Question

We want to construct a solenoid with a resistance of 4.30 Ω and generate a magnetic field of 3.70 × 10−2 T at its center when applying 4.60 A of electrical current. We want to use copper wire with a diameter of 0.500 mm. If we need the solenoid's radius to be 1.00 cm, how many turns of wire will be need?

Answer 1

N=780 turns will be need to be radius solenoid's 1.00 cm

Step-by-step explanation:

Length of wire to give 4.30Ω obtained from

R = ρL/A

ρ(Cu) = 1.72^-8Ωm

L = RA/ρ

L = 4.3 * π(0.25^-3m)² / 1.72^-8Ωm

L = 49.0m

No# of turns N with radius 0.01m obtained from

N * (2π*r) = L

N=L/(2π*r)

N=49.0m/(2π*0.01m)

N = 780 turns

Using solenoid field

B = μₒ Ni/L

L = μₒ Ni/B

L = (4π^-7)(780)(4.60A) / (0.037T)

L = 0.122m (12.2cm)

It has been assumed that 780 turns formed into a solenoid [ie spiral] will retain the radius 0.01m there will be a slight radius shrinkage.