Question

What diameter should the nichrome wire in FIGURE P27.63 be in order for the electric field strength to be the same in both wires?

Short answer

The diameter of the nichrome wire, for the electric field strength to be the same in both wires, is $7.29\mathrm{mm}$.

Step-by-step solution

Given information

The electric field $E$ is determined by using, $E=\frac{V}{d}$where E indicates the magnitude of the electric field, V denotes thepotential difference or voltage, and d displays the separation.

Explanation

The electric field $E$denotes, $E=\frac{V}{d}$
The voltage $V$or potential difference between two points is directly proportional to the current $I$,$V=IR$
Hence same value of separation$d$and current $l$
$E\propto V\propto R$(a)
But,$R$ resistance of the wire is,$R=\frac{\rho L}{A}$
So, for same values of $L$
$R\propto \frac{\rho }{A}$(b)
Merge the two equations (a), (b)
$E\propto \frac{\rho }{A}$(c)
but, area of cross-section of the wire,
$A=\frac{\pi d^2}{2}$

$A\alpha d^2$(d)
Blend the equations (c) , (d)
$E\propto \frac{p}{d^2}$(e)

Explanation

Apply equation (e) ,
$\frac{E_A}{E_N}=\frac{{\rho }_A}{d_A^2}\times \frac{d_N^2}{{\rho }_N}$
As,
$E_A=E_N$
$d_A^2{\rho }_N={\rho }_A{d}_N^2$

$d_N^2=\frac{d_A^2{\rho }_N}{{\rho }_A}$

$d_N=d_A\sqrt{\frac{{\rho }_N}{{\rho }_A}}$

Then,

${\rho }_A=2.82\times {10}^{-8}\mathrm{\Omega m}$

${\rho }_N=150\times {10}^{-8}\mathrm{\Omega m}$

so,$d_N=\left(1.0\mathrm{mm}\right)\sqrt{\frac{150\times {10}^{-8}\mathrm{\Omega m}}{2.82\times {10}^{-8}\mathrm{\Omega m}}}$

$=7.29\mathrm{mm}$

The diameter of the nichrome wire, for the electric field strength to be the same in both wires, is$7.2mm$