Magnetism

Given current $i = 10\text{ A}$
Length element $d\vec{l} = (3\hat{i} + 4\hat{j}) \times 10^{-3}\text{ m}$
Position vector $\vec{r} = 5\hat{i} - 12\hat{j}\text{ m}$
Magnitude of position vector $r = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = 13\text{ m}$
Cube of magnitude $r^3 = 13^3 = 2197\text{ m}^3$
Cross product $(3\hat{i} + 4\hat{j}) \times (5\hat{i} - 12\hat{j}) = -36\hat{k} - 20\hat{k} = -56\hat{k}$
Vector product $d\vec{l} \times \vec{r} = -56 \times 10^{-3} \hat{k}\text{ m}^2$
Value of constant $\frac{\mu_0}{4\pi} = 10^{-7}\text{ T}\cdot\text{m/A}$
Substitute values into Biot-Savart Law $d\vec{B} = \frac{10^{-7} \times 10 \times (-56 \times 10^{-3} \hat{k})}{2197}$
Simplifying gives $d\vec{B} = -\frac{56}{2197} \times 10^{-9} \hat{k}\text{ T}$

Given angles subtended at point P are $\theta_1 = 37^\circ$ and $\theta_2 = 53^\circ$
Formula for magnetic field due to finite straight wire is $B_P = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2) (-\hat{k})$
Value of $\sin 37^\circ = \frac{3}{5}$
Value of $\sin 53^\circ = \frac{4}{5}$
Sum of sines is $\sin 37^\circ + \sin 53^\circ = \frac{3}{5} + \frac{4}{5} = \frac{7}{5}$
Substituting the values gives $B_P = \frac{K i}{r} \left(\frac{7}{5}\right) (-\hat{k})$
Simplifying gives $B_P = \frac{7 K i}{5 r} (-\hat{k})$

Given angle subtended by upper end is $\theta_1 = 45^\circ$
Given angle subtended by lower end is $\theta_2 = 45^\circ$
Formula for magnetic field at point P is $B_P = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2) \otimes$
Value of $\sin 45^\circ = \frac{1}{\sqrt{2}}$
Substituting values into the formula gives $B_P = \frac{K i}{r} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) \otimes$
Adding terms inside brackets gives $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$
Simplifying the expression yields $B_P = \frac{\sqrt{2} K i}{r} \otimes$

Given upper angle $\theta_1 = 53^\circ$
Given lower angle $\theta_2 = 0^\circ$
Formula for magnetic field is $B_P = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2) \otimes$
Value of $\sin 53^\circ = \frac{4}{5}$
Value of $\sin 0^\circ = 0$
Substituting values gives $B_P = \frac{K i}{r} \left(\frac{4}{5} + 0\right) \otimes$
Simplifying yields $B_P = \frac{4 K i}{5 r} \otimes$

Given upper angle extending to infinity is $\theta_1 = 90^\circ$
Given lower end opposite to point P has angle $\theta_2 = 0^\circ$
Formula for magnetic field is $B_P = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2) \otimes$
Value of $\sin 90^\circ = 1$
Value of $\sin 0^\circ = 0$
Substituting values gives $B_P = \frac{K i}{r} (1 + 0) \otimes$
Simplifying yields $B_P = \frac{K i}{r} \otimes = \frac{\mu_0 i}{4\pi r} \otimes$

Given upper angle extending to infinity is $\theta_1 = 90^\circ$
Given lower end angle is $\theta_2 = 37^\circ$
Formula for magnetic field at point P is $B_P = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2) (-\hat{k})$
Value of $\sin 90^\circ = 1$
Value of $\sin 37^\circ = \frac{3}{5}$
Substituting values into the formula gives $B_P = \frac{K i}{r} \left(1 + \frac{3}{5}\right) (-\hat{k})$
Sum of terms inside brackets is $1 + \frac{3}{5} = \frac{8}{5}$
Final magnetic field is $B_P = \frac{8 K i}{5 r} (-\hat{k})$

