Question

The radius of current carrying coil is $R$. If fractional decreases in field value with respect to the center of the coil for a nearby axial point is $1\%$, then find the axial position of that point.
$[$Take, $\frac{1}{{0.99}^{2/3}}=1.0067]$

• A. $\pm 0.28R$
• B. $\pm 0.82R$
• C. $\pm 0.028R$
• D. $\pm 0.082R$

Answer 1

The correct option is D $\pm 0.082R$

Given that, the fractional decrease in magnetic field along the axis of the circular loop at a distance $x$ from the center is $1\%$.

The magnetic field due to circular loop on the axis is given by formula,

$B_{axis}=\frac{{\mu }_{0}i{R}^2}{2(R^2+x^2{)}^{3/2}}$

And magnetic field at the center of the circular loop will be

$B_{center}=\frac{{\mu }_{0}i}{2R}$

From question,

$B_{center}-B_{axis}=1\%B_{center}$

$B_{center}-B_{axis}=\frac{1}{100}{B}_{center}$

$B_{axis}=\frac{99}{100}{B}_{center}$

$\frac{{\mu }_{0}i{R}^2}{2(R^2+x^2{)}^{3/2}}=\frac{99}{100}\frac{{\mu }_{0}i}{2R}$

$\frac{R^3}{0.99}=(R^2+x^2{)}^{3/2}$

$x^2=(\frac{R^3}{0.99}{)}^{2/3}-R^2$

$x^2=1.0067R^2-R^2=0.0067R^2$

$\therefore x=\pm 0.082R$

Hence, option $\left(D\right)$ is the correct answer.