High School

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A solenoid with 1961 turns has a radius of 92.6 mm and is 62.2 cm long. If this solenoid carries a current of 62.8 A, what is the magnitude of the magnetic field near the center of the solenoid?

Answer :

In this case, the magnitude of the magnetic field near the center of the solenoid carrying a current of 62.8 A is approximately 2.50 x 10^-4 Tesla.

To calculate the magnitude of the magnetic field near the center of the solenoid with the given parameters, we can use the formula for the magnetic field inside a solenoid.

1. Given:

- Number of turns, N = 1.961

- Radius, r = 92.6 mm = 0.0926 m

- Length, L = 62.2 cm = 0.622 m

- Current, I = 62.8 A

2. Calculate the total number of turns per unit length of the solenoid:

- Number of turns per unit length, n = N / L

3. Calculate the magnetic field inside the solenoid using the formula:

- Magnetic field inside a solenoid, B = μâ‚€ n I

- where μâ‚€ is the permeability of free space (4π x 10^-7 T·m/A)

4. Calculate the magnetic field near the center of the solenoid:

- Magnetic field near the center = B = μâ‚€ n I

5. Substitute the values into the formula:

- n = 1.961 / 0.622 ≈ 3.153 turns/m

- B = (4π x 10^-7 T·m/A) 3.153 turns/m 62.8 A

6. Calculate the magnitude of the magnetic field near the center of the solenoid using the calculated values:

- B ≈ 2.50 x 10^-4 T

Therefore, the magnitude of the magnetic field near the center of the solenoid carrying a current of 62.8 A is approximately 2.50 x 10^-4 Tesla.

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Rewritten by : Jeany