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sergeinik [125]

4 years ago

10

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?

1 answer:

KIM [24]4 years ago

7 0

<h2>Answer:</h2>

0.31425 Tesla

<h2>Explanation:</h2>

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.

<em>From the question, </em>

L = 20cm = 0.2m

N = 5000 turns

I = 10A

<em>Substitute these values into equation (i) as follows;</em>

B x 0.2 = 1.257 x 10⁻⁶ x 5000 x 10

0.2B = 6.285 x 10⁻²

<em>Solve for B;</em>

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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