Question

A student is experimenting with some 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 = 90.0 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 7.40 mT at its center? (The

Answer 1

Answer:

$P=214.7187\,W$

Explanation:

Given that:

Diameter of the solenoid, $D=10\,cm=0.1\,m$

length of the solenoid, $L=90\,cm=0.9\,m$

diameter of the wire, $d=0.1\,cm=10^{-3}\,m$

magnetic field at the center of the solenoid, $B=7.4\times 10^{-3}\,T$

Now we need the no. of turns incorporated in the length of 90 cm:

$N=\frac{Length\,\,of\,\,solenoid}{diameter\,\,of\,\, wire}$

$N=\frac{L}{d}$

$N=\frac{0.9}{10^{-3}}$

$N=900\,\,turns$

For solenoids we have:

$B=\mu.n.I$ ...............................(1)

where:

$\mu=$permeability of the medium

n = no. of turns per unit length

I = current in the coil

So,

$n=\frac{900}{0.9}$

$n=1000\,turns\,.\,m^{-1}$

Now putting the respective values in the eq. (1)

$7.4\times 10^{-3}=4\pi\times10^{-7}\times 1000\times I$

$I=5.8887\,A$

• For copper we have resistivity:

• $\rho=1.72\times 10^{-8}\, \Omega.m$

We know that resistance is given by:

$R=\rho.\frac{l}{a}$ .....................................(2)

where:

l = length of the conducting wire

a = cross sectional area of the conducting wire

Now we need the length (l) of the wire:

Circumference of the solenoid,

$C=\pi.D$

$C=0.1\pi\,m$

$\therefore l=C\times N$

$l=90\pi\,m$

&

Cross-sectional area of wire:

$a=\pi.\frac{d^2}{4}$

$a=\pi. \frac{(10^{-3})^2}{4}\,m^2$

Resistance from eq. (2):

$R=1.72\times 10^{-8}\times \frac{90\pi}{\pi. \frac{(10^{-3})^2}{4}}$

$R=6.192 \,\Omega$

• For power we have:

$P=I^2.R$

$P=5.8887^2 \times 6.192$

$P=214.7187\,W$