As a new electrical technician, you are designing a large solenoid to produce a uniform 0.1 T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 1.0 m long and 3.0 cm in diameter. What current will you need to produce the necessary field?

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Teebhabzie Teebhabzie · Answer 1 · 2020-04-17T04:19:38+02:00

Answer:

1.6 mA

Explanation:

Parameters given:

Magnetic field strength, B = 0.1 T

Number of possible turns, N = 4000

Length of solenoid, L = 1 m

Diameter of solenoid = 3 cm = 0.03 m

Radius of solenoid, r = 0.03/2 = 0.015 m

Magnetic field strength, B, is given as:

[tex]B = \frac{NrI}{L}[/tex]

where I = current

Making current, I, subject of formula, we have that:

[tex]BL = NrI\\\\\\=> I = \frac{BL}{Nr}[/tex]

Therefore:

[tex]I = \frac{0.1 * 1}{4000 * 0.015}\\ \\\\I = 0.0016 A = 1.6 mA[/tex]

The current needed to produce the magnetic field is 1.6 mA

aabdulsamir aabdulsamir · Answer 2 · 2020-04-17T04:22:00+02:00

Answer:

19.89A

Explanation:

B = μI (N/L)

B = magnetic field strength = 0.1T

μ = 4Π * 10⁻⁷ T.A/m

I = current passing through the solenoid = ?

N = number of turns = 4000

L = length of the solenoid = 1m

I = B / μ (N/L)

I = 0.1 / [4Π*10⁻⁷ * (4000/1)]

I = 0.1 / 5.03*10⁻³

I = 19.89A

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