Question

A circular coil of 200 turns has a radius of 10 cm and carries a current of 2.0 A. (a) Find the magnitude of the magnetic field $\overrightarrow{B}$ at the centre of the coil. (b) At what distance from the centre along the axis of the coil will the field B drop to half its value at the centre?
$(\sqrt[3]{4}=1·5874...)$

Answer 1

Number of turns, n = 200
Radius of the coil, r = 10 cm
Current in the coil, i = 2A
(a) Let the magnetic field at the centre of the coil is B.
As the relation for magnetic field at the centre of a circular coil is given by
$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{n\mu }}_0\mathrm{i}}{2\mathrm{r}} \\ =\frac{200\times 4\mathrm{\pi }\times {10}^{-7}\times 2}{2\times 10\times {10}^{-2}} \\ =2.51\times {10}^{-3} \\ =12.56\mathrm{mT} \end{gathered}$$
(b) As magnetic field at any point P (say) on the axis of the circular coil is given by
$B_P=n\frac{{\mu }_{0}i{r}^2}{2(x^2+r^2{)}^{{\displaystyle \frac{3}{2}}}}$
Where x is the distance of the point from the centre of the coil.
As per the question
$$\begin{gathered} \frac{1}{2}{B}_{centre}=B_P \\ \Rightarrow \frac{1}{2}\frac{n{\mu }_{0}i}{2r}=\frac{n{\mu }_{0}i{r}^2}{2(x^2+r^2{)}^{{\displaystyle \frac{3}{2}}}} \\ \Rightarrow (x^2+r^2{)}^{\frac{3}{2}}=2r^3 \\ \Rightarrow (x^2+r^2)=4^{1/3}{r}^2 \\ \Rightarrow x^2+r^2=1.58r^2 \\ \Rightarrow x=0.766\left|r\right| \\ \Rightarrow x=\pm 7.66cm \end{gathered}$$
Magnetic field will drop to half of its value at the centre if the distance of that point from the centre of the coil along the axis of coil is equal to 7.66 cm.