Consider a circular coil of radius a and carrying current I in the direction shown in Figure. Suppose the loop lies in the plane of paper. It is desired to find the magnetic field at the centre 
of the coil. Suppose the entire circular coil is divided into a large number of current elements, each of length di. According to Biot-Savart law, the magnetic field 
at the centre 
of the coil due to current element I 
is given by,
![\[\overrightarrow{\mathrm{dB}}=\frac{\mu_{0} \mathrm{I}(\overrightarrow{\mathrm{dl}} \times \overrightarrow{\mathrm{I}})}{4 \pi r^{3}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cf027f77423e747f78fd7e3a27fc8ca5_l3.png "Rendered by QuickLaTeX.com")
where 
is the position vector of point from the current element. The magnitude of at the centre is
![\[\begin{array}{l} \mathrm{dB}=\frac{\mu_{0} \mathrm{I} \mathrm{dlasin} \theta}{4 \pi \mathrm{a}^{3}} \\ \therefore \mathrm{dB}=\frac{\mu_{0} \mathrm{Idl} \sin \theta}{4 \pi \mathrm{a}^{2}} \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4f977c36bc1ca5e79384102444769006_l3.png "Rendered by QuickLaTeX.com")
The direction of is perpendicular to the plane of the coil and is directed inwards. Since each current element contributes to the magnetic field in the same direction, the total magnetic field B at the center can be found by integrating the above equation around the loop i.e.
![\[\therefore \mathrm{B}=\int \mathrm{dB}=\int \frac{\mu_{0} \mathrm{Idlsin} \theta}{4 \pi \mathrm{a}^{2}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-016ba78483873b7e318c5288efb64d89_l3.png "Rendered by QuickLaTeX.com")
For each current element, angle between 
and 
is 
Also distance of each current element from the center is a.
![\[\therefore \mathrm{B}=\frac{\mu_{0} \mathrm{I} \sin 90^{\circ}}{4 \pi \mathrm{a}^{2}} / \mathrm{dl}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-afa5d28e7349f242290b09ecce0500bf_l3.png "Rendered by QuickLaTeX.com")
But 
total length of the coil
![\[\begin{array}{l} \therefore \mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}^{2}} 2 \pi \mathrm{a} \\ \therefore \mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{a}} \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b4af7715e632e4b123ca7aa7fe939600_l3.png "Rendered by QuickLaTeX.com")
.
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west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points above the horizontal. The horizontal component of the earth’s field is measured to be 
Specify the direction and magnitude of the earth’s field at the location.At a certain location in Africa, a compass points 
and 
carrying currents of 
and 
in the same direction are separated by a distance of 
. Estimate the force on a 
section of wire A.
wire carrying a current of 
is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 
. What is the magnetic force on the wire?
and making an angle of 
with the direction of a uniform magnetic field of 
. What is the magnitude of the field 
at a point 
from the wire?
-turns carrying current 
.
is at distance 
from the wire. Find the force acting on the electron if its velocity is directed towards the wire.