Given upper angle extending to infinity is $\theta_1 = 90^\circ$
Given lower angle extending to infinity is $\theta_2 = 90^\circ$
Formula for magnetic field is $B_P = \frac{K i}{r} (\sin\theta_1 + \sin\theta_2) (-\hat{k})$
Value of $\sin 90^\circ = 1$
Substituting values gives $B_P = \frac{K i}{r} (1 + 1) (-\hat{k})$
Simplifying yields $B_P = \frac{2 K i}{r} (-\hat{k}) = \frac{\mu_0 i}{2\pi r} (-\hat{k})$

Points P and Q are located along the same straight line as the current-carrying wire
Angle between current element vector $d\vec{l}$ and position vector $\vec{r}$ for point P is $\theta = 0^\circ$
Angle between current element vector $d\vec{l}$ and position vector $\vec{r}$ for point Q is $\theta = 180^\circ$
Cross product $d\vec{l} \times \vec{r} = d l \cdot r \sin\theta \hat{n}$
Since $\sin 0^\circ = 0$ and $\sin 180^\circ = 0$, the cross product is zero for both points
Therefore, magnetic field at point P is $B_P = 0$
Therefore, magnetic field at point Q is $B_Q = 0$

The given wire arrangement consists of two distinct wire sections
First section is the horizontal wire lying on the x-axis carrying current $i$ towards point P
Since point P $(0,0)$ lies directly on the axis of this horizontal wire, its magnetic field contribution is $\vec{B}_1 = 0$
Second section is the vertical semi-infinite wire located at $x = -a$ carrying current $i$ upwards along the y-direction
Perpendicular distance from this vertical wire to point P is $a$
Upper end of the vertical wire extends to infinity giving $\theta_1 = 90^\circ$
Lower end of the vertical wire is level with point P giving $\theta_2 = 0^\circ$
By the right-hand rule, current flowing upwards creates a magnetic field into the page at point P, which corresponds to the $(-\hat{k})$ direction
Magnetic field due to the vertical wire is $\vec{B}_2 = \frac{K i}{a} (\sin 90^\circ + \sin 0^\circ) (-\hat{k})$
Substituting values gives $\vec{B}_2 = \frac{K i}{a} (1 + 0) (-\hat{k}) = \frac{K i}{a} (-\hat{k})$
Total magnetic field at point P is $\vec{B}_P = \vec{B}_1 + \vec{B}_2 = 0 + \frac{K i}{a} (-\hat{k}) = \frac{K i}{a} (-\hat{k})$

First wire on the left carries current $i$ upwards extending from $-\infty$ to $+\infty$ at a distance $r_1$ from point P
Magnetic field magnitude due to the left wire is $B_1 = \frac{2 K i}{r_1}$
By the right-hand thumb rule, the direction of magnetic field $\vec{B}_1$ at point P is into the page, i.e., $(-\hat{k})$
Second wire on the right carries current $i$ downwards extending from $+\infty$ to $-\infty$ at a distance $r_2$ from point P
Magnetic field magnitude due to the right wire is $B_2 = \frac{2 K i}{r_2}$
By the right-hand thumb rule, the direction of magnetic field $\vec{B}_2$ at point P is also into the page, i.e., $(-\hat{k})$
Total magnetic field vector $\vec{B}_A$ at point P is the vector sum $\vec{B}_1 + \vec{B}_2$
Substituting values gives $\vec{B}_A = \frac{2 K i}{r_1}(-\hat{k}) + \frac{2 K i}{r_2}(-\hat{k})$
Factoring out common terms yields $\vec{B}_A = 2 K i \left[\frac{1}{r_1} + \frac{1}{r_2}\right] (-\hat{k})$

First wire on the left carries current $i_1$ upwards at a distance $r_1$ from point P
By the right-hand thumb rule, the magnetic field vector due to the left wire at point P is directed into the page, i.e., $\vec{B}_1 = \frac{2 K i_1}{r_1} (-\hat{k})$
Second wire on the right carries current $i_2$ upwards at a distance $r_2$ from point P
By the right-hand thumb rule, the magnetic field vector due to the right wire at point P is directed out of the page, i.e., $\vec{B}_2 = \frac{2 K i_2}{r_2} (+\hat{k})$
Net magnetic field vector at point P is the vector sum $\vec{B}_P = \vec{B}_1 + \vec{B}_2$
Combining both components yields $\vec{B}_P = \frac{2 K i_1}{r_1} (-\hat{k}) + \frac{2 K i_2}{r_2} (+\hat{k})$

