Question

Radius of the current carrying coil is ‘R’. If the fractional decrease 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.

Answer 1

Hint: In this question let the distance of the axial point with respect to the center of the coil be x. Use the direct formula of magnetic field at this point w.r.t the center or at the axis $B = \dfrac{{{\mu _o}i{R^2}}}{{2{{\left( {{x^2} + {R^2}} \right)}^{\dfrac{3}{2}}}}}$ and the magnetic field at the center of the coil that is ${B_{center}} = \dfrac{{{\mu _o}i}}{{2R}}$. Use the constraints of the question to reach the right answer.

Complete step-by-step solution -
Given data:
Radius of the current carrying coil is R.
Let the distance of the axial point w.r.t the center of the coil be x.
So the magnetic field at this point w.r.t the center or at the axis is given as
$\Rightarrow B = \dfrac{{{\mu _o}i{R^2}}}{{2{{\left( {{x^2} + {R^2}} \right)}^{\dfrac{3}{2}}}}}$...................... (1)
Where, B = magnetic field at the axis
i = current in the coil
R = radius of the coil
${\mu _o}$= permeability of the free space.
Now the magnetic field at the center of the coil is given as
$\Rightarrow {B_{center}} = \dfrac{{{\mu _o}i}}{{2R}}$........................ (2)
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 that the fractional decrease in the field w.r.t the center of the coil for a nearby axial point is 1%.
$\Rightarrow \Delta B = \dfrac{1}{{100}}{B_{center}}$
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,
$\Rightarrow \dfrac{{{\mu _o}i}}{{2R}} - \dfrac{1}{{100}}\dfrac{{{\mu _o}i}}{{2R}} = \dfrac{{{\mu _o}i{R^2}}}{{2{{\left( {{x^2} + {R^2}} \right)}^{\dfrac{3}{2}}}}}$
$\Rightarrow \dfrac{{99}}{{100}}\dfrac{{{\mu _o}i}}{{2R}} = \dfrac{{{\mu _o}i{R^2}}}{{2{{\left( {{x^2} + {R^2}} \right)}^{\dfrac{3}{2}}}}}$
Now cancel out the common terms we have,
$\Rightarrow \dfrac{{99}}{{100}} = \dfrac{{{R^3}}}{{{{\left( {{x^2} + {R^2}} \right)}^{\dfrac{3}{2}}}}}$
$\Rightarrow {\left( {{x^2} + {R^2}} \right)^{\dfrac{3}{2}}} = \dfrac{{100}}{{99}}{R^3}$
$\Rightarrow \left( {{x^2} + {R^2}} \right) = {\left( {\dfrac{{100}}{{99}}{R^3}} \right)^{\dfrac{2}{3}}}$
$\Rightarrow \left( {{x^2} + {R^2}} \right) = {\left( {\dfrac{{100}}{{99}}} \right)^{\dfrac{2}{3}}}{R^2} = 1.0067{R^2}$
$\Rightarrow {x^2} = \left( {1.0067 - 1} \right){R^2} = 0.0067{R^2}$
Now take square root on both sides we have,
$\Rightarrow x = \sqrt {0.0067{R^2}} = \pm 0.0818R$
So at distance $\pm 0.0818R$ the fractional decrease in the field value w.r.t the center of the coil for a nearby axial point is 1%.
So, this is the required answer.

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