Biot-Savart's Law and its application Questions in English

(B) For a long cylindrical wire of radius '$a$' carrying current '$I$':
$1$. Inside the wire $(r < a)$, the magnetic field is given by $B = \frac{\mu_0 I r}{2\pi a^2}$. This shows that $B$ is directly proportional to $r$ $(B \propto r)$, resulting in a linear increase from the center to the surface.
$2$. Outside the wire $(r > a)$, the magnetic field is given by $B = \frac{\mu_0 I}{2\pi r}$. This shows that $B$ is inversely proportional to $r$ $(B \propto 1/r)$, resulting in a hyperbolic decrease as distance increases.
$3$. Combining these, the graph shows a linear increase up to $r = a$ and a hyperbolic decrease for $r > a$. This corresponds to Graph $B$.

(A) The magnetic field $B$ due to a long straight current-carrying wire at a distance $r$ is given by the formula $B = \frac{\mu_0 I}{2\pi r}$.
Given values are current $I = 90 \text{ A}$ and distance $r = 1.5 \text{ m}$.
The permeability of free space is $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
Substituting these values into the formula:
$B = \frac{4\pi \times 10^{-7} \times 90}{2\pi \times 1.5}$
$B = \frac{2 \times 10^{-7} \times 90}{1.5}$
$B = 2 \times 10^{-7} \times 60 = 120 \times 10^{-7} = 1.2 \times 10^{-5} \text{ T}$.
According to the right-hand thumb rule, if the current flows from east to west, the magnetic field at a point above the wire points towards the north.

(B) Let the two wires be parallel and separated by a distance $d = 30 \text{ cm} = 0.3 \text{ m}$.
The current in each wire is $I = 8 \text{ A}$.
Since the currents flow in opposite directions, the magnetic fields produced by both wires at the midpoint (distance $r = d/2 = 0.15 \text{ m}$) will point in the same direction according to the Right-Hand Thumb Rule.
The magnetic field due to a long straight wire is $B = \frac{\mu_0 I}{2\pi r}$.
For the first wire, $B_1 = \frac{\mu_0 I}{2\pi (d/2)} = \frac{2 \times 10^{-7} \times 8}{0.15} = \frac{16 \times 10^{-7}}{0.15} = \frac{1600}{15} \times 10^{-7} \approx 106.67 \mu \text{T}$.
Since both wires contribute equally in the same direction, the total magnetic field $B_{total} = B_1 + B_2 = 2 \times B_1 = 2 \times 106.67 \mu \text{T} = 213.33 \mu \text{T}$.
Given the options provided, $213.33 \mu \text{T}$ is closest to $200 \mu \text{T}$, but based on standard physics problem sets where $d$ is often adjusted to yield integer results (e.g.,if $d = 0.2 \text{ m}$), $213.33 \mu \text{T}$ is the calculated value. Among the choices, $213.33$ is not present, but $213.33$ is mathematically correct.

(C) For wire $I$: The magnetic field at center $P$ is the sum of the fields due to the two straight segments and the semicircular arc. The field due to a semi-infinite wire at distance $r$ is $B_{straight} = \frac{\mu_0 I}{4\pi r}$. Since there are two such segments, their contribution is $2 \times \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{2\pi r}$. The field due to the semicircular arc is $B_{arc} = \frac{\mu_0 I}{4r}$. Thus, $B_1 = \frac{\mu_0 I}{2\pi r} + \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4r} (\frac{2}{\pi} + 1) = \frac{\mu_0 I}{4r} (\frac{2+\pi}{\pi})$.
For wire $II$: The magnetic field at center $Q$ is the difference between the field due to the semicircular arc and the field due to the straight segments. The straight segments contribute $B_{straight} = \frac{\mu_0 I}{4\pi r}$ each. However, based on the geometry, the net field is $B_2 = \frac{\mu_0 I}{4r} - \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{4r} (1 - \frac{1}{\pi}) = \frac{\mu_0 I}{4r} (\frac{\pi-1}{\pi})$.
Taking the ratio $\frac{B_1}{B_2} = \frac{\frac{2+\pi}{\pi}}{\frac{\pi-1}{\pi}} = \frac{2+\pi}{\pi-1}$.