First wire on the left carries current $i$ upwards at a distance $r$ from point P
Magnetic field magnitude due to the left wire is $B_1 = \frac{2 K i}{r}$
By the right-hand thumb rule, the magnetic field $\vec{B}_1$ at point P is directed into the page ($\otimes$)
Second wire on the right carries current $i$ upwards at a distance $r$ from point P
Magnetic field magnitude due to the right wire is $B_2 = \frac{2 K i}{r}$
By the right-hand thumb rule, the magnetic field $\vec{B}_2$ at point P is directed out of the page ($\odot$)
Net magnetic field at point P is $B_P = B_1 - B_2$
Substituting values gives $B_P = \frac{2 K i}{r} - \frac{2 K i}{r} = 0$

First wire on the left carries current $i$ upwards extending from $-\infty$ to $+\infty$ at a distance $r$ from point P
Magnetic field magnitude due to the left wire is $B_1 = \frac{2 K i}{r}$
By the right-hand thumb rule, the direction of magnetic field $\vec{B}_1$ at point P is directed into the page ($\otimes$)
Second wire on the right carries current $i$ downwards extending from $+\infty$ to $-\infty$ at a distance $r$ from point P
Magnetic field magnitude due to the right wire is $B_2 = \frac{2 K i}{r}$
By the right-hand thumb rule, the direction of magnetic field $\vec{B}_2$ at point P is also directed into the page ($\otimes$)
Since both magnetic field vectors point in the same direction, the total magnitude is $B_P = B_1 + B_2$
Substituting values gives $B_P = \frac{2 K i}{r} + \frac{2 K i}{r}$
Adding both terms yields $B_P = \frac{4 K i}{r}$

First semi-infinite wire on the left carries current $i$ upwards extending to infinity at a perpendicular distance $r$ from point P
Magnetic field magnitude due to the left wire at point P is $B_1 = \frac{K i}{r}$
By the right-hand thumb rule, the direction of magnetic field $\vec{B}_1$ at point P is directed into the page ($\otimes$)
Second semi-infinite wire on the right carries current $i$ downwards extending to infinity at a perpendicular distance $r$ from point P
Magnetic field magnitude due to the right wire at point P is $B_2 = \frac{K i}{r}$
By the right-hand thumb rule, the direction of magnetic field $\vec{B}_2$ at point P is directed out of the page ($\odot$)
Net magnetic field at point P is $B_P = B_1 \otimes + B_2 \odot$
Substituting values gives $B_P = \frac{K i}{r} \otimes + \frac{K i}{r} \odot = 0$

Perpendicular distance from the center P to each side of the square loop is $r = \frac{L}{2}$
Angles subtended by each side at the center P are $\theta_1 = 45^\circ$ and $\theta_2 = 45^\circ$
Magnetic field due to one side of the square is $B_1 = \frac{K i}{L/2} (\sin 45^\circ + \sin 45^\circ)$
Value of $\sin 45^\circ = \frac{1}{\sqrt{2}}$
Substituting values for one side gives $B_1 = \frac{2 K i}{L} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = \frac{2 K i}{L} \left(\frac{2}{\sqrt{2}}\right) = \frac{2\sqrt{2} K i}{L}$
Since all 4 sides contribute equally in the same direction, total magnetic field at center P is $B_P = 4 \times B_1$
Substituting $B_1$ yields $B_P = 4 \times \frac{2\sqrt{2} K i}{L} = \frac{8\sqrt{2} K i}{L} \otimes$

Perpendicular distance from centroid P to each side of the equilateral triangle is $r = \frac{l}{2\sqrt{3}}$
Angles subtended by each side at centroid P are $\theta_1 = 60^\circ$ and $\theta_2 = 60^\circ$
Magnetic field due to one side of the triangle is $B_1 = \frac{K i}{r} (\sin 60^\circ + \sin 60^\circ)$
Value of $\sin 60^\circ = \frac{\sqrt{3}}{2}$
Sum of sines is $\sin 60^\circ + \sin 60^\circ = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}$
Substituting $r$ and sine values into $B_1$ gives $B_1 = \frac{K i}{l / (2\sqrt{3})} (\sqrt{3}) = \frac{2\sqrt{3} K i \cdot \sqrt{3}}{l} = \frac{6 K i}{l}$
Since all 3 sides contribute equally in the same direction, total magnetic field at center P is $B_P = 3 \times B_1$
Substituting $B_1$ yields $B_P = 3 \times \frac{6 K i}{l} = \frac{18 K i}{l} \otimes$

