Question 14 Marks
Derive an expression for the magnetic field at any point on the axis of a current carrying loop from Bio-Savart's law. Draw the necessry fig.
Answer
View full question & answer→When the observation point is located at the centre of the coil-we consider a circular coil whose radius is $a$, in which current I is flowing. When the observation point is located at the center of the coil, then the distance of the observation point from each part of the coil $a$ as shown in the figure is equal to the radius $r$ of the coil. Apart from this, the position vector $P$ relative to each current lop makes an angle of $90^{\circ}$ with the direction of the current due to which $\sin \theta=\sin 90^{\circ}=1$.
Hence, according to Biot-Savart law, the magnetic field at the observation point due to loop of the coil is
$ \delta B =\frac{\mu_0}{4 \pi} \frac{I / l}{a^2} $
Due to all the loops, the direction of the magnetic field is perpendicular to the plane of the coil as per the clockwise screw rule. Hence, the total magnetic field at the center of the coil due to complete coil is
$B =\Sigma \delta B =\frac{\mu_0}{4 \pi} \frac{1}{a^2} \Sigma \delta l$
If the coil has $n$ turns of equal radius then
$=\frac{\mu_0}{4 \pi} \cdot \frac{l}{a^2}(2 \pi a)=\frac{\mu_0 I }{2 a}$
$\delta l=2 \pi n a$
$B =\frac{\mu_0 n I }{2 a}$ Tesla
The direction of the magnetic field $B$ is perpendicular to the plane of the coil and this direction is known by the right hand palm rule. If in the coil of 1 metre radius having 1 turn, 1 ampere current is passed then the value of mgnetic induction at the centre of the coil will be as follows:
$B =\frac{4 \pi \times 10^{-7} \times 1 \times 1}{2 \times 1}$
$=2 \pi \times 10^{-7}$ Tesla.
Therefore, 1 ampere is the current which, when flows in a coil of 1 metre radius having 1 turn produces $2 \times 10^{-7}$ Weber $/ m ^2$ (Tesla) magnetic induction is produced at the centre of the coil.
Hence, according to Biot-Savart law, the magnetic field at the observation point due to loop of the coil is
$ \delta B =\frac{\mu_0}{4 \pi} \frac{I / l}{a^2} $
Due to all the loops, the direction of the magnetic field is perpendicular to the plane of the coil as per the clockwise screw rule. Hence, the total magnetic field at the center of the coil due to complete coil is
$B =\Sigma \delta B =\frac{\mu_0}{4 \pi} \frac{1}{a^2} \Sigma \delta l$
If the coil has $n$ turns of equal radius then
$=\frac{\mu_0}{4 \pi} \cdot \frac{l}{a^2}(2 \pi a)=\frac{\mu_0 I }{2 a}$
$\delta l=2 \pi n a$
$B =\frac{\mu_0 n I }{2 a}$ Tesla
The direction of the magnetic field $B$ is perpendicular to the plane of the coil and this direction is known by the right hand palm rule. If in the coil of 1 metre radius having 1 turn, 1 ampere current is passed then the value of mgnetic induction at the centre of the coil will be as follows:
$B =\frac{4 \pi \times 10^{-7} \times 1 \times 1}{2 \times 1}$
$=2 \pi \times 10^{-7}$ Tesla.
Therefore, 1 ampere is the current which, when flows in a coil of 1 metre radius having 1 turn produces $2 \times 10^{-7}$ Weber $/ m ^2$ (Tesla) magnetic induction is produced at the centre of the coil.