# (a) State and explain the law used to determine magnetic field at a point due to a current element. Derive the expression for the magnetic field due to a circular current carrying loop of radius r at its centre.(b) A long wire with a small current element of length 1 cm is placed at the origin and carries a current of 10 A along the X-axis. Find out the magnitude and direction of the magnetic field due to the element on the Y-axis at a distance 0·5 m from it.

Biot-Savart's Law: Determines the magnetic field at any point due to current-carrying conductor.

According to Biot-Savart law,  the magnetic field at a point P which is at a distance r from the current element is given by

$\overrightarrow{dB}=K \tfrac{id\vec l\times \vec r }{r^{3}}$

$\overrightarrow{dB}=K \tfrac{idl\sin \theta }{r^{2}} \\\Rightarrow \overrightarrow{B} =\int \overrightarrow{dB} = \frac{\mu _{0}}{4\pi } \int i\frac{dl\sin \theta }{^{r^{2}}}$

Consider a circular loop of radius r and a small current element dl, then the magnetic field due to the current element at the centre o of the circle is given by

$\overrightarrow{dB}=K \tfrac{id\vec l\times \vec r }{r^{3}}$

$\overrightarrow{dB}=K \tfrac{idl\sin \theta }{r^{2}}$

Total magnetic field
$\overrightarrow{B} =\int \overrightarrow{dB} = \frac{\mu _{0}}{4\pi } \int i\frac{dl\sin 90 }{^{r^{2}}}=\frac{\mu _{0}i}{4\pi r^2} \int dl\\=\frac{\mu _{0}i}{4\pi r^2} \times 2\pi r=\frac{\mu _{0}i}{2 r}$

(b) Given:  A length of wire - 1 cm

current carry =10A

Now, the direction of  magnetic fields:

$\left | d\vec{B} \right |=\frac{\mu _{0}I}{4\pi }\frac{dl\sin \theta }{\left [ \vec{r} \right ]^{2}}$

$=\frac{4\pi \times 10^{-7}}{4\pi }\times \frac{10\times (1\times 10^{-2})\times \sin 90^{\circ}}{(0.5)^{2}}$

$=4\times 10^{-8}T$ in z-direction

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