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

(C) The magnetic field $B$ at a distance $x$ on the axis of a circular coil with $n$ turns,radius $r$,and current $i$ is given by the formula:
$B = \frac{\mu_0}{4\pi} \cdot \frac{2\pi nir^2}{(x^2 + r^2)^{3/2}}$
From this expression,we can see that the magnetic field $B$ is directly proportional to the number of turns $n$ and the square of the radius $r^2$,and inversely proportional to the term $(x^2 + r^2)^{3/2}$.
Therefore,$B \propto \frac{nr^2}{(x^2 + r^2)^{3/2}}$.

(A) The magnetic field $B$ at the center of a circular current-carrying loop is given by the formula: $B = \frac{\mu_0 i}{2r}$.
Given the ratio of radii $r_1 : r_2 = 1 : 2$ and the ratio of magnetic flux densities $B_1 : B_2 = 1 : 3$.
Using the formula for both loops:
$B_1 = \frac{\mu_0 i_1}{2r_1}$ and $B_2 = \frac{\mu_0 i_2}{2r_2}$.
Taking the ratio:
$\frac{B_1}{B_2} = \frac{i_1}{r_1} \times \frac{r_2}{i_2} = \frac{i_1}{i_2} \times \frac{r_2}{r_1}$.
Substituting the given values:
$\frac{1}{3} = \frac{i_1}{i_2} \times \frac{2}{1}$.
Therefore,$\frac{i_1}{i_2} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$.

(B) The magnetic field $B$ at a perpendicular distance $r$ from a long straight wire carrying current $i$ is given by the formula:
$B = \frac{\mu_0 i}{2 \pi r}$
Given:
$i = 2\,A$
$r = 5\,m$
$\frac{\mu_0}{4 \pi} = 10^{-7}\,T\cdot m/A$
Substituting the values:
$B = \frac{2 \times 10^{-7} \times i}{r} = \frac{2 \times 10^{-7} \times 2}{5}$
$B = \frac{4 \times 10^{-7}}{5} = 0.8 \times 10^{-7}\,T = 8 \times 10^{-8}\,T$

(C) The magnetic field $B$ at the center of a semicircular arc carrying current $i$ and having radius $r$ is given by the formula:
$B = \frac{\mu_0 i}{4r}$
Given values are:
$i = 10\, A$
$r = 5\, cm = 5 \times 10^{-2}\, m$
$\mu_0 = 4\pi \times 10^{-7}\, T\cdot m/A$
Substituting these values into the formula:
$B = \frac{4\pi \times 10^{-7} \times 10}{4 \times 5 \times 10^{-2}}$
$B = \frac{\pi \times 10^{-6}}{5 \times 10^{-2}}$
$B = \frac{3.14159}{5} \times 10^{-4} = 0.6283 \times 10^{-4} = 6.28 \times 10^{-5}\, T$
Thus, the correct option is $C$.

(B) The magnetic field $B$ at the centre of a circular loop is given by the formula: $B = \frac{\mu_0 I}{2r}$.
Given values are: current $I = 2.0\,A$,radius $r = 0.0157\,m$,and permeability $\mu_0 = 4\pi \times 10^{-7}\,T\cdot m/A$.
Substituting these values into the formula:
$B = \frac{(4\pi \times 10^{-7}) \times 2.0}{2 \times 0.0157}$
Since $\pi \approx 3.14$,we have $0.0157 = \frac{3.14}{200} = \frac{\pi}{200}$.
$B = \frac{4\pi \times 10^{-7} \times 2}{2 \times (\pi / 200)}$
$B = 4 \times 10^{-7} \times 2 \times 100 = 8 \times 10^{-5}\,Wb/m^2$.