Point P has coordinates $(3, 4)$, so its perpendicular distance from the vertical y-axis wire is $r_1 = 3$ and from the horizontal x-axis wire is $r_2 = 4$
Formula for magnetic field due to an infinitely long wire is $B = \frac{2 K i}{r}$
For the vertical wire carrying current $i_1 = 12\text{ A}$ upward, distance is $r_1 = 3$
Magnetic field magnitude due to vertical wire is $B_1 = \frac{2 K \times 12}{3} = 8 K$
By the right-hand thumb rule, direction of $B_1$ at point P $(3,4)$ is out of the page ($\odot$)
For the horizontal wire carrying current $i_2 = 4\text{ A}$ leftward, distance is $r_2 = 4$
Magnetic field magnitude due to horizontal wire is $B_2 = \frac{2 K \times 4}{4} = 2 K$
By the right-hand thumb rule, direction of $B_2$ at point P $(3,4)$ is into the page ($\otimes$)
Net magnetic field at point P is $B_P = B_1 \odot - B_2 \otimes$
Substituting values gives $B_P = 8 K - 2 K = 6 K \odot$

The given configuration consists of four wire segments carrying current $i$
First horizontal segment lies along the negative x-axis towards $(-r, 0)$ with point O on its axis, so $B_1 = 0$
Second vertical semi-infinite segment at $x = -r$ extends upwards to $+\infty$ at distance $r$ from point O, contributing $B_2 = \frac{K i}{r} (-\hat{k})$
Third horizontal segment lies along the positive x-axis extending from $(r, 0)$ with point O on its axis, so $B_3 = 0$
Fourth vertical semi-infinite segment at $x = 0$ extends downwards to $-\infty$ from $(0, -r)$ at distance $r$ from point O, contributing $B_4 = \frac{K i}{r} (-\hat{k})$
Total magnetic field vector at point O is $\vec{B}_O = B_1 + B_2 + B_3 + B_4$
Substituting the individual contributions gives $\vec{B}_O = 0 + \frac{K i}{r} (-\hat{k}) + 0 + \frac{K i}{r} (-\hat{k})$
Combining the non-zero terms yields $\vec{B}_O = \frac{2 K i}{r} (-\hat{k})$

The point A lies at a perpendicular distance $r$ from the extended line of wire segment PQ
Upper end P subtends an angle $\theta_1 = 53^\circ$ with the perpendicular
Lower end Q subtends an angle $\theta_2 = 37^\circ$ with the perpendicular on the same side
Formula for magnetic field when both ends are on the same side of the perpendicular is $B_A = \frac{K i}{r} (\sin 53^\circ - \sin 37^\circ)$
Value of $\sin 53^\circ = \frac{4}{5}$
Value of $\sin 37^\circ = \frac{3}{5}$
Difference of sines is $\sin 53^\circ - \sin 37^\circ = \frac{4}{5} - \frac{3}{5} = \frac{1}{5}$
Wait, let us re-evaluate $\frac{K i}{r} \left(\frac{4}{5} - \frac{3}{5}\right) = \frac{K i}{5 r}$
Let us re-check: $\frac{K i}{5 r}$ corresponds to Option A
Substituting values into the formula gives $B_A = \frac{K i}{r} \left(\frac{4}{5} - \frac{3}{5}\right)$
Simplifying the expression yields $B_A = \frac{K i}{5 r}$

