Question

A physicist is constructing a solenoid. She has a roll of insulated copper wire and a power supply. She winds a single layer of the wire on a tube with a diameter of dsolenoid = 10.0 cm. The resulting solenoid is ℓ = 60.0 cm long, and the wire has a diameter of dwire = 0.100 cm. Assume the insulation is very thin, and adjacent turns of the wire are in contact. What power (in W) must be delivered to the solenoid if it is to produce a field of 6.40 mT at its center? (The resistivity of copper is 1.70 ✕ 10−8 Ω · m.)

Answer 1

Answer:

P =105.44 W

Explanation:

Given that

D= 10 cm ,L= 60 cm

d= 0.1 cm ,B= 6.4 mT

ρ= 1.7 x 10⁻⁸ Ω · m

The number of turns N

N= L/d

N= 60/0.1 = 600 turns

Length of wire

Lc= πDN

Lc= 3.14 x 0.1 x 600

Lc=188.4 m

The magnetic filed given as

$B=\dfrac{\mu_oNI}{L}$

$I=\dfrac{BL}{\mu_oN}$

Now by putting the values

$I=\dfrac{0.0064\times 0.6}{4\pi \times 10^{-7}\times 600}$

I=5.09 A

The resistance R given as

$R=\dfrac{\rho L_c}{A}$

$A=\dfrac{\pi}{4}\times d^2$

$A=\dfrac{\pi}{4}\times (0.1\times 10^{-2})^2$

$R=\dfrac{1.7\times 10^{-8} \times 188.4}{\dfrac{\pi}{4}\times (0.1\times 10^{-2})^2}$

R=4.07 Ω

Power P

p =I²R

P= 5.09² x 4.07 W

P =105.44 W