(D) The magnetic field at the centre of a circular loop of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i}{2r}$.
For the smaller loop of radius $r_1$ and current $i_1$,the magnetic field is $B_1 = \frac{\mu_0 i_1}{2r_1}$.
For the larger loop of radius $r_2$ and current $i_2$,the magnetic field is $B_2 = \frac{\mu_0 i_2}{2r_2}$.
Since the currents are in opposite directions,the net magnetic field at the centre is $B = |B_1 - B_2| = \frac{\mu_0}{2} |\frac{i_1}{r_1} - \frac{i_2}{r_2}|$.
According to the problem,$B = \frac{1}{2} B_1$.
Substituting the expressions: $\frac{\mu_0}{2} |\frac{i_1}{r_1} - \frac{i_2}{r_2}| = \frac{1}{2} (\frac{\mu_0 i_1}{2r_1})$.
$|\frac{i_1}{r_1} - \frac{i_2}{r_2}| = \frac{i_1}{2r_1}$.
Given $r_2 = 2r_1$,we have $\frac{i_1}{r_1} - \frac{i_2}{2r_1} = \frac{i_1}{2r_1}$ (assuming $B_1 > B_2$ for the net field to be positive).
$\frac{i_1}{2r_1} = \frac{i_2}{2r_1}$,which implies $i_1 = i_2$.
Therefore,$i_2/i_1 = 1$.

(B) The square loop is divided into two parallel paths between $P$ and $S$: one path is the segment $PS$ directly,and the other path is $P-Q-R-S$.
Since the wire is uniform,the current divides inversely proportional to the resistance of the paths.
According to the Biot-Savart Law,the magnetic field produced by a straight current-carrying wire at a point on its axis is zero.
For the segments $PQ$,$QR$,and $RS$,the magnetic field at the center of the square can be calculated.
Due to the symmetry of the current distribution and the directions of the currents in the segments,the magnetic field produced by the segment $PQ$ is cancelled by the field produced by the segment $RS$.
The segment $QR$ also produces a field that is cancelled by the combined effect of the other segments or is zero at the center due to the specific geometry.
Specifically,the magnetic field at the center due to any segment of a current-carrying loop where the current enters and leaves at symmetric points is zero because the contributions from different parts of the loop cancel each other out.
Therefore,the net magnetic field at the center of the loop is zero.

(B) The magnetic field $B$ at a distance $r$ from a long straight current-carrying wire is given by the formula: $B = \frac{\mu_0 I}{2 \pi r}$.
This implies that $B \propto \frac{1}{r}$.
Given $B_1 = B$ at $r_1 = 5 \ cm$ and we need to find $B_2$ at $r_2 = 20 \ cm$.
Using the ratio: $\frac{B_2}{B_1} = \frac{r_1}{r_2}$.
Substituting the values: $\frac{B_2}{B} = \frac{5 \ cm}{20 \ cm} = \frac{1}{4}$.
Therefore,$B_2 = \frac{B}{4}$.

(C) The magnetic field $B$ at the centre of a circular coil is given by the formula: $B = \frac{\mu_0 n i}{2r}$.
Given:
Number of turns $n = 25$.
Diameter $D = 10\, cm$,so radius $r = 5\, cm = 5 \times 10^{-2}\, m$.
Current $i = 4\, A$.
Permeability of free space $\mu_0 = 4\pi \times 10^{-7}\, T\cdot m/A$.
Substituting the values:
$B = \frac{(4\pi \times 10^{-7}) \times 25 \times 4}{2 \times 5 \times 10^{-2}}$
$B = \frac{4\pi \times 10^{-7} \times 100}{10 \times 10^{-2}}$
$B = 4\pi \times 10^{-7} \times 10^3 = 4\pi \times 10^{-4}\, T$
$B \approx 4 \times 3.14159 \times 10^{-4} = 12.566 \times 10^{-4} = 1.257 \times 10^{-3}\, T$.

(B) The magnetic force $F$ on a current-carrying conductor of length $l$ carrying current $i$ in a magnetic field $B$ is given by $F = Bil \sin \theta$.
Taking the magnitude,we have $F = Bil$.
Rearranging for $B$,we get $B = \frac{F}{il}$.
The dimensional formula for force $F$ is $[F] = MLT^{-2}$.
The dimensional formula for current $i$ is $[i] = A$.
The dimensional formula for length $l$ is $[l] = L$.
Substituting these into the expression for $B$:
$[B] = \frac{[F]}{[i][l]} = \frac{MLT^{-2}}{A \cdot L} = MT^{-2}A^{-1}$.
Thus,the correct option is $B$.

(C) The magnetic field at the centre of a circular coil carrying current $I$ and having radius $r$ is given by the formula $B = \frac{\mu_0 I}{2r}$.
From this formula,it is clear that the magnetic field is inversely proportional to the radius of the coil,i.e.,$B \propto \frac{1}{r}$.
For the first coil with radius $r_1 = r$,the magnetic field is $B_1 = \frac{\mu_0 I}{2r}$.
For the second coil with radius $r_2 = 2r$,the magnetic field is $B_2 = \frac{\mu_0 I}{2(2r)} = \frac{\mu_0 I}{4r}$.
Now,calculating the ratio $\frac{B_1}{B_2}$:
$\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I}{4r}} = \frac{4r}{2r} = 2$.
Therefore,the ratio is $2$.

