Two identical loops are placed coaxially as shown. Radius of both loops is r

Correct Answer: Option 1: $\frac{3{\mu }_{0}i}{4\sqrt{2}r}$ towards P

Step-by-step Explanation:

1. Magnetic Field Formula for a Circular Loop
The magnetic field at a point on the axis of a circular current-carrying loop of radius $r$ at a distance $x$ from its center is given by the formula:

$B=\frac{{\mu }_{0}I{r}^2}{2{(r^2+x^2)}^{3/2}}$
Where:

• ${\mu }_0$ is the permeability of free space.
• $I$ is the current in the loop.
• $r$ is the radius of the loop.
• $x$ is the distance of the point on the axis from the center of the loop.

2. Calculate the Fields at the Midpoint O
Looking at the provided diagram:

• The loop centered at Q has a radius $r$ and carries a current $i$. The distance from Q to O is $x=r$.
• The loop centered at P has a radius $r$ and carries a current $4i$. The distance from P to O is $x=r$.

Let us calculate the magnitude of the magnetic fields produced by each loop at O:

For the loop at Q (current $I=i$, distance $x=r$):
$B_Q=\frac{{\mu }_{0}i{r}^2}{2{(r^2+r^2)}^{3/2}}=\frac{{\mu }_{0}i{r}^2}{2{\left(2r^2\right)}^{3/2}}=\frac{{\mu }_{0}i}{4\sqrt{2}r}$

For the loop at P (current $I=4i$, distance $x=r$):
$B_P=\frac{{\mu }_0\left(4i\right)r^2}{2{(r^2+r^2)}^{3/2}}=\frac{4{\mu }_{0}i}{4\sqrt{2}r}$

3. Determine the Directions of the Fields
Using the right-hand grip rule:

• For the loop at Q, the current is directed upwards on the front/left side (clockwise when viewed from the right). Therefore, the magnetic field ${\overrightarrow{B}}_Q$ at point O points to the left (towards Q).
• For the loop at P, the current is directed downwards on the front/left side (counter-clockwise when viewed from the right). Therefore, the magnetic field ${\overrightarrow{B}}_P$ at point O points to the right (towards P).

4. Net Magnetic Field at Point O
Since the two magnetic fields point in opposite directions, the magnitude of the net magnetic field is the difference between the two fields, pointing in the direction of the stronger field:

$B_{\text{net}}=B_P-B_Q$

$B_{\text{net}}=\frac{4{\mu }_{0}i}{4\sqrt{2}r}-\frac{{\mu }_{0}i}{4\sqrt{2}r}=\frac{3{\mu }_{0}i}{4\sqrt{2}r}$

Since $B_P>B_Q$, the direction of the net magnetic field is the same as that of ${\overrightarrow{B}}_P$, which is towards P.