Answer:
The magnitude of the magnetic field is 55μT
Explanation:
Given;
number of turns of the solenoid per length, n = N/L = 1500 turns/m
current in the solenoid, I = 20 mA = 20 x 10⁻³ A
diameter of the solenoid, d = 4 cm = 0.04 m
The magnetic field at a point that is inside the solenoid;
B₁ = μ₀nI
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³
B₁ = 3.77 x 10⁻⁵ T
Given;
current in the wire, I = 2 A
distance of magnetic field from the wire, r = 1 cm = 0.01 m
The magnetic field at 1.0 cm from the wire;
[tex]B_2 = \frac{\mu_0I}{2\pi r} \\\\B_2 = \frac{4\pi*10^{-7}*2}{2\pi *0.01}\\\\B_2 = 4 *10^{-5} \ T[/tex]
The magnitude of the magnetic field;
[tex]B = \sqrt{B_1^2 +B_2^2} \\\\B = \sqrt{(3.77*10^{-5})^2 + (4*10^{-5})^2} \\\\B = 5.5 *10^{-5} \ T\\\\B = 55 \mu T[/tex]
Therefore, the magnitude of the magnetic field is 55μT
The magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]
Given the following parameters from the question
The magnetic field at a point that is inside the solenoid is expressed according to the formula;
Where;
μ₀ is the permeability of free space = 4π x 10⁻⁷ m/A
B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³
B₁ = 3.77 x 10⁻⁵ T
Next is to get the magnetic field strength in the second wire.
Substitute into the formula:
[tex]B_2=\dfrac{\mu_0 I}{2 \pi r} \\B_2=\frac{4\pi \times 10^{-7}\times 2}{2 \times 3.14\times 0.01} \\B_2 =4.0 \times 10^{-5}T[/tex]
Get the resultant magnetic field:
[tex]B = \sqrt{(0.00003771)^2+(0.00004)^2} \\B =5.5 \times 10^{-7}T[/tex]
Therefore the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]
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