Respuesta :
Answer:
d = 1.047 mm
Explanation:
given,
diameter of the wire = 2.0-cm
length of solenoid = 15 cm = 0.15
Current in the wire = I = 2.5 A
magnetic field = B = 3.0 mT
Magnetic field inside the solenoid
[tex]B = \dfrac{\mu_0 N I}{L}[/tex]
[tex]B = \dfrac{\mu_0 N I}{L}[/tex]
N x d = l
[tex]N = \dfrac{l}{d}[/tex]
[tex]B = \dfrac{\mu_0 \dfrac{l}{d} I}{L}[/tex]
[tex]3 \times 10^{-3} = \dfrac{4\pi \times 10^{-7}\times \dfrac{0.15}{d}\times 2.5}{0.15}[/tex]
[tex]0.45 \times 10^{-3}\ d = 4\pi \times 10^{-7}\times 0.15\times 2.5[/tex]
[tex]\ d = \dfrac{4\pi \times 10^{-7}\times 0.15\times 2.5}{0.45 \times 10^{-3}}[/tex]
d = 1.047 x 10⁻³ m
d = 1.047 mm
diameter of the wire is d = 1.047 mm
The diameter of the wire, which is tightly wound over the solenoid, is 1.0472 mm.
What is magnetic field?
The magnetic field is the field in the space and around the magnet in which the magnetic field can be fill.
The magnetic field generate inside the solenoid to current flowing wire can be given as,
[tex]B=\dfrac{(4\pi 10^{-7})I}{d}[/tex]
Here, (I) is the current,and (d) is the diameter of the wire.
A 2.0-cm-diameter, 15-cm-long solenoid is tightly wound with one layer of wire. A 2.5.A current through the wire generates a 3.0 mT magnetic field inside the solenoid.
Put the values in the above formula as,
[tex]3\times10^{-3}=\dfrac{(4\pi 10^{-7})2.5}{d}\\d=1.0472\times10^{-3}\rm m\\d=1.0472 mm[/tex]
Hence, the diameter of the wire, which is tightly wound over the solenoid, is 1.0472 mm.
Learn more about magnetic field here;
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