(A) The magnetic field at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2R}$.
Substituting the given values: $B = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} = 4\pi \times 10^{-6} \text{ T}$.
The magnetic energy density $u$ is given by $u = \frac{B^2}{2\mu_0}$.
$u = \frac{(4\pi \times 10^{-6})^2}{2 \times 4\pi \times 10^{-7}} = \frac{16\pi^2 \times 10^{-12}}{8\pi \times 10^{-7}} = 2\pi \times 10^{-5} \text{ J/m}^3$.
The volume of the cube $V$ is $(1 \text{ mm})^3 = (10^{-3} \text{ m})^3 = 10^{-9} \text{ m}^3$.
The magnetic energy stored $U$ is $U = u \times V$.
$U = (2\pi \times 10^{-5}) \times 10^{-9} = 2\pi \times 10^{-14} \text{ J}$.
Using $\pi = 3.14$, $U = 2 \times 3.14 \times 10^{-14} = 6.28 \times 10^{-14} \text{ J}$.
Comparing this with $\alpha \times 10^{-14} \text{ J}$, we get $\alpha = 6.28$.

(A) Inside the conductor $(r < a)$, the magnetic field is given by $B = \frac{\mu_0 I r}{2 \pi a^2}$, which shows that $B \propto r$. This represents a linear increase in the magnetic field from the axis to the surface of the conductor.
Outside the conductor $(r > a)$, the magnetic field is given by $B = \frac{\mu_0 I}{2 \pi r}$, which shows that $B \propto \frac{1}{r}$. This represents a reciprocal decrease in the magnetic field as the distance from the conductor increases.
Graph $(A)$ correctly displays a linear increase followed by a reciprocal decrease. Therefore, option $(A)$ is correct.

(A) Let $R_1$ be the resistance of segment $ABC$ and $R_2$ be the resistance of segment $ADC$. Given $R_1 = \frac{1}{2} R_2$, or $R_2 = 2R_1$.
Since the segments are in parallel, the potential difference across them is the same, so $I_1 R_1 = I_2 R_2$, where $I_1$ and $I_2$ are currents in segments $ABC$ and $ADC$ respectively.
Substituting $R_2 = 2R_1$, we get $I_1 R_1 = I_2 (2R_1)$, which implies $I_1 = 2I_2$.
Also, $I_1 + I_2 = I_0$. Substituting $I_1 = 2I_2$, we get $3I_2 = I_0$, so $I_2 = \frac{I_0}{3}$ and $I_1 = \frac{2I_0}{3}$.
Assuming the loop is a semicircle, the magnetic field at the center due to a circular arc of angle $\theta$ is $B = \frac{\mu_0 I \theta}{4\pi r}$.
For segment $ABC$ (semicircle, $\theta = \pi$), $B_1 = \frac{\mu_0 I_1 \pi}{4\pi r} = \frac{\mu_0 I_1}{4r} = \frac{\mu_0 (2I_0/3)}{4r} = \frac{\mu_0 I_0}{6r}$ (directed into the page).
For segment $ADC$ (semicircle, $\theta = \pi$), $B_2 = \frac{\mu_0 I_2 \pi}{4\pi r} = \frac{\mu_0 I_2}{4r} = \frac{\mu_0 (I_0/3)}{4r} = \frac{\mu_0 I_0}{12r}$ (directed out of the page).
The net magnetic field $B_{net} = |B_1 - B_2| = |\frac{\mu_0 I_0}{6r} - \frac{\mu_0 I_0}{12r}| = \frac{\mu_0 I_0}{12r}$.

