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A long solenoid has a length of 0.50 m and contains 1400 turns of wire. There is a current of 7.4 A in the wire. What is the magnitude of the magnetic field within the solenoid? Number Units

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Final answer:

The magnitude of the magnetic field within the given solenoid is approximately 0.0260 Tesla, calculated using the formula B = μ_0 * (N/L) * I with the known values of the solenoid's properties and the current.

Step-by-step explanation:

To find the magnitude of the magnetic field within a solenoid, you can use the formula: B = μ_0 * (N/L) * I. where B is the magnetic field, μ_0 is the permeability of free space (4π x 10-7 T·m/A), N is the number of turns, L is the length of the solenoid, and I is the current. For the given solenoid with 1400 turns, a length of 0.50 m, and a current of 7.4 A: B = (4π x 10-7 T·m/A) * (1400 turns / 0.50 m) * 7.4 A. Plugging the numbers: B = (4π x 10-7 T·m/A) * 2800 turns/m * 7.4 A. B = (4π x 7.4 x 2800 x 10-7) T. Calculating the value gives: B ≈ 0.0260. The magnitude of the magnetic field within the solenoid is approximately 0.0260 Tesla (T).

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