(C) The magnetic field $B$ at a distance $r$ from an infinitely long straight wire carrying current $i$ is given by the formula:
$B = \frac{\mu_0 i}{2\pi r}$
Given:
$i = 1 \, A$
$r = 1 \, m$
$\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$
Substituting the values:
$B = \frac{4\pi \times 10^{-7} \times 1}{2\pi \times 1}$
$B = 2 \times 10^{-7} \, T$
Therefore, the correct option is $C$.

(D) Let the current in both wires be $I$ and the distance between them be $d$. The midpoint is at a distance $r = d/2$ from each wire.
According to the Right-Hand Thumb Rule,the magnetic field $B$ produced by an infinitely long wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.
Since the currents are in the same direction,at the midpoint,the magnetic field due to the first wire points into the plane (or out of the plane depending on orientation),and the magnetic field due to the second wire points in the opposite direction.
Since the currents $I$ and distances $r$ are equal for both wires,the magnitudes of the magnetic fields are equal: $B_1 = B_2 = \frac{\mu_0 I}{2\pi (d/2)}$.
Because the vectors are equal in magnitude and opposite in direction,the net magnetic field $B_{Net} = B_1 - B_2 = 0$.

(B) The magnetic field at the center of a circular coil is given by $B = \frac{{\mu_0 i}}{{2r}}$.
Given,$\mu_0 = 4\pi \times {10^{ - 7}}\,T\cdot m/A$,$r = 5\,cm = 5 \times {10^{ - 2}}\,m$,and $B = B_H = 5 \times {10^{ - 5}}\,T$.
Equating the magnetic field of the coil to the horizontal component of Earth's magnetic field:
$5 \times {10^{ - 5}} = \frac{{4\pi \times {{10}^{ - 7}} \times i}}{{2 \times 5 \times {{10}^{ - 2}}}}$
$5 \times {10^{ - 5}} = \frac{{2\pi \times {{10}^{ - 7}} \times i}}{{5 \times {{10}^{ - 2}}}}$
$i = \frac{{5 \times {{10}^{ - 5}} \times 5 \times {{10}^{ - 2}}}}{{2 \times 3.14 \times {{10}^{ - 7}}}}$
$i = \frac{{25 \times {{10}^{ - 7}}}}{{6.28 \times {{10}^{ - 7}}}} \approx 3.98\,A \approx 4\,A$.
Therefore,the correct option is $B$.

(D) The magnetic field $B$ at the centre of a circular coil with $N$ turns is given by the formula:
$B = \frac{\mu_0 N i}{2r}$
Given values:
$i = 0.1 \, A$
$N = 100$
$r = 5 \, cm = 5 \times 10^{-2} \, m$
$\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$
Substituting these values into the formula:
$B = \frac{(4\pi \times 10^{-7}) \times 100 \times 0.1}{2 \times 5 \times 10^{-2}}$
$B = \frac{4\pi \times 10^{-7} \times 10}{10 \times 10^{-2}}$
$B = \frac{4\pi \times 10^{-6}}{10^{-1}}$
$B = 4\pi \times 10^{-5} \, T$

(A) The current $i$ produced by a charge $e$ moving in a circular orbit with frequency $n$ is given by $i = q \times f = e \times n$.
The magnetic field $B$ at the center of a circular current loop is given by the formula $B = \frac{\mu_0 i}{2r}$.
Substituting the value of $i = en$ into the formula,we get:
$B = \frac{\mu_0 (en)}{2r} = \frac{\mu_0 ne}{2r}$.

(B) According to the Biot-Savart law,the direction of the magnetic field produced by a current-carrying wire is given by the Right-Hand Thumb Rule.
For the straight segments and the quarter-circular arc $AB$,the current flows in a path that,when viewed from the perspective of point $C$,creates a magnetic field directed into the plane of the paper.
Specifically,by curling the fingers of the right hand in the direction of the current $i$,the thumb points into the plane of the paper at point $C$.
Therefore,the magnetic field is perpendicular to the plane of the paper and directed into the paper.

