Question

hey

i am with another question

a copper wire has a diameter of 0.5mm and a resistivity of .
$1.6 \times {10}^{ - 6} ohm \times cm$
How much of this wire would be required to make a ten ohm coil

Answer 1

$\huge{hey}$

$\huge{ \mathfrak{here \: is \: answer}}$


$\bf{given}$


$\sf{resistivity = 1.6 \times {10}^{ - 6} ohm \times cm}$
diameter of. wire= 0.5 mm

so,

$\sf{area \: of \: cross \: section }$


$\sf{a = \pi {r}^{2} }$

$\pi {( \frac{d}{2} )}^{2}$
=
$3.141 \times ( \frac{0.5mm}{2} )^{2}$


$\sf{3.141 \times (0.025cm) ^{2}}$

using the. formula ,

$\sf{r = p\frac{l}{a} }$


$\sf{l = \frac{r}{p} \times a}$

$\sf{l = \frac{10ohm}{1.6 \times {10}^{ - 6}ohm \times cm } \times 3.141 \times ( {0.025}^{2} {cm}}^{2}$


$\bf{l = 12270cm}$

$\bf {l = 122.7cm}$


if diameter doubled then

D=2*0.5mm

=1.0mm
=0.1 cm

$\sf{area \: of \: cross \: section}$


$\pi \: {r}^{2}$

$3.141 \times ( \frac{0.1cm}{2} )^{2}$

=3.141*(0.05 cm)2

$\huge \boxed{hope \: it \: hels}$

$\huge{ \mathbb{THANKS}}$

Answer 2

$\bf\boxed{Solution}$

Diameter of copper wire, D = 0.5mm

Therefore Area of cross section A
= nD^2/4 3.14 (0.5)^2/4 = 0.1963 mm^2

Resistivity of the copper wire p
1.6 ohm cm = 16 × 10^-6 ohm mm

Let the required length be l_________

Then ,
l = R ×A/P________________
l = 10 x 0.1963/16 × 10^-6
= 0.12269 × 10^-6mm
= 122.7 mm Answer.......

HOPE IT'S HELP YOU A LOT.....!!