Question

A solenoid is designed to produce a magnetic field of 2.30×10−2 T at its center. It has a radius of 1.40 cm and a length of 30.0 cm , and the wire can carry a maximum current of 14.0 A . You may want to review (Pages 939 - 944) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Field of a solenoid. Part A What minimum number of turns per unit length must the solenoid have? nn = m−1 Previous AnswersRequest Answer Incorrect; Try Again; 5 attempts remaining Part B What total length of wire is required? LL = nothing m

Answer 1

Answer:

1308 turns per meter

34.57 m

Explanation:

N = Number of turns

L = Length of wire = 0.3 m

r = Radius of wire = 0.014 m

B = Magnetic field = $2.3\times 10^{-2}\ T$

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

Magnetic field at the center of the solenoid is given by

$B=\frac{\mu_0 Ni}{L}\\\Rightarrow \frac{N{L}=\frac{B}{\mu_0 i}\\\Rightarrow \frac{N}{L}=\frac{2.3\times 10^{-2}}{4\pi\times 10^{-7}\times 14}\\\Rightarrow \frac{N}{L}=1307.34417$

The number of turns will be 1308 per meter

Number of turns would be

$N=1307.34417\times L\\\Rightarrow N=1307.34417\times 0.3\\\Rightarrow N=392.2$

The total number of turns is 393.

Length of the wire is given by

$l=2\pi r\times N\\\Rightarrow l=2\pi \times 0.014\times 393\\\Rightarrow l=34.57\ m$

The total length of the wire of the solenoid is 34.57 m

Answer 2

although this is not the answer, the formula is correct and if you mirror it, you will get the correct answer.