Answer:
The magnitude of the magnetic field is [tex]B=3.39*10^{-3} T[/tex]
Explanation:
current:[tex]I=3.0A[/tex]
radius: [tex]0.03m[/tex]
[tex]n=900 turns/m[/tex]
[tex]u_{0} =4\pi *10^{-7}Tm/A[/tex]
The find number of turns:
[tex]n=\frac{turn }{length } \\turn=n*length\\turn=(900turns/m)*(0.02m)\\turn=18 turns[/tex]
so the solenoid has 18 turns
Now to find magnetic field of the solenoid:
[tex]B=u_{0} nI\\B=(4\pi *10^{-7}Tm/A)*(900turns/m)*(3.0A)\\B=3.39*10^{-3} T[/tex]
Answer:
B = 63.93 μT
Explanation:
Current carried by the wire, I₁ = 3.0 A
Number of turns, n = 900 turns/ meter
Radius of the solenoid, R = 3 cm = 0.03 m
r = 2 cm = 0.02 m
Current due to the solenoid, I₂ = 30 mA = 0.03 A
The magnetic field is due to the current carrying conductor and the solenoid
Magnetic field due to the current carrying conductor:
[tex]B_{1} = \frac{\mu I_{1} }{2\pi r} \\B_{1} = \frac{\mu 3}{2\pi *0.02}\\ B_{1} = \frac{\mu 3}{0.04\pi }[/tex]
Magnetic field due to the solenoid:
[tex]B_{2} = \mu N I_{2} \\B_{2} = \mu * 900 * 0.03\\B_{2} = 27 \mu[/tex]
Add B₁ and B₂ together
B = B₁ and B₂
[tex]B = \frac{\mu 3}{0.04\pi } + 27 \mu\\B = 50.87 \mu\\\mu = 4\pi * 10^{-7} \\B = 50.87 * (4\pi * 10^{-7})\\B = 63.93 * 10^{-6} T\\B = 63.93 \mu T[/tex]
