Question

A long wire carries a current of 3.0A along the central axis of a long solenoid (radius=3.0cm, n=900 turns per meter, current=30.0mA). What is the magnitude of the magnetic field at a point 2.0cm from the axis of the solenoid (neglecting any end effects)?

Answer 1

Answer:

The magnitude of the magnetic field is $B=3.39*10^{-3} T$

Explanation:

current:$I=3.0A$

radius: $0.03m$

$n=900 turns/m$

$u_{0} =4\pi *10^{-7}Tm/A$

The find number of turns:

$n=\frac{turn }{length } \\turn=n*length\\turn=(900turns/m)*(0.02m)\\turn=18 turns$

so the solenoid has 18 turns

Now to find magnetic field of the solenoid:

$B=u_{0} nI\\B=(4\pi *10^{-7}Tm/A)*(900turns/m)*(3.0A)\\B=3.39*10^{-3} T$

Answer 2

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:

$B_{1} = \frac{\mu I_{1} }{2\pi r} \\B_{1} = \frac{\mu 3}{2\pi *0.02}\\ B_{1} = \frac{\mu 3}{0.04\pi }$

Magnetic field due to the solenoid:

$B_{2} = \mu N I_{2} \\B_{2} = \mu * 900 * 0.03\\B_{2} = 27 \mu$

Add B₁ and B₂ together

B = B₁ and B₂

$B = \frac{\mu 3}{0.04\pi } + 27 \mu\\B = 50.87 \mu\\\mu = 4\pi * 10^{-7} \\B = 50.87 * (4\pi * 10^{-7})\\B = 63.93 * 10^{-6} T\\B = 63.93 \mu T$