General formula for magnetic field at the center of a circular arc of radius $R$ subtending angle $\theta$ is $B_O = \frac{K i}{R} \theta$
For case (a), the loop is a full circle subtending an angle $\theta = 2\pi$ radians at its center O
Substituting $\theta = 2\pi$ gives $B_O = \frac{K i}{R} (2\pi) = \frac{\mu_0}{4\pi} \cdot \frac{i (2\pi)}{R} = \frac{\mu_0 i}{2 R}$
For case (b), the loop is a semi-circular arc subtending an angle $\theta = \pi$ radians at its center O
Substituting $\theta = \pi$ gives $B_O = \frac{K i}{R} \pi = \frac{\mu_0}{4\pi} \cdot \frac{i \pi}{R} = \frac{\mu_0 i}{4 R}$
Therefore, the magnetic fields for cases (a) and (b) are $B_O = \frac{\mu_0 i}{2 R}$ and $B_O = \frac{K i}{R} \pi$ respectively

General formula for the magnetic field at the center of a circular arc is $B_O = \frac{K i}{R} \theta$
For case (a), the quarter-circular arc subtends an angle of $90^\circ$, which equals $\theta = \frac{\pi}{2}$ radians
Substituting $\theta = \frac{\pi}{2}$ into the formula gives $B_O = \frac{K i}{R} \left(\frac{\pi}{2}\right) = \frac{\mu_0 i}{8 R}$
For case (b), the circular arc subtends an angle of $60^\circ$, which equals $\theta = \frac{\pi}{3}$ radians
Substituting $\theta = \frac{\pi}{3}$ into the formula gives $B_O = \frac{K i}{R} \left(\frac{\pi}{3}\right) = \frac{\mu_0 i}{12 R}$
Therefore, the magnetic fields at center O for cases (a) and (b) are $B_O = \frac{K i}{R} \left(\frac{\pi}{2}\right)$ and $B_O = \frac{K i}{R} \left(\frac{\pi}{3}\right)$ respectively

The given configuration consists of four distinct wire segments carrying current $i$
Subtended angle for both circular arcs is $\theta = 60^\circ = \frac{\pi}{3}\text{ radians}$
Segment 1 is the inner arc of radius $R_1$ where current produces a magnetic field $B_1 = \frac{K i}{R_1} \frac{\pi}{3} (-\hat{k})$
Segment 2 is the radial straight section extending outward whose extension passes through center O, so $B_2 = 0$
Segment 3 is the outer arc of radius $R_2$ where current produces a magnetic field $B_3 = \frac{K i}{R_2} \frac{\pi}{3} (+\hat{k})$
Segment 4 is the radial straight section extending inward whose extension passes through center O, so $B_4 = 0$
Total magnetic field vector at center O is $\vec{B}_O = \vec{B}_1 + \vec{B}_2 + \vec{B}_3 + \vec{B}_4$
Substituting values gives $\vec{B}_O = \frac{K i}{R_1} \frac{\pi}{3} (-\hat{k}) + 0 + \frac{K i}{R_2} \frac{\pi}{3} (+\hat{k}) + 0$
Factoring common terms yields $\vec{B}_O = K i \frac{\pi}{3} \left[ \frac{1}{R_1} (-\hat{k}) + \frac{1}{R_2} (+\hat{k}) \right]$

The wire arrangement consists of three distinct segments carrying current $i$
Segment 1 is the left horizontal straight wire whose line of action passes directly through point O, contributing $B_1 = 0$
Segment 2 is the upper semi-circular arc of radius $R$ subtending an angle $\theta = \pi$ radians at center O
Magnetic field magnitude due to segment 2 is $B_2 = \frac{K i}{R} \pi$
By the right-hand thumb rule, current flowing clockwise around center O creates a magnetic field directed into the page ($\otimes$)
Segment 3 is the right horizontal straight wire whose line of action passes directly through point O, contributing $B_3 = 0$
Total magnetic field vector at point O is $\vec{B}_O = B_1 + B_2 + B_3$
Substituting the individual contributions yields $\vec{B}_O = 0 + \frac{K i}{R} \pi \otimes + 0 = \frac{K i}{R} \pi \otimes$

Wire segment 1 is a horizontal straight wire along the x-axis whose line of action passes directly through origin O, contributing $B_1 = 0$
Wire segment 2 is a quarter-circular arc of radius $R$ subtending an angle $\theta = \frac{\pi}{2}$ radians at center O
Magnetic field magnitude due to segment 2 is $B_2 = \frac{K i}{R} \left(\frac{\pi}{2}\right)$
By the right-hand thumb rule, current flowing counter-clockwise around center O produces a magnetic field directed out of the page ($\odot$)
Wire segment 3 is a vertical straight wire along the y-axis extending to infinity whose line of action passes directly through origin O, contributing $B_3 = 0$
Total magnetic field vector at point O is $\vec{B}_O = B_1 + B_2 + B_3$
Substituting individual contributions yields $\vec{B}_O = 0 + \frac{K i}{R} \left(\frac{\pi}{2}\right) \odot + 0 = \frac{K i}{R} \frac{\pi}{2} \odot$

