Question

It is desired to construct a solenoid that will have a resistance of 4.10 Ω (at 20°C) and produce a magnetic field of 4.00 ✕ 10^−2 T at its center when it carries a current of 3.95 A. The solenoid is to be constructed from copper wire having a diameter of 0.500 mm. If the radius of the solenoid is to be 1.00 cm, determine the following. (The resistivity of copper at 20°C is 1.7 ✕ 10^−8 Ω · m.) (a) the number of turns of wire needed to build the solenoid turns (b) the length the solenoid should have.

Answer 1

(a) 753.68

(b) 9.35 cm

Step-by-step explanation:

R = 4.1 ohm

B = 4 x 10^-2 T

i = 3.95 A

diameter of copper wire = 0.5 mm

radius of copper wire, r = 0.5 / 2 = 0.25 mm = 0.25 x 10^-3 m

Radius of solenoid, R = 1 cm = 0.01 m

Resistivity of copper, ρ = 1.7 x 10^-8 ohm - m

(a) Let N be the total number of turns

use the formula for the resistance

Let L be the length of wire

R = ρ L / A

L = R x A / ρ = (4.10 x 3.14 x 0.25 x 10^-3 x 0.25 x 10^-3) / (1.7 x 10^-8)

L = 47.33 m

N = L / 2 π R = 47.33 / ( 2 x 3.14 x 0.01) = 753.68

(b) Let the length of the solenoid is L'

So, n = N / L' = 753.68 / L'

B = μo x n x i

4 x 10^-2 = 4 x 3.14 x 10^-7 x 753.68 x 3.95 / L'

4 x 10^-2 = 3.74 x 10^-3 / L'

L' = 0.0935 m = 9.35 cm

Answer 2

Part a)

`N = 754 turns`

Part b)

`L = 0.376 m`

Step-by-step explanation:

Resistance of the solenoid is given as

R = 4.10 ohm

now we have

`R = \rho (L)/(A)`

here we know that

`\rho = 1.7 * 10^(-8) ohm-m`

`A = \pi r^2 = \pi ((0.5 * 10^(-3))/(2))^2`

`A = 1.96 * 10^(-7) m^2`

now from above formula

`4.10 = (1.7 * 10^(-8))(L)/(1.96 * 10^(-7))`

`L = 47.35 m`

Part a)

now we know that it is wounded over the solenoid

so here we can say

`N(2\pi R) = L`

`N(2 \pi (0.01)) = 47.35`

`N = 754 turns`

Part b)

now length of solenoid is given as

`L = N(diameter\: of\: the\: wire)`

`L = 754(0.5 * 10^(-3))`

`L = 0.376 m`