(C) Let the wires be at $A$ and $B$ with currents $I_1 = 8 \text{ A}$ and $I_2 = 15 \text{ A}$ in opposite directions. The distance $AB = 7 \text{ cm}$.
Point $P$ is equidistant from $A$ and $B$, so $PA = PB = r$. Since $\angle APB = 90^\circ$, in $\triangle APB$, $AB^2 = PA^2 + PB^2 = 2r^2$.
$7^2 = 2r^2 \implies 49 = 2r^2 \implies r^2 = 24.5 \implies r = \sqrt{24.5} \text{ cm} = \sqrt{24.5} \times 10^{-2} \text{ m}$.
The magnetic field due to wire $A$ is $B_1 = \frac{\mu_0 I_1}{2\pi r}$ and due to wire $B$ is $B_2 = \frac{\mu_0 I_2}{2\pi r}$.
Since the currents are in opposite directions, the fields $B_1$ and $B_2$ at $P$ are perpendicular to each other.
The resultant magnetic field $B = \sqrt{B_1^2 + B_2^2} = \frac{\mu_0}{2\pi r} \sqrt{I_1^2 + I_2^2}$.
$B = \frac{2 \times 10^{-7}}{r} \sqrt{8^2 + 15^2} = \frac{2 \times 10^{-7}}{\sqrt{24.5} \times 10^{-2}} \times 17 = \frac{34 \times 10^{-5}}{\sqrt{24.5}} \approx \frac{34 \times 10^{-5}}{4.95} \approx 6.86 \times 10^{-5} \text{ T}$.
Given $B = X \times 10^{-6} \text{ T}$, so $X \approx 68.6 \approx 68$.

(B) The distance between the two wires is $d = 3 \text{ m}$. The midpoint is at a distance $r = d/2 = 1.5 \text{ m}$ from each wire.
For a long straight wire, the magnetic field is given by $B = \frac{\mu_0 I}{2\pi r}$.
For the first wire $(I_1 = 3 \text{ A})$, the magnetic field at the midpoint is $B_1 = \frac{\mu_0 \times 3}{2\pi \times 1.5} = \frac{3\mu_0}{3\pi} = \frac{\mu_0}{\pi} = \frac{2\mu_0}{2\pi}$.
For the second wire $(I_2 = 4.5 \text{ A})$, the magnetic field at the midpoint is $B_2 = \frac{\mu_0 \times 4.5}{2\pi \times 1.5} = \frac{4.5\mu_0}{3\pi} = \frac{1.5\mu_0}{\pi} = \frac{3\mu_0}{2\pi}$.
Since the currents are in opposite directions, the magnetic fields at the midpoint due to both wires will be in the same direction (by the Right-Hand Thumb Rule).
Therefore, the resultant magnetic field is $B_{net} = B_1 + B_2 = \frac{2\mu_0}{2\pi} + \frac{3\mu_0}{2\pi} = \frac{5\mu_0}{2\pi}$.

(C) The magnetic field $B$ inside a long solenoid is given by $B = \mu_n I = \mu_0 \mu_r n I$, where $n = N/L$ is the number of turns per unit length.
Given: Area $A = 2 \text{ cm}^2 = 2 \times 10^{-4} \text{ m}^2$, Length $L = 40 \text{ cm} = 0.4 \text{ m}$, Number of turns $N = 400$, Current $I = 0.4 \text{ A}$, Magnetic flux $\phi = 4\pi \times 10^{-6} \text{ Wb}$.
The magnetic flux is $\phi = B \cdot A = \mu_0 \mu_r (N/L) I \cdot A$.
Substituting the values: $4\pi \times 10^{-6} = (4\pi \times 10^{-7}) \cdot \mu_r \cdot (400 / 0.4) \cdot 0.4 \cdot (2 \times 10^{-4})$.
$4\pi \times 10^{-6} = (4\pi \times 10^{-7}) \cdot \mu_r \cdot 1000 \cdot 0.4 \cdot 2 \times 10^{-4}$.
$4\pi \times 10^{-6} = (4\pi \times 10^{-7}) \cdot \mu_r \cdot 8 \times 10^{-2}$.
$10^{-6} = 10^{-7} \cdot \mu_r \cdot 8 \times 10^{-2}$.
$10 = \mu_r \cdot 8 \times 10^{-2}$.
$\mu_r = 10 / (8 \times 10^{-2}) = 1000 / 8 = 125$.

