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?

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.

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)

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$7.2mm$