Question

A circular loop of radius 4.0 cm is placed in a horizontal plane and carries an electric current of 5.0 A in the clockwise direction as seen from above. Find the magnetic field (a) at a point 3.0 cm above the centre of the loop (b) at a point 3.0 cm below the centre of the loop.

Answer 1

Given:
Magnitude of current, I = 5.0 A
Radius of the loop, r = 4.0 cm



(a) The magnetic field intensity B on point O at a distance x on the axial line is given by
$B=\frac{{\mathrm{\mu }}_0}{2}\frac{i{r}^2}{(x^2+r^2{)}^{3/2}}$
$$\begin{gathered} =\frac{4\mathrm{\pi }\times {10}^{-7}\times 5\times 16\times {10}^{-4}}{2\left[\right(9\times 16)\times {10}^{-4}{]}^{3/2}} \\ =\frac{4\mathrm{\pi }\times 80\times {10}^{-11}}{2\times 125\times {10}^{-6}} \\ =4.019\times {10}^{-5}\mathrm{T}(\mathrm{in}\mathrm{downward}\mathrm{direction}) \end{gathered}$$

(b) The magnetic field intensity B on point O' at a distance x on the axial line is given by
$B=\frac{{\mathrm{\mu }}_0}{2}\frac{i{r}^2}{(x^2+r^2{)}^{3/2}}$
$$\begin{gathered} =\frac{4\mathrm{\pi }\times {10}^{-7}\times 5\times 16\times {10}^{-4}}{2\left[\right(9\times 16)\times {10}^{-4}{]}^{3/2}} \\ =\frac{4\mathrm{\pi }\times 80\times {10}^{-11}}{2\times 125\times {10}^{-6}} \\ =4.019\times {10}^{-5}\mathrm{T}(\mathrm{in}\mathrm{downward}\mathrm{direction}) \end{gathered}$$