An ideal solenoid 20 cm long is wound with 5000 turns of very thin wire. What strength magnetic field is produced at the center of the solenoid when a current of 10 A flows through it?

0.31425 Tesla

The magnetic field strength of a solenoid can be found by using the Ampere's law as follows;

BL = μ₀ x N x I      (i)

Where;

B = magnetic field strength

L = length of the solenoid

μ₀ = magnetic constant = 1.257 x 10⁻⁶H/m

N = number of turns in the coil of the solenoid

I = current flowing through the coil of the solenoid.

From the question,

L = 20cm = 0.2m

N = 5000 turns

I = 10A

Substitute these values into equation (i) as follows;

B x 0.2 = 1.257 x 10⁻⁶ x 5000 x 10

0.2B = 6.285 x 10⁻²

Solve for B;

B = 6.285 x 10⁻² / 0.2

B = 31.425 x 10⁻²

B = 0.31425 T

Therefore, the magnetic field strength is 0.31425 Tesla

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