Question

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 the final axial position of that point.

Answer 1

Hint: In this question let the distance of the axial point with respect to the center of the coil be x. We have to use the formula of magnetic field at the axis and the magnetic field at center of the coil. Then we can evaluate the axial position of the point.
Formula used:
Use the formula of the magnetic field at the axis.
$B=\dfrac{{{\mu }_{o}}i{{R}^{2}}}{2{{\left( {{X}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}$
And magnetic field at the center of coil is
$B=\dfrac{{{\mu }_{0}}i}{2R}$

Complete answer:
Given data:
Radius of the current carrying coil is R. Let the distance of the axial point w.r.t center of the coil be x. We know, magnetic field at this point w.r.t the center or at the axis is given as
$B=\dfrac{{{\mu }_{o}}i{{R}^{2}}}{2{{\left( {{X}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}........\left( 1 \right)$
Where, B = magnetic field at the axis
‘i’ = current in the coil
R = radius of the coil
${{\mu }_{o}}$ = permeability of free space
As we know magnetic field at the center of the coil is given as,
${{B}_{center}}=\dfrac{{{\mu }_{o}}i}{2R}.......\left( 2 \right)$
Where, ${{B}_{center}}$ = magnetic field at the center
‘i’ = current in the coil
R = radius of the coil
\[{{\mu }_{o}}\]= permeability of the free space
Now it is given to us in question that the fractional decreases in the field w.r.t. the center of the coil for a nearby axial point is 1%
$\Delta B=\dfrac{1}{100}{{B}_{center}}$
Where $\Delta B$ is the difference of equation (2) and (1) i.e. difference of the magnetic fields. Now substitute the values we have,
$\Rightarrow \dfrac{{{\mu }_{o}}i}{2R}-\dfrac{{{\mu }_{o}}i{{R}^{2}}}{2{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}=\dfrac{1}{100}\dfrac{{{\mu }_{o}}i}{2R}$
Now simplify this we have,
$\begin{align}
 & \dfrac{{{\mu }_{o}}i}{2R}-\dfrac{1}{100}\dfrac{{{\mu }_{o}}i}{2R}=\dfrac{{{\mu }_{o}}i{{R}^{2}}}{2{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}} \\
& \implies \dfrac{99}{100}\dfrac{{{\mu }_{o}}i}{2R}=\dfrac{{{\mu }_{o}}i{{R}^{2}}}{2{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}} \\
\end{align}$
Now cancel out the common terms, we have,
$\begin{align}
  & \dfrac{99}{100}=\dfrac{{{R}^{3}}}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}} \\
 & \implies {{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}=\dfrac{100}{99}{{R}^{3}} \\
 & \implies \left( {{x}^{2}}+{{R}^{2}} \right)={{\left( \dfrac{100}{99}{{R}^{3}} \right)}^{\dfrac{2}{3}}} \\
 & \implies \left( {{x}^{2}}+{{R}^{2}} \right)={{\left( \dfrac{100}{99} \right)}^{\dfrac{2}{3}}}{{R}^{2}} \\
 & \implies {{x}^{2}}+{{R}^{2}}=1.0067{{R}^{2}} \\
 & \implies {{x}^{2}}=\left( 1.0067-1 \right){{R}^{2}} \\
 & \implies {{x}^{2}}=0.0067{{R}^{2}} \\
\end{align}$

Now take square root on both sides we have,
$x=\sqrt{0.0067{{R}^{2}}}=\pm 0.0818R$
So at the distance $\pm 0.0818R$ the fractional decrease in field value w.r.t center of the coil for nearby axial point is 1% hence, this is the required answer.

Note:
It is advised to remember the formula for magnetic field at any point x due to circular current carrying coil of radius R and magnetic field at the center of coil the trick was that magnetic permeability is taken as ${{\mu }_{o}}$ because coil is placed in air however suppose if this would have not been the case then it would have taken in accordance to medium’s permeability involved.