Question

State Biot - Savart law. Deduce the expression for the magnetic field at a point on the axis of a current carrying circular loop of radius 'R' distant 'x' from the center. Hence, write the magnetic field at the center of a loop.

Answer 1

Biot- Savart law states that for charge in rest produce electric field, but when in motion current flow and it induces magnetic fields.

Biot-Savart law helps in calculating force(magnetic) at any point.

Let us consider a circular loop of

Radius r with center C, P be a,

Point at distance x from C

Now considering small section

$LP\perp dl\text{and}MP\perp dl$

$LP=MP=\sqrt{x^2+r^2}$ (1)

Using Biot-Savart law

$dB=\frac{{\mu }_0}{4\pi }\frac{Idl\sin 90°}{(r^2+x^2})$ (2)

$dB\cos \theta$ components balance each other and

$B=\int dB\sin \theta$

$\int \frac{{\mu }_0}{4\pi }\left[\frac{Idl}{(r^2+x^2)}\right]\frac{r}{\sqrt{r^2+x^2}}$

$\sin \theta =\frac{r}{\sqrt{r^2+x^2}}$

$|thereforeB=\frac{{\mu }_0}{4\pi }\frac{rI}{{(r^2+x^2)}^{\frac{3}{2}}}\int dl$

$\frac{{\mu }_0}{4\pi }\frac{{\mu }_{0}Ir}{4\pi {(r^2+x^2)}^{\frac{3}{2}}}\times 2\pi r$

$\therefore \overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{{\mu }_{0}Ir}{4\pi {(r^2+x^2)}^{\frac{3}{2}}}\times 2\pi r$

$=\frac{{\mu }_{0}Ir}{2\pi {(r^2+x^2)}^{\frac{3}{2}}}$

And magnetic field at center of the loop is

$\overrightarrow{B}=\frac{{\mu }_{0}Ir}{2\pi {(r^2+x^2)}^{\frac{3}{2}}}$

Putting $x=0$

$\overrightarrow{B}=\frac{{\mu }_{0}Ir}{2r^3}$