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A solenoid of 200 turns carrying a current of 2 A has a length of 25 cm. What is the magnitude of the magnetic field at the center of the solenoid?

User Renraku
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Final answer:

The magnitude of the magnetic field at the center of the solenoid is approximately 5.03 x 10^-4 T.

Step-by-step explanation:

The magnitude of the magnetic field at the center of 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 (4π × 10-7 T·m/A), n is the number of turns per unit length and I is the current passing through the solenoid.

In this case, the solenoid has 200 turns and a length of 25 cm, which is equivalent to 0.25 m. So the number of turns per unit length is:

n = 200 / 0.25 = 800 turns/m

Substituting the values into the formula:

B = (4π × 10-7 T·m/A) * (800 turns/m * 2 A)

The magnetic field at the center of the solenoid is approximately 5.03 x 10-4 T.

User Catquas
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