Inner semi-circular arc has radius $R$ and subtends an angle $\theta = \pi$ radians at center O
Current in the inner arc flows counter-clockwise, creating a magnetic field $\vec{B}_R = \frac{K i}{R} \pi (+\hat{k})$ directed out of the page
Outer semi-circular arc has radius $2R$ and subtends an angle $\theta = \pi$ radians at center O
Current in the outer arc flows clockwise, creating a magnetic field $\vec{B}_{2R} = \frac{K i}{2 R} \pi (-\hat{k})$ directed into the page
Straight wire segments connecting the arcs have lines of action passing through center O, contributing zero magnetic field
Net magnetic field vector at center O is $\vec{B}_O = \vec{B}_R + \vec{B}_{2R}$
Substituting values yields $\vec{B}_O = \frac{K i \pi}{R} (+\hat{k}) + \frac{K i \pi}{2 R} (-\hat{k})$
Factoring common terms yields $\vec{B}_O = \frac{K i \pi}{R} \left(1 - \frac{1}{2}\right) (+\hat{k}) = \frac{K i \pi}{2 R} (+\hat{k})$
Substituting $K = \frac{\mu_0}{4\pi}$ gives $\vec{B}_O = \frac{\mu_0}{4\pi} \cdot \frac{i \pi}{2 R} (+\hat{k}) = \frac{\mu_0 i}{8 R} \odot$

The wire arrangement consists of three distinct segments carrying current $i$
Segment 1 is the first straight radial wire whose extension passes directly through point O, contributing zero magnetic field
Segment 2 is a circular arc of radius $R$ subtending an angle of $60^\circ$ at center O
Converting the angle to radians gives $\theta = 60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3}\text{ radians}$
Magnetic field magnitude due to the arc is $B_{\text{arc}} = \frac{K i}{R} \left(\frac{\pi}{3}\right)$
By the right-hand thumb rule, current in the arc produces a magnetic field directed into the page ($\otimes$)
Segment 3 is the second straight radial wire whose extension passes directly through point O, contributing zero magnetic field
Total magnetic field vector at point O is $B_O = 0 + \frac{K i}{R} \frac{\pi}{3} \otimes + 0$
Therefore, the net magnetic field at point O is $B_O = \frac{K i}{R} \frac{\pi}{3} \otimes$

The given wire loop forms a major circular arc of radius $R$ carrying current $i$
Subtended angle at center O is given as $\theta = \frac{3\pi}{2}\text{ radians}$
General formula for magnetic field at the center of a circular arc is $B_O = \frac{K i}{R} \theta$
Expressing $K$ in terms of permeability of free space gives $K = \frac{\mu_0}{4\pi}$
Substituting $K$ into the magnetic field formula yields $B_O = \frac{\mu_0 i}{4\pi R} \theta$
Substituting the angle $\theta = \frac{3\pi}{2}$ gives $B_O = \frac{K i}{R} \cdot \frac{3\pi}{2} = \frac{\mu_0 i}{4\pi R} \frac{3\pi}{2}$
Simplifying further yields $B_O = \frac{3\mu_0 i}{8 R}$

The given wire configuration consists of three distinct segments carrying current $i$
Segment 1 is a horizontal semi-infinite straight wire whose line of action passes directly through point O, contributing $B_1 = 0$
Segment 2 is a circular arc of radius $R$ subtending an angle $\theta = \frac{3\pi}{2}$ radians at center O
Magnetic field magnitude due to segment 2 is $B_2 = \frac{K i}{R} \frac{3\pi}{2}$
By the right-hand thumb rule, current flowing counter-clockwise around center O produces a magnetic field directed out of the page ($\odot$)
Segment 3 is a vertical semi-infinite straight wire whose line of action passes directly through point O, contributing $B_3 = 0$
Total magnetic field vector at point O is $\vec{B}_O = B_1 + B_2 + B_3$
Substituting individual contributions yields $\vec{B}_O = 0 + \frac{K i}{R} \frac{3\pi}{2} \odot + 0 = \frac{K i}{R} \frac{3\pi}{2} \odot$