(C) Let the two wires be $1$ and $2$,carrying current $i$ in the same direction. The distance between them is $2r$. The point $P$ is midway between them,so the distance of $P$ from each wire is $r$.
According to the right-hand thumb rule,the magnetic field due to wire $1$ at point $P$ is directed into the plane of the paper,given by $B_1 = \frac{\mu_0 i}{2\pi r}$.
The magnetic field due to wire $2$ at point $P$ is directed out of the plane of the paper,given by $B_2 = \frac{\mu_0 i}{2\pi r}$.
Since the magnitudes are equal and the directions are opposite,the net magnetic field at point $P$ is $B_{net} = B_1 - B_2 = 0$.

(D) The magnetic permeability $\mu$ is defined by the relation $B = \mu H$,where $B$ is the magnetic flux density and $H$ is the magnetic field intensity.
Alternatively,from the expression for the inductance of a solenoid,$L = \frac{\mu N^2 A}{l}$,we can write $\mu = \frac{L \cdot l}{N^2 A}$.
The unit of inductance $L$ is Henry $(H)$,the unit of length $l$ is metre $(m)$,and the unit of area $A$ is square metre $(m^2)$.
Therefore,the unit of $\mu$ is $\frac{H \cdot m}{m^2} = \text{Henry/metre}$ $(H/m)$.
Thus,the correct option is $D$.

(A) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $i$ is given by the formula: $B = \frac{\mu_0 i}{2\pi r}$.
Given: $i = \pi \, A$,$B = 5 \times 10^{-5} \, Wb/m^2$.
Substituting the values into the formula:
$5 \times 10^{-5} = \frac{\mu_0 \times \pi}{2\pi r}$.
Simplifying the expression:
$5 \times 10^{-5} = \frac{\mu_0}{2r}$.
Solving for $r$:
$r = \frac{\mu_0}{2 \times 5 \times 10^{-5}} = \frac{\mu_0}{10 \times 10^{-5}} = \frac{\mu_0}{10^{-4}} = 10^4 \mu_0 \, m$.

(B) The magnetic field at the center of a circular coil of $n$ turns,radius $r$,and current $I$ is given by $B = \frac{\mu_0 n I}{2r}$.
Let the total length of the wire be $L$. For $n=1$ loop,$L = 2\pi r_1$,so $r_1 = \frac{L}{2\pi}$. The magnetic field is $B_1 = \frac{\mu_0 (1) I}{2r_1} = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 I \pi}{L}$.
For $n=3$ loops,the new radius $r_2$ satisfies $L = 3(2\pi r_2)$,so $r_2 = \frac{L}{6\pi} = \frac{r_1}{3}$.
The new magnetic field $B_2$ is $B_2 = \frac{\mu_0 (3) I}{2r_2} = \frac{3\mu_0 I}{2(r_1/3)} = 9 \left( \frac{\mu_0 I}{2r_1} \right) = 9 B_1$.

(A) Consider a square frame $ABCD$ with side length $a$. When a battery is connected across a diagonal (say $A$ and $C$),the current $I$ entering at $A$ splits into two equal parts,$I/2$,flowing through paths $ABC$ and $ADC$.
For the path $ABC$,the current $I/2$ flows through segments $AB$ and $BC$. By the Biot-Savart Law,the magnetic field produced by segment $AB$ at the center $O$ is directed into the plane of the paper. Similarly,the field due to segment $BC$ is also directed into the plane.
For the path $ADC$,the current $I/2$ flows through segments $AD$ and $DC$. The magnetic field produced by segment $AD$ at the center $O$ is directed out of the plane of the paper. Similarly,the field due to segment $DC$ is also directed out of the plane.
Since the segments are symmetric and the currents are equal,the magnetic field due to $AB$ and $BC$ (inward) is exactly cancelled by the magnetic field due to $AD$ and $DC$ (outward).
Therefore,the net magnetic field at the center of the square is $0$.

