Question

Two very small identical circular loops, (1) and (2), carrying equal currents $I$ are placed vertically ( with respect to the plane of the paper) with their geometrical axes perpendicular to each other as shown in the figure. Find the magnitude and direction of the net magnetic field produced at point O.

Answer 1

Hint: Magnetic field due to current carrying circular loop is given by
\[B = \dfrac{{{\mu _0}i{R^2}}}{{2{{({x^2} + {R^2})}^{\dfrac{3}{2}}}}}\] where ${\mu _0}$ is called permeability of free space and the value of ${\mu _0}$ is \[4\pi \times {10^{ - 7}}\dfrac{{Tm}}{A}\] and $R$ is the radius of the loop and $I$ is the current through the circular loop.
The direction of the magnetic field can be calculated by using the Right hand thumb rule.

Here is an explanation of Right hand thumb rule in simple words: If a current carrying conductor is imagined to be held in the right hand such that the thumb points in the direction of current then curled fingers of the right hand tell us the direction of the magnetic field.
First of all, find out the magnetic field at point O due to loop 1, then at point O due to loop 2, then find out their resultant so we will get the magnitude of the net magnetic field & then with the help of right hand thumb rule we will get the direction of the net magnetic field.

Complete solution:
Magnetic field at point O due to circular loop 1 is given by \[{B_1} = \dfrac{{{\mu _0}i{R^2}}}{{2{{({x^2} + {R^2})}^{\dfrac{3}{2}}}}}\] and the direction of ${B_1}$ would be towards left.
Magnetic field at point O due to circular loop 2 is given by ${B_2} = \dfrac{{{\mu _0}i{R^2}}}{{2{{({x^2} + {R^2})}^{\dfrac{3}{2}}}}}$ and the direction of ${B_2}$ is acting vertically upwards.
Therefore, the resultant magnetic field at point O will become, $B = \sqrt {B_1^2 + B_2^2} $… … …(i)
Here, we know that the value of \[{B_1}\] and ${B_2}$is same so we can write, ${B_1} = {B_2}$
Now, put ${B_1} = {B_2}$ in equation (i) so we will get $B = \sqrt {2} {B_1} $ … … …(ii)
By substituting the value of ${B_1}$ in equation (ii) we will get \[\dfrac{{{\mu _0}i{R^2}}}{{\sqrt 2 {{({x^2} + {R^2})}^{\dfrac{3}{2}}}}}\]

The direction of the net magnetic field would be at an angle $45^\circ $ with the axis of circular loop 1 as shown in the above figure.

Note: When we want to find out the direction of the magnetic field due to a current carrying wire this rule “Right hand thumb rule” can be used. This rule also can be used to find out the direction of current if you know the magnetic field around the current carrying wire. The most important point to learn from this rule: If the current flows in the upward direction, the direction of the magnetic field.