The given wire configuration consists of three distinct segments carrying current $i$
Segment 1 is a horizontal semi-infinite straight wire at perpendicular distance $r = R$ from point O
For segment 1, subtended angles are $\theta_1 = 90^\circ$ and $\theta_2 = 0^\circ$
Magnetic field magnitude due to segment 1 is $B_1 = \frac{K i}{R} (\sin 90^\circ + \sin 0^\circ) = \frac{K i}{R} (1 + 0) = \frac{K i}{R}$
By the right-hand thumb rule, current in segment 1 produces a magnetic field at point O directed out of the page ($\odot$)
Segment 2 is a circular arc of radius $R$ subtending an angle $\theta = \frac{3\pi}{2}$ radians at center O
Magnetic field magnitude due to segment 2 is $B_2 = \frac{K i}{R} \left(\frac{3\pi}{2}\right)$
By the right-hand thumb rule, counter-clockwise current in segment 2 produces a magnetic field directed out of the page ($\odot$)
Segment 3 is a horizontal semi-infinite straight wire whose line of action passes directly through point O, contributing $B_3 = 0$
Total magnetic field vector at point O is $\vec{B}_O = B_1 + B_2 + B_3$
Substituting individual contributions gives $\vec{B}_O = \frac{K i}{R} \odot + \frac{K i}{R} \left(\frac{3\pi}{2}\right) \odot + 0$
Factoring common terms yields $\vec{B}_O = \frac{K i}{R} \left( \frac{3\pi}{2} + 1 \right) \odot$

Perpendicular distance from the center O to each side of the square loop is $r = \frac{L}{2}$
Angles subtended by the endpoints of each side at the center O are $\theta_1 = 45^\circ$ and $\theta_2 = 45^\circ$
Formula for the magnetic field due to a straight wire segment of one side is $B_1 = \frac{K i}{L/2} (\sin 45^\circ + \sin 45^\circ)$
Value of $\sin 45^\circ = \frac{1}{\sqrt{2}}$
Sum of sines is $\sin 45^\circ + \sin 45^\circ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$
Substituting the sum of sines into $B_1$ gives $B_1 = \frac{K i}{L/2} (\sqrt{2}) \otimes$
Total magnetic field at the center O is due to the 3 current-carrying sides of the square loop, so $B_O = 3 \times B_1$
Substituting $B_1$ yields $B_O = \frac{K i}{L/2} (\sin 45^\circ + \sin 45^\circ) 3 \otimes = \frac{3\sqrt{2} K i}{L/2} \otimes$

The wire configuration consists of three distinct segments carrying current $i$
Segment 1 is a semi-infinite straight wire contributing a magnetic field vector $\vec{B}_1 = \frac{K i}{R} (0 + 1) (-\hat{j}) = \frac{K i}{R} (-\hat{j})$
Segment 2 is a semi-circular arc of radius $R$ subtending an angle of $\pi$ radians in the xy-plane
Magnetic field vector due to segment 2 is $\vec{B}_2 = \frac{K i}{R} \pi (-\hat{k})$
Segment 3 is a semi-infinite straight wire contributing a magnetic field vector $\vec{B}_3 = \frac{K i}{R} (1 + 0) (-\hat{j}) = \frac{K i}{R} (-\hat{j})$
Total magnetic field vector at point O is $\vec{B}_O = \vec{B}_1 + \vec{B}_2 + \vec{B}_3$
Substituting individual contributions yields $\vec{B}_O = \frac{K i}{R} (-\hat{j}) + \frac{K i}{R} \pi (-\hat{k}) + \frac{K i}{R} (-\hat{j})$
Factoring common terms gives $\vec{B}_O = \frac{K i}{R} [-\hat{j} - \pi\hat{k} - \hat{j}]$
Simplifying the vector sum gives $\vec{B}_O = \frac{K i}{R} [-2\hat{j} - \pi\hat{k}]$

