Question

A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the magnetic induction will be ${{\left( \dfrac{1}{8} \right)}^{th}}$ of its value at the centre of the coil is:
A. $\dfrac{R}{\sqrt{3}}$
B. $R\sqrt{3}$
C. $2R\sqrt{3}$
D. $\left( \dfrac{2}{\sqrt{3}} \right)R$

Answer 1

Hint: This question can be solved by applying the formula for the magnitude of magnetic induction for a current carrying circular coil at a distance from its axis. We will then plug in the required value given in the question and solve the equation to get the value of the distance.

Formula used: $B=\dfrac{{{\mu }_{0}}I}{2}\dfrac{{{R}^{2}}}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}$
Where I is the current in the coil, R is the radius of the coil, x is the distance on the axis from its centre and ${{\mu }_{0}}$is the magnetic permeability of free space (vacuum) equal to ${{\mu }_{0}}=4\pi \times {{10}^{-7}}N.{{A}^{-2}}$

Complete Step-by-Step solution:
A current carrying circular coil behaves as a magnet with two poles. The value of the magnetic field at a distance x from the centre on its axis is given by,
$B=\dfrac{{{\mu }_{0}}I}{2}\dfrac{{{R}^{2}}}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}$ -----------------(1)
Where I is the current in the coil, R is the radius of the coil, x is the distance on the axis from its centre and ${{\mu }_{0}}$is the magnetic permeability of free space (vacuum) equal to ${{\mu }_{0}}=4\pi \times {{10}^{-7}}N.{{A}^{-2}}$
Now, let us analyse the question. It is given that the value of magnetic induction at a distance is ${{\left( \dfrac{1}{8} \right)}^{th}}$ of its value at the centre for the same coil. We have to find the distance.
Let the distance be x.
Now, using (1)
For magnetic induction at the centre, $x=0$. Putting this in (1), we get,
$B(\text{centre})=\dfrac{{{\mu }_{0}}I}{2}\dfrac{{{R}^{2}}}{{{\left( {{0}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}=\dfrac{{{\mu }_{0}}I}{2}\dfrac{{{R}^{2}}}{{{\left( {{R}^{2}} \right)}^{\dfrac{3}{2}}}}=\dfrac{{{\mu }_{0}}I}{2}\dfrac{{{R}^{2}}}{{{R}^{3}}}=\dfrac{{{\mu }_{0}}I}{2R}$ ------------------(2)
where $B\left( \text{centre} \right)$ is the value of the magnetic induction at the centre.
Using (1), magnetic induction at the distance x on the axis is given by,
$B\left( \text{at x} \right)=\dfrac{{{\mu }_{0}}I}{2}\dfrac{{{R}^{2}}}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}$---------------(3)
Now, by the problem $\dfrac{B\left( \text{at x} \right)}{B\left( \text{at centre} \right)}=\dfrac{1}{8}$ -----------------(4)
Therefore, putting (2) and (3) in (4), we get,
$\dfrac{\dfrac{{{\mu }_{0}}I}{2}\dfrac{{{R}^{2}}}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}}{\dfrac{{{\mu }_{0}}I}{2R}}=\dfrac{1}{8}$
$\therefore \dfrac{{{R}^{3}}}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}={{\left( \dfrac{1}{2} \right)}^{3}}$
Putting cube root on both sides,
$\therefore \dfrac{R}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{1}{2}}}}=\dfrac{1}{2}$
Squaring both sides,
$\dfrac{{{R}^{2}}}{\left( {{x}^{2}}+{{R}^{2}} \right)}=\dfrac{1}{4}$
$\therefore 4{{R}^{2}}={{x}^{2}}+{{R}^{2}}$
$\therefore 3{{R}^{2}}={{x}^{2}}$
Putting square root on both sides,
$\therefore x=R\sqrt{3}$
Hence the required distance is $R\sqrt{3}$ .
Therefore, the correct option is B) $R\sqrt{3}$.

Note: To check that a student has written formula (1) correctly or not, he can put the value $x=0$ to get the familiar equation for the magnetic induction at the centre that is $\dfrac{{{\mu }_{0}}I}{2R}$. This is a good way of checking that you are not going wrong in an exam.
Hence, a circular current carrying coil behaves similarly to a bar magnet with two poles. The poles of a circular current carrying coil can be determined by looking at it from the front, if the current is going anticlockwise, it is the north pole and if it is going clockwise, it is the south pole and opposite side of the coil will be the north pole.