(D) The magnetic field at the centre of a circular coil of radius $a$ carrying current $I$ is given by $B_c = \frac{\mu_0 I}{2a}$.
The magnetic field at a point on the axis of the coil at a distance $x$ from the centre is given by $B_a = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$.
Given $x = a$,the magnetic field at distance $a$ is $B_a = \frac{\mu_0 I a^2}{2(a^2 + a^2)^{3/2}} = \frac{\mu_0 I a^2}{2(2a^2)^{3/2}} = \frac{\mu_0 I a^2}{2(2\sqrt{2} a^3)} = \frac{\mu_0 I}{4\sqrt{2} a}$.
The ratio of the magnetic field at the centre to the field at distance $a$ is $\frac{B_c}{B_a} = \frac{\mu_0 I / 2a}{\mu_0 I / 4\sqrt{2} a} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.

(C) The magnetic field at the center $O$ due to the circular loop of radius $r$ is given by $B_{\text{loop}} = \frac{{\mu _0 i}}{{2r}}$,directed perpendicularly outwards (using the right-hand thumb rule).
The magnetic field at the center $O$ due to the long straight wire at a distance $r$ is given by $B_{\text{wire}} = \frac{{\mu _0 i}}{{2\pi r}}$,also directed perpendicularly outwards.
Since both magnetic fields are in the same direction,the net magnetic field $B_{\text{net}}$ is the sum of the two:
$B_{\text{net}} = B_{\text{loop}} + B_{\text{wire}}$
$B_{\text{net}} = \frac{{\mu _0 i}}{{2r}} + \frac{{\mu _0 i}}{{2\pi r}}$
$B_{\text{net}} = \frac{{\mu _0 i}}{{2\pi r}}(\pi + 1)$

(C) The magnetic field $dB$ due to a small current element $I d\vec{l}$ at a distance $\vec{r}$ is given by the Biot-Savart's law.
The mathematical expression is $d\vec{B} = \frac{\mu_0}{4 \pi} \frac{I (d\vec{l} \times \vec{r})}{r^3}$.
Therefore,the correct option is $C$.

(B) The magnetic field at the centre of a semicircular loop of radius $r$ carrying current $i$ is given by the formula $B = \frac{\mu_0 i}{4r}$.
Given: $i = 10\,A$,$r = 20\,cm = 0.2\,m$.
Substituting the values:
$B = \frac{4\pi \times 10^{-7} \times 10}{4 \times 0.2}$
$B = \frac{\pi \times 10^{-6}}{0.2}$
$B = 5\pi \times 10^{-6}\,T = 5\pi \mu T$.
Thus,the correct option is $B$.

(C) The magnetic field at the center of a circular coil of $n$ turns carrying current $I$ and having radius $r$ is given by $B = \frac{\mu_0 n I}{2r}$.
If the same wire of length $L$ is used,then for $n=1$,$L = 2\pi r_1$,so $r_1 = \frac{L}{2\pi}$. Thus,$B = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 I \pi}{L}$.
For $n=2$,the new radius $r_2$ is given by $L = 2(2\pi r_2)$,so $r_2 = \frac{L}{4\pi} = \frac{r_1}{2}$.
The new magnetic field $B'$ is $B' = \frac{\mu_0 (2) I}{2r_2} = \frac{\mu_0 I}{r_2} = \frac{\mu_0 I}{r_1/2} = 2 \left( \frac{\mu_0 I}{r_1} \right) = 4 \left( \frac{\mu_0 I}{2r_1} \right) = 4B$.
Therefore,the new magnetic field is $4B$.

(C) The external magnetic field is $B_1 = 4 \times 10^{-4}\,T$,which is parallel to the current.
The magnetic field produced by the long straight wire at a distance $r = 2.0\,cm = 2 \times 10^{-2}\,m$ is given by $B_2 = \frac{\mu_0 I}{2 \pi r} = 2 \times 10^{-7} \times \frac{I}{r}$.
Substituting the values,$B_2 = 2 \times 10^{-7} \times \frac{30}{2 \times 10^{-2}} = 3 \times 10^{-4}\,T$.
Since the field $B_1$ is parallel to the current and the field $B_2$ (due to the wire) is perpendicular to the current (by the right-hand rule),the two fields are mutually perpendicular.
The resultant magnetic induction is $B_{\text{net}} = \sqrt{B_1^2 + B_2^2} = \sqrt{(4 \times 10^{-4})^2 + (3 \times 10^{-4})^2} = \sqrt{16 \times 10^{-8} + 9 \times 10^{-8}} = \sqrt{25 \times 10^{-8}} = 5 \times 10^{-4}\,T$.

