You have a 6-m-long copper wire. You want to make an N-turn current loop that generates a 4.085 mT magnetic field at the center when the current is 1.513 A. You must use the entire wire. What will be the diameter, in cm, of your coil?

Respuesta :

ANSWER:

2.98 cm

STEP-BY-STEP EXPLANATION:

Given:

Length wire (L) = 6 m

Current (i) = 1.513 A

magnetic field (β) = 4.085 mT = 4.085 x 10^-3 T

Magnetic field at the center of circular coil is given for the following formula:

[tex]\beta=\frac{\mu_0\cdot N\cdot i}{2R}=\frac{\mu_0\cdot N\cdot\imaginaryI}{d}[/tex]

The total length of wire in the coil is given by:

[tex]\begin{gathered} L=\pi\cdot d\cdot N \\ \\ N=\frac{L}{\pi d} \end{gathered}[/tex]

We replacing:

[tex]\begin{gathered} \beta=\frac{\mu_0\cdot\frac{L}{\pi d}\cdot\imaginaryI}{d} \\ \\ \beta=\frac{\mu_0\cdot L\cdot\imaginaryI}{\pi\cdot d^2} \\ \\ d^2=\frac{\mu_0\cdot L\cdot\imaginaryI}{\pi\beta} \\ \\ d=\sqrt{\frac{\mu_0L\imaginaryI}{\pi\beta}} \\ \\ \text{ Therefore:} \\ \\ d=\sqrt{\frac{4\pi\cdot10^{-7}\cdot6\cdot1.513}{\pi\cdot4.085\cdot10^{-3}}} \\ \\ d=0.0298\text{ m}=2.98\text{ cm} \end{gathered}[/tex]

Therefore, the diameter in centimeters is equal to 2.98 cm