Question

A long insulated copper wire is closely wound as a spiral of $10$ turns, the spiral has inner radius $20\text{cm}$ & outer radius $25\text{cm}$. The spiral lies in the $xy-$ plane, & a steady current $2\text{A}$ flows through the wire. The $z-$component of the magnetic field at the centre of the spiral is
(Take $\ln \left(1.25\right)=0.223$)

• A. $5.6\times {10}^{-5}\text{T}$
• B. $56.5\times {10}^{-5}\text{T}$
• C. $0.5\times {10}^{-5}\text{T}$
• D. $565\times {10}^{-5}\text{T}$

Answer 1

The correct option is A $5.6\times {10}^{-5}\text{T}$

Let us consider an elementary ring of radius $r$ & thickness $dr$ in which current $I$ is flowing.

Number of turns in this elementary ring is given by,

$dN=\left(\frac{N}{b-a}\right)dr......\left(1\right)$

Thus, the magnetic field at the centre $O$ due to the elementary ring is,

$dB=\frac{{\mu }_{0}IdN}{2r}......\left(2\right)$

Integrating the eq $\left(2\right)$ with proper limits we get,

${\int }_0^{B}dB={\int }_a^b\frac{{\mu }_{0}INdr}{2(b-a)r}$

$\Rightarrow B=\frac{{\mu }_{0}IN}{2(b-a)}{\int }_a^b\frac{dr}{r}$

$=\frac{{\mu }_{0}IN}{2(b-a)}[\ln \left(r\right){]}_a^b$

$=\frac{{\mu }_{0}IN}{2(b-a)}\ln \left(\frac{b}{a}\right)$

Given, $N=10,a=20\text{cm},b=25\text{cm}$ and $I=2\text{A}$

$B=\frac{(4\pi \times {10}^{-7}\times 2\times 10)}{2\times (25-20)\times {10}^{-2}}\ln \left(\frac{25}{20}\right)$

$=\frac{4\pi \times {10}^{-7}\times 10}{5\times {10}^{-2}}\ln \left(\frac{5}{4}\right)$

$=5.6\times {10}^{-5}\text{T}$

Hence, option (a) is the correct answer.