(A) The ring acts as a circular current loop when it rotates.
The current $I$ is defined as the rate of flow of charge, $I = \frac{q}{T}$, where $T$ is the time period of one revolution.
Since the frequency is $f$, the time period is $T = \frac{1}{f}$.
Therefore, the equivalent current is $I = qf$.
The magnetic field $B$ at the centre of a circular loop of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2R}$.
Substituting the value of $I$, we get $B = \frac{\mu_0 (qf)}{2R} = \frac{\mu_0 qf}{2R}$.

(C) Let the length of the wire be $L$.
For a single turn of radius $R_1$, the circumference is $2\pi R_1 = L$, so $R_1 = L / (2\pi)$.
The magnetic field at the centre is $B_1 = \frac{\mu_0 I}{2 R_1} = \frac{\mu_0 I}{2 (L / 2\pi)} = \frac{\mu_0 I \pi}{L}$.
For $n$ turns of radius $R_n$, the total length of wire is $n(2\pi R_n) = L$, so $R_n = L / (2\pi n)$.
The magnetic field at the centre is $B_n = \frac{n \mu_0 I}{2 R_n} = \frac{n \mu_0 I}{2 (L / 2\pi n)} = \frac{n^2 \mu_0 I \pi}{L}$.
The ratio of the magnetic field for one turn to that of $n$ turns is $B_1 / B_n = (\frac{\mu_0 I \pi}{L}) / (\frac{n^2 \mu_0 I \pi}{L}) = 1 / n^2$.

(B) The magnetic field '$B$' at a perpendicular distance '$r$' from a long straight wire carrying current '$I$' is given by the formula:
$B = \frac{\mu_0 I}{2 \pi r}$
From this expression, it is clear that the magnetic field is inversely proportional to the distance '$r$':
$B \propto \frac{1}{r}$
Let '$B_1$' be the magnetic field at distance '$r_1 = r$' and '$B_2$' be the magnetic field at distance '$r_2 = 2r$'.
Then, $\frac{B_2}{B_1} = \frac{r_1}{r_2}$
Substituting the values:
$\frac{B_2}{B} = \frac{r}{2r} = \frac{1}{2}$
Therefore, $B_2 = \frac{B}{2}$.

(A) The magnetic field at the center of a circular coil of $N$ turns, radius $R$, and current $I$ is given by $B = \frac{\mu_0 NI}{2R}$.
For a single turn $(N=1)$ of wire of length $L$, the circumference is $2\pi R = L$, so $R = \frac{L}{2\pi}$.
Thus, $B = \frac{\mu_0 (1) I}{2(L/2\pi)} = \frac{\mu_0 \pi I}{L}$.
When the same wire is bent into $N'=2$ turns, the new radius $R'$ satisfies $2\pi R' = L/2$, so $R' = \frac{L}{4\pi} = \frac{R}{2}$.
The new magnetic field $B'$ is $B' = \frac{\mu_0 N' I}{2R'} = \frac{\mu_0 (2) I}{2(R/2)} = 4 \left( \frac{\mu_0 I}{2R} \right) = 4B$.

(C) The magnetic field at point '$O$' is the sum of the magnetic fields due to three segments: the straight wire $AB$, the circular arc $BC$, and the straight wire $CD$.
$1$. For the straight wire $AB$, point '$O$' lies on the line extending from the wire. Therefore, the magnetic field due to $AB$ at '$O$' is $B_1 = 0$.
$2$. For the circular arc $BC$ of radius $r$ and angle $\theta = \pi/2$, the magnetic field at the center is given by $B_2 = \frac{\mu_0 I \theta}{4\pi r} = \frac{\mu_0 I (\pi/2)}{4\pi r} = \frac{\mu_0 I}{8r}$.
$3$. For the straight wire $CD$, point '$O$' lies on the line extending from the wire. Therefore, the magnetic field due to $CD$ at '$O$' is $B_3 = 0$.
Wait, re-evaluating the geometry: The wire segments $AB$ and $CD$ are semi-infinite wires ending at $B$ and starting at $C$ respectively. The distance from '$O$' to the line of $AB$ is $r$, and from '$O$' to the line of $CD$ is $r$.
For a semi-infinite wire, the magnetic field at a perpendicular distance $r$ from the end is $B = \frac{\mu_0 I}{4\pi r}$.
- Magnetic field due to $AB$ at '$O$': $B_{AB} = \frac{\mu_0 I}{4\pi r}$.
- Magnetic field due to arc $BC$ at '$O$': $B_{arc} = \frac{\mu_0 I}{4\pi r} \cdot \frac{\pi}{2} = \frac{\mu_0 I}{8r}$.
- Magnetic field due to $CD$ at '$O$': $B_{CD} = \frac{\mu_0 I}{4\pi r}$.
Total magnetic field $B = B_{AB} + B_{arc} + B_{CD} = \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{8r} + \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{4\pi r} (1 + \frac{\pi}{2} + 1) = \frac{\mu_0 I}{4\pi r} (2 + \frac{\pi}{2})$.
Thus, the correct option is $C$.