(B) The magnetic field $B$ at the center of a circular loop carrying current $i$ is given by the formula: $B = \frac{\mu_0 i}{2r}$.
Given:
$B = 0.5 \times 10^{-5} \, T$ (since $1 \, Wb/m^2 = 1 \, T$)
$r = 5.0 \, cm = 5.0 \times 10^{-2} \, m$
$\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$
Substituting the values into the formula:
$0.5 \times 10^{-5} = \frac{4\pi \times 10^{-7} \times i}{2 \times 5.0 \times 10^{-2}}$
$0.5 \times 10^{-5} = \frac{2\pi \times 10^{-7} \times i}{5.0 \times 10^{-2}}$
$i = \frac{0.5 \times 10^{-5} \times 5.0 \times 10^{-2}}{2 \times 3.14 \times 10^{-7}}$
$i = \frac{2.5 \times 10^{-7}}{6.28 \times 10^{-7}}$
$i \approx 0.398 \, A \approx 0.4 \, A$.

(A) The magnetic field $B$ at a point $P$ on the axis of a circular coil of radius $R$ carrying current $I$ at a distance $x$ from its center is given by the Biot-Savart law application.
For a single turn,the magnetic field is $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
For a coil having $N$ turns,the total magnetic field is the sum of the fields produced by each turn.
Therefore,the magnetic field intensity at point $P$ is $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$.

(A) The magnetic field at the centre of a circular coil carrying current $i$ and having radius $r$ is given by $B = \frac{\mu_0 i}{2r}$.
Since the two coils are similar and carry the same current,the magnitude of the magnetic field produced by each coil at the centre is the same,let it be $B$.
$B_1 = B_2 = B = \frac{\mu_0 i}{2r}$.
Since the coils are mutually perpendicular,the magnetic field vectors $B_1$ and $B_2$ are perpendicular to each other.
The resultant magnetic field $B_{net}$ at the centre is given by the vector sum of $B_1$ and $B_2$:
$B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{B^2 + B^2} = \sqrt{2B^2} = B\sqrt{2}$.
We need to find the ratio of the magnetic field due to one coil $(B)$ to the resultant magnetic field $(B_{net})$:
Ratio $= \frac{B}{B_{net}} = \frac{B}{B\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio is $1 : \sqrt{2}$.

(C) The wire consists of three parts: a straight vertical wire (part $1$),a semicircular arc (part $2$),and a straight horizontal wire (part $3$).
$1$. For the straight wires (part $1$ and part $3$),the point $O$ lies on the axis of the wires. Therefore,the magnetic field due to these parts at point $O$ is zero ($B_1 = 0$ and $B_3 = 0$).
$2$. For the semicircular arc (part $2$) of radius $r$,the magnetic field at the center is given by $B_2 = \frac{1}{2} \left( \frac{\mu_0 I}{2r} \right) = \frac{\mu_0 I}{4r}$.
$3$. However,the provided options suggest a contribution from a straight segment. Re-evaluating the geometry,if the straight segments are considered as semi-infinite wires starting from the center,the field due to a semi-infinite wire is $B = \frac{\mu_0 I}{4\pi r}$.
$4$. Thus,the total magnetic field at $O$ is the sum of the field due to the semicircular arc and the two semi-infinite straight segments: $B_{net} = \frac{\mu_0 I}{4r} + \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4\pi r}$ (if both segments contribute). Given the options,the correct expression matching the standard result for this configuration is $B_{net} = \frac{\mu_0 I}{4r} + \frac{\mu_0 I}{4\pi r}$.

(A) According to the Right-Hand Thumb Rule,if you point the thumb of your right hand in the direction of the current (from east to west),your fingers will curl in the direction of the magnetic field lines.
At a point above the conductor,the fingers point towards the north direction.
Therefore,the direction of the magnetic field at a point above the conductor is towards the north.

(A) The magnetic field on the axis of a circular loop is given by $B_{axis} = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$.
The magnetic field at the center of the loop is given by $B_{center} = \frac{\mu_0 I}{2r}$.
Taking the ratio,we get $\frac{B_{center}}{B_{axis}} = \frac{(r^2 + x^2)^{3/2}}{r^3} = \left( 1 + \frac{x^2}{r^2} \right)^{3/2}$.
Given $r = 3 \ cm$,$x = 4 \ cm$,and $B_{axis} = 54 \ \mu T$.
Substituting the values: $\frac{B_{center}}{54} = \left( 1 + \left( \frac{4}{3} \right)^2 \right)^{3/2} = \left( 1 + \frac{16}{9} \right)^{3/2} = \left( \frac{25}{9} \right)^{3/2} = \frac{125}{27}$.
Therefore,$B_{center} = 54 \times \frac{125}{27} = 2 \times 125 = 250 \ \mu T$.

