Question

Two identical circular loops $P$ and $Q$, each of radius $r$ and carrying equal currents are kept in the parallel planes having a common axis passing through$O$ . The direction of current in $P$ is clockwise and in $Q$ is anti-clockwise as seen from $O$, which is equidistant from the loops P and $Q$. Find the magnitude of the net magnetic field at $O$.

Answer 1

All magnetic fields are due to currents and due to intrinsic magnetic moments of particles. The relation between current and magnetic field produced by the current is given by the Biot Savart law. Know the relationship between the magnetic field, current, and radius by using Biot Savart law. Apply the logic magnetic field at $P$ and $Q$. then add the two magnetic fields to obtain the net magnetic field.

Formula used:
$B = \dfrac{{{\mu _0}i{a^2}}}{{2{{\left( {{a^2} + {d^2}} \right)}^{\dfrac{3}{2}}}}}$
$B$ is the magnetic field, $i$ is current, $a$ is the distance from the center of the loop.

Complete step to step solution:
The magnetic field produced due to a current-carrying segment is given by Biot Savart law. It is a vector quantity. In order to know the magnetic field produced at a point due to this small element, one can use Biot-Savart’s Law. According to Biot Savart law, the magnetic field depends is directly proportional to the current, is directly proportional to the length, and is inversely proportional to the square of the distance of the point from the element.
The formula to find the magnetic field at an axial point:
$B = \dfrac{{{\mu _0}i{a^2}}}{{2{{\left( {{a^2} + {d^2}} \right)}^{\dfrac{3}{2}}}}}$

Now let us consider current is flowing in $P$ a loop. By using the right-hand thumb rule the direction of the magnetic field is towards left and the current in $Q$ the loop, then the direction of the magnetic field is towards left. So, the net magnetic field at $O$ is the sum of the magnetic field due to $Q$ and $P$ loops
$B = \dfrac{{{\mu _0}I{r^2}}}{{2{{\left( {{r^2} + {{\left( {\dfrac{{2r}}{2}} \right)}^2}} \right)}^{\dfrac{3}{2}}}}}$
Since the field produced is at the same distance as$O$ ,${B_P} = {B_Q}$
Then the net magnetic field at $O$:
$B = {B_p} + {B_Q} = 2\dfrac{{{\mu _0}I}}{{{{\left( 2 \right)}^{\dfrac{5}{3}}}r}} = \dfrac{{{\mu _0}I}}{{{{\left( 2 \right)}^{\dfrac{3}{2}}}r}}$.

Note: Biot-Savart’s Law for the magnetic field has certain similarities as well as the difference with the coulomb's law. In Biot-Savart’s Law, the magnetic field is directly proportional to the sine of the angle. The magnetic field produced due to a current-carrying segment is given by Biot Savart law. Biot-Savart’s Law is a vector quantity.