenoid?

Answer :

Final answer:

The magnetic field inside a solenoid carrying a 1.69 A current, with 4,464 turns, and a length of 0.24 m, can be calculated using the formula B = μonI. After substituting the given values into this formula, it is determined that the magnetic field is approximately 0.04 Tesla (T).

Explanation:

In physics, the magnetic field inside a solenoid (a coil of wire) can be calculated using Ampere's Law. The magnetic field inside a solenoid is given by the formula B = μonI. Where B is the magnetic field, μo is the permeability of free space (4π × 10^-7 T m/A), n is the number of turns per unit length, and I is the current.

In your case, the number of turns per unit length can be found by dividing the total number of turns, 4,464, by the length of the solenoid, 0.24 m. Hence, n = 4,464 / 0.24 = 18,600 turns/m. Substituting these values into the equation, we find B = μonI = (4π × 10^-7 T m/A) × 18,600 turns/m × 1.69 A = 0.04 Tesla (T). Hence, the magnetic field inside your solenoid is approximately 0.04 T.

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