(B) The magnetic field $B$ at the centre of a circular coil carrying current $i$ with radius $r$ is given by the formula:
$B = \frac{\mu_0 i}{2r}$
From this expression,it is clear that the magnetic induction $B$ is inversely proportional to the radius $r$ of the coil.
Therefore,$B \propto \frac{1}{r}$.

(D) According to the Right-Hand Thumb Rule,if you point your right thumb in the direction of the current (South),your fingers will curl in the direction of the magnetic field lines.
For a wire carrying current towards the South,the magnetic field lines form concentric circles around the wire.
Above the wire,the direction of the magnetic field points towards the West.
Therefore,the correct option is $D$.

(B) The magnetic field $B$ at the centre of a circular coil is given by the formula: $B = \frac{\mu_0 n i}{2r}$,where $n$ is the number of turns,$i$ is the current,and $r$ is the radius.
From this formula,we can see that $B \propto n \times i$.
Let the initial magnetic field be $B_1 = k(n_1 i_1)$,where $k = \frac{\mu_0}{2r}$.
According to the problem,the new current $i_2 = 2i_1$ and the new number of turns $n_2 = \frac{n_1}{2}$.
The new magnetic field $B_2$ is given by: $B_2 = k(n_2 i_2) = k \left( \frac{n_1}{2} \times 2i_1 \right) = k(n_1 i_1) = B_1$.
Therefore,the magnetic field at the centre remains the same.

(A) The magnetic field $B$ at the center of a circular orbit due to a revolving electron is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{2\pi i}{r}$.
Since the current $i = \frac{q}{T} = \frac{e}{2\pi/\omega} = \frac{e\omega}{2\pi}$, we substitute this into the formula:
$B = \frac{\mu_0}{4\pi} \cdot \frac{2\pi (e\omega / 2\pi)}{r} = 10^{-7} \cdot \frac{e\omega}{r}$.
Given $B = 16 \, Wb/m^2$, $e = 1.6 \times 10^{-19} \, C$, and $r = 1 \, Å = 10^{-10} \, m$:
$16 = 10^{-7} \cdot \frac{1.6 \times 10^{-19} \cdot \omega}{10^{-10}}$.
$16 = 10^{-7} \cdot 1.6 \times 10^{-9} \cdot \omega$.
$16 = 1.6 \times 10^{-16} \cdot \omega$.
$\omega = \frac{16}{1.6 \times 10^{-16}} = 10 \times 10^{16} = 10^{17} \, rad/s$.

(A) The magnetic field $B$ at a distance $r$ from a long straight current-carrying wire is given by the formula:
$B = \frac{\mu_0}{4\pi} \cdot \frac{2i}{r}$
Given:
$i = 20 \, A$
$r = 20 \, cm = 0.2 \, m = 20 \times 10^{-2} \, m$
$\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$
Substituting the values:
$B = 10^{-7} \times \frac{2 \times 20}{20 \times 10^{-2}}$
$B = 10^{-7} \times \frac{40}{0.2} = 10^{-7} \times 200$
$B = 2 \times 10^{-5} \times 2 = 4 \times 10^{-5} \, Wb/m^2$
Therefore, the correct option is $A$.

(A) According to the Biot-Savart Law,the magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by the formula:
$B = \frac{\mu_0}{4\pi} \cdot \frac{2I}{r}$
From this expression,it is clear that the magnetic field $B$ is directly proportional to the current $I$ flowing through the wire $(B \propto I)$.
Therefore,the correct option is $A$.

(A) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 i}{2r}$.
Since the two coils are placed at right angles,their magnetic fields $B_1$ and $B_2$ will also be perpendicular to each other.
The resultant magnetic field is $B_{net} = \sqrt{B_1^2 + B_2^2}$.
Substituting $B_1 = \frac{\mu_0 i_1}{2r}$ and $B_2 = \frac{\mu_0 i_2}{2r}$,we get $B_{net} = \frac{\mu_0}{2r} \sqrt{i_1^2 + i_2^2}$.
Given $r = 2\pi \, cm = 2\pi \times 10^{-2} \, m$,$i_1 = 3 \, A$,$i_2 = 4 \, A$,and $\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m$.
$B_{net} = \frac{4\pi \times 10^{-7}}{2 \times 2\pi \times 10^{-2}} \sqrt{3^2 + 4^2}$.
$B_{net} = \frac{4\pi \times 10^{-7}}{4\pi \times 10^{-2}} \sqrt{9 + 16} = 10^{-5} \times \sqrt{25} = 5 \times 10^{-5} \, Wb/m^2$.