Let $R$ be the resistance of each side of the square loop of side length $L$
The bottom branch consists of $1$ side with resistance $R_1 = R$
The remaining top branch consists of $3$ sides in series with resistance $R_2 = 3R$
Total current $i$ divides into $i_1$ through the bottom side and $i_2$ through the top three sides
Using parallel voltage division, $i_1 R = i_2 (3R)$, giving $i_1 = 3 i_2$
Since $i_1 + i_2 = i$, we obtain $i_2 = \frac{i}{4}$ and $i_1 = \frac{3i}{4}$
Magnetic field at center O due to the bottom side carrying current $i_1$ is $B_1 = \frac{2\sqrt{2} K i_1}{L} = \frac{2\sqrt{2} K (3i/4)}{L} = \frac{3\sqrt{2} K i}{2 L}$ directed out of the page ($\odot$)
Magnetic field at center O due to each of the three top sides carrying current $i_2$ is $B_2 = \frac{2\sqrt{2} K (i/4)}{L} = \frac{\sqrt{2} K i}{2 L}$ directed into the page ($\otimes$)
Total magnetic field from the three top sides is $B_{\text{top}} = 3 \times \frac{\sqrt{2} K i}{2 L} = \frac{3\sqrt{2} K i}{2 L}$ directed into the page ($\otimes$)
Net magnetic field at center O is $B_O = B_1 - B_{\text{top}} = \frac{3\sqrt{2} K i}{2 L} - \frac{3\sqrt{2} K i}{2 L} = 0$

The current configuration consists of three parts carrying current
Segment 1 is the straight inlet wire whose line of action passes directly through center O, contributing $B_1 = 0$
The total current $i$ splits equally into two semi-circular paths, each carrying current $I = \frac{i}{2}$
Segment 2 is the upper semi-circular arc of radius $R$ carrying current $I = \frac{i}{2}$ clockwise, producing $B_2 = \frac{K (i/2)}{R} \pi \otimes$
Segment 3 is the lower semi-circular arc of radius $R$ carrying current $I = \frac{i}{2}$ counter-clockwise, producing $B_3 = \frac{K (i/2)}{R} \pi \odot$
Segment 4 is the straight outlet wire whose line of action passes directly through center O, contributing $B_4 = 0$
Total magnetic field vector at center O is $\vec{B}_O = \vec{B}_1 + \vec{B}_2 + \vec{B}_3 + \vec{B}_4$
Substituting individual contributions yields $\vec{B}_O = 0 + \frac{K i \pi}{2 R} \otimes + \frac{K i \pi}{2 R} \odot + 0$
Since the magnetic fields of equal magnitude are in opposite directions, $\vec{B}_O = \frac{K i \pi}{2 R} - \frac{K i \pi}{2 R} = 0$

The system consists of a circular loop and an infinitely long straight wire carrying current $i$
For the circular loop of radius $R$, counter-clockwise current produces a magnetic field out of the page ($\odot$)
Magnetic field magnitude due to the loop is $B_{\text{loop}} = \frac{2 K i}{R} = \frac{\mu_0 i}{2 R}$
Vector contribution of the loop is $\vec{B}_{\text{loop}} = \frac{2 K i}{R} \odot = \frac{2 K i}{R} (+\hat{k})$
For the infinite straight wire at perpendicular distance $R$ below center O, rightward current produces a magnetic field into the page ($\otimes$)
Magnetic field magnitude due to the infinite wire is $B_{\text{wire}} = \frac{\mu_0 i}{2\pi R}$
Vector contribution of the wire is $\vec{B}_{\text{wire}} = \frac{\mu_0 i}{2\pi R} \otimes = \frac{\mu_0 i}{2\pi R} (-\hat{k})$
Total magnetic field vector at center O is $\vec{B}_O = \vec{B}_{\text{loop}} + \vec{B}_{\text{wire}}$
Substituting individual contributions yields $\vec{B}_O = \frac{2 K i}{R} (+\hat{k}) + \frac{\mu_0 i}{2\pi R} (-\hat{k})$
Combining terms gives $\vec{B}_O = \left( \frac{2 K i}{R} - \frac{\mu_0 i}{2\pi R} \right) \hat{k}$