Find magnetic field at point ' 𝑂 ' in the given figure shown.

The correct answer is:

$\frac{{\mu }_{0}I}{2R}(1-\frac{1}{\pi })$

Step-by-step Explanation:

From the given diagram, the system consists of two parts:
1. A circular current loop of radius R carrying a current I in the clockwise direction.
2. A long straight wire placed tangentially to the circular loop at a perpendicular distance R from center O, carrying current I towards the right.

1. Magnetic Field due to the Circular Loop (B₁):
The magnitude of the magnetic field at the center O due to a circular current-carrying loop of radius R is given by:

$B_1=\frac{{\mu }_{0}I}{2R}$

Using the Right-Hand Thumb Rule, since the current flows clockwise around the center O, the direction of magnetic field B₁ is directed into the page (⊗).

2. Magnetic Field due to the Straight Wire (B₂):
The magnitude of the magnetic field at point O located at a distance R from an infinitely long straight wire carrying current I is given by:

$B_2=\frac{{\mu }_{0}I}{2\pi R}$

Using the Right-Hand Palm Rule, since the current in the straight wire flows towards the right and point O lies above the wire, the magnetic field B₂ is directed out of the page (⊙).

3. Net Magnetic Field at Point O (B_net):
Since B₁ (into the page) and B₂ (out of the page) act in opposite directions, the net magnetic field magnitude at point O is:

$B_{\text{net}}=B_1-B_2$

Substituting the expressions for B₁ and B₂:

$B_{\text{net}}=\frac{{\mu }_{0}I}{2R}-\frac{{\mu }_{0}I}{2\pi R}$

Factoring out $\frac{{\mu }_{0}I}{2R}$:

$B_{\text{net}}=\frac{{\mu }_{0}I}{2R}(1-\frac{1}{\pi })$

Thus, the magnitude of the net magnetic field at point O is $\frac{{\mu }_{0}I}{2R}(1-\frac{1}{\pi })$ (directed into the page).