(D) The magnetic field $B$ at a perpendicular distance $r$ from a long straight wire carrying current $I$ is given by the formula:
$B = \frac{\mu_0 I}{2 \pi r}$
From this formula, we can see that $B \propto \frac{1}{r}$.
Let $B_1$ be the magnetic field at distance $r_1 = 2 \text{ cm}$ and $B_2$ be the magnetic field at distance $r_2 = 5 \text{ cm}$.
Then, the ratio $\frac{B_1}{B_2} = \frac{r_2}{r_1}$.
Substituting the given values:
$\frac{B_1}{B_2} = \frac{5}{2} = 5 : 2$.
Therefore, the ratio of the magnetic field at $2 \text{ cm}$ and $5 \text{ cm}$ is $5 : 2$.

(D) The magnetic field $B$ at the centre of a circular arc of radius $r$ carrying current $i$ that subtends an angle $\theta$ at the centre is given by the formula:
$B = \frac{\mu_0 i \theta}{4 \pi r}$
Given, $\theta = \frac{\pi}{5}$ radians.
Substituting the value of $\theta$ in the formula:
$B = \frac{\mu_0 i (\frac{\pi}{5})}{4 \pi r}$
$B = \frac{\mu_0 i \pi}{20 \pi r}$
$B = \frac{\mu_0 i}{20 r}$
Therefore, the correct option is $D$.

(C) The magnetic field at the center of a circular loop of radius $R$ carrying current $I$ is given by: $B_{center} = \frac{\mu_0 I}{2R}$.
The magnetic field at a point on the axis of the loop at a distance $x$ from the center is given by: $B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
Given $x = R$, the magnetic field on the axis is: $B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(2R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(2\sqrt{2} R^3)} = \frac{\mu_0 I}{4\sqrt{2} R}$.
Now, the ratio of the magnetic field at the center to the magnetic field on the axis is: $\frac{B_{center}}{B_{axis}} = \frac{\mu_0 I / 2R}{\mu_0 I / 4\sqrt{2} R} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} : 1$.

(D) The arrangement consists of two semi-infinite wires and two quarter-infinite segments. However, looking at the geometry, the magnetic field at point '$O$' due to the segments $AB$ and $BC$ can be calculated using the Biot-Savart Law.
For a semi-infinite wire, the magnetic field at a perpendicular distance $a$ is $B = \frac{\mu_0 I}{4\pi a}$.
In this configuration, the segments $AB$ and $BC$ create a magnetic field at '$O$' directed into the page. Similarly, the segments $TE$ and the other horizontal segment create a magnetic field at '$O$' also directed into the page.
By symmetry and applying the right-hand rule, the contributions from the four segments add up. Specifically, each segment acts as a semi-infinite wire relative to point '$O$'.
The total magnetic field $B_{total} = 4 \times (\frac{\mu_0 I}{4\pi a}) = \frac{\mu_0 I}{\pi a}$.