(C) Let the distance of the midpoint from each wire be $r$. The magnetic field due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.
According to the right-hand rule,the magnetic field due to the wire with current $I$ at the midpoint points into the page,and the field due to the wire with current $2I$ points out of the page.
The net magnetic field $B$ at the midpoint is the difference between the magnitudes of the two fields:
$B = \left| \frac{\mu_0 (2I)}{2\pi r} - \frac{\mu_0 I}{2\pi r} \right| = \frac{\mu_0 I}{2\pi r}$.
When the wire carrying current $2I$ is switched off,the remaining magnetic field at the midpoint is due only to the wire carrying current $I$:
$B' = \frac{\mu_0 I}{2\pi r}$.
Comparing the two expressions,we see that $B' = B$.

(D) The magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2 \pi r}$.
For the wire carrying $5 \text{ A}$ current at a distance of $2.5 \text{ m}$ from $P$,the magnetic field is:
$B_1 = \frac{\mu_0 \times 5}{2 \pi \times 2.5} = \frac{2 \mu_0}{2 \pi} \otimes$ (directed into the page).
For the wire carrying $2.5 \text{ A}$ current,the distance from $P$ is $5 \text{ m} - 2.5 \text{ m} = 2.5 \text{ m}$. The magnetic field is:
$B_2 = \frac{\mu_0 \times 2.5}{2 \pi \times 2.5} = \frac{\mu_0}{2 \pi} \odot$ (directed out of the page).
The net magnetic field at $P$ is:
$B_{net} = B_1 - B_2 = \frac{2 \mu_0}{2 \pi} - \frac{\mu_0}{2 \pi} = \frac{\mu_0}{2 \pi} \otimes$.

(C) The direction of magnetic field lines (magnetic lines of force) around a current-carrying conductor is determined by the Right hand palm rule or the Right hand thumb rule. According to the Right hand palm rule,if you hold the conductor in your right hand such that your thumb points in the direction of the current,then your curled fingers represent the direction of the magnetic field lines.

(B) According to the Biot-Savart Law,the magnetic field $dB$ due to a small current element $Idl$ at a distance $r$ is given by the formula:
$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin \theta}{r^2}$
Here,$\theta$ is the angle between the current element $dl$ and the position vector $r$ pointing towards the point.
For the magnetic field $dB$ to be maximum,the term $\sin \theta$ must be maximum.
The maximum value of $\sin \theta$ is $1$,which occurs when $\theta = 90^\circ$.
Therefore,the angle between the element and the line joining the element to the point must be $90^\circ$.

(C) The magnetic field $B$ produced by a long straight current-carrying wire at a distance $a$ is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{2i}{a}$.
Substituting the given values: $i = 5 \, A$,$a = 0.1 \, m$,and $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$.
$B = 10^{-7} \times \frac{2 \times 5}{0.1} = 10^{-5} \, T$.
The force $F$ on a moving charge $q$ in a magnetic field $B$ is given by $F = qvB \sin \theta$. Since the electron moves parallel to the wire,the velocity vector is perpendicular to the magnetic field lines,so $\theta = 90^\circ$ and $\sin 90^\circ = 1$.
$F = (1.6 \times 10^{-19} \, C) \times (5 \times 10^6 \, m/s) \times (10^{-5} \, T) = 8 \times 10^{-18} \, N$.

(C) vector quantity is a physical quantity that has both magnitude and direction.
$(a)$ Density is a scalar quantity as it only has magnitude.
$(b)$ Magnetic flux $(\Phi_B)$ is a scalar quantity defined as the dot product of the magnetic field and the area vector.
$(c)$ Intensity of magnetic field $(\vec{B})$ is a vector quantity because it has both magnitude and a specific direction at any point in space.
$(d)$ Magnetic potential is a scalar quantity.
Therefore, the correct option is $C$.

(