Question

A solenoid is designed to produce a magnetic field of 3.50×10^−2 T at its center. It has a radius of 1.80 cm and a length of 46.0 cm , and the wire can carry a maximum current of 13.0 A.

Required:
a. What minimum number of turns per unit length must the solenoid have?
b. What total length of wire is required?

Answer 1

Answer:

a. 2143 turns/m

b. 111.5 m

Explanation:

a. The minimum number of turns per unit length (N/L) can be found using the following equation:

$B = \frac{\mu_{0}NI}{L}$

$\frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m$

Hence, the minimum number of turns per unit length is 2143 turns/m.

b. The total length of wire is the following:

$N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns$

Since each turn has length 2πr of wire, the total length is:

$L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m$

Therefore, the total length of wire required is 111.5 m.

I hope it helps you!