(B) The magnetic field $B$ at a distance $r$ from a long straight current-carrying wire is given by the formula:
$B = \frac{\mu_0 I}{2\pi r}$
From this formula, it is clear that the magnetic field is inversely proportional to the distance $r$, i.e.,$B \propto \frac{1}{r}$.
Given that at distance $r$, $B_1 = 0.4 \text{ tesla}$.
When the distance is doubled, $r_2 = 2r$.
The new magnetic field $B_2$ will be:
$B_2 = \frac{\mu_0 I}{2\pi (2r)} = \frac{1}{2} \times \left( \frac{\mu_0 I}{2\pi r} \right) = \frac{1}{2} B_1$
$B_2 = \frac{0.4}{2} = 0.2 \text{ tesla}$.

(C) The magnetic field on the axis of a circular coil of radius $R$ at a distance $x$ from its center is given by $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
For coil $X$: Radius $R_x = r$, distance $x_x = d/2$. Since $d \gg r$, $B_x \approx \frac{\mu_0 I r^2}{2(d/2)^3} = \frac{\mu_0 I r^2}{2(d^3/8)} = \frac{4 \mu_0 I r^2}{d^3}$.
For coil $Y$: Radius $R_y = 2r$, distance $x_y = d$. Since $d \gg r$, $B_y \approx \frac{\mu_0 I (2r)^2}{2d^3} = \frac{4 \mu_0 I r^2}{2d^3} = \frac{2 \mu_0 I r^2}{d^3}$.
Comparing $B_x$ and $B_y$, we get $B_x = 2 B_y$.

(B) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B_2 = \frac{\mu_0 I}{2R}$.
The magnetic field at an axial point $P$ at a distance $x$ from the centre is given by $B_1 = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
Given the condition $B_2 = 8B_1$, we substitute the expressions:
$\frac{\mu_0 I}{2R} = 8 \times \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
Simplifying the equation:
$\frac{1}{R} = \frac{8R^2}{(R^2 + x^2)^{3/2}}$.
$(R^2 + x^2)^{3/2} = 8R^3$.
Taking the cube root of both sides:
$(R^2 + x^2)^{1/2} = 2R$.
Squaring both sides:
$R^2 + x^2 = 4R^2$.
$x^2 = 3R^2$.
$x = \sqrt{3} R$.

(B) The magnetic field $\vec{B}$ produced by a moving charge $q$ is given by the expression $\vec{B} = \frac{\mu_0}{4\pi} \frac{q(\vec{v} \times \vec{r})}{r^3}$.
Since $\vec{B}$ is proportional to the cross product of $\vec{v}$ and $\vec{r}$ (i.e., $\vec{B} \propto \vec{v} \times \vec{r}$), the vector $\vec{B}$ must be perpendicular to both $\vec{v}$ and $\vec{r}$ by the definition of the cross product.
Therefore, $\vec{B}$ is always perpendicular to $\vec{v}$.

(A) The magnetic field on the axis of a circular loop of radius $R$ at a distance $d$ from its center is given by $B = \frac{\mu_0 I R^2}{2(R^2 + d^2)^{3/2}}$.
Given $d = \sqrt{3}R$, we have $B = \frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(4R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(8R^3)} = \frac{\mu_0 I}{16R}$.
Let the first loop be in the $YZ$-plane with its axis along the $X$-axis. Its magnetic field at $P$ is $\vec{B}_1 = \frac{\mu_0 I}{16R} \hat{i}$.
The second loop's axis is inclined at $45^\circ$ to the $X$-axis in the $XY$-plane. Its magnetic field vector $\vec{B}_2$ will be at $45^\circ$ to the $X$-axis: $\vec{B}_2 = \frac{\mu_0 I}{16R} (\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = \frac{\mu_0 I}{16R} (\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})$.
The resultant field is $\vec{B} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0 I}{16R} [(1 + \frac{1}{\sqrt{2}})\hat{i} + \frac{1}{\sqrt{2}}\hat{j}] = \frac{\mu_0 I}{16R} [(\frac{\sqrt{2} + 1}{\sqrt{2}})\hat{i} + \frac{1}{\sqrt{2}}\hat{j}] = \frac{\mu_0 I}{16\sqrt{2}R} [(\sqrt{2} + 1)\hat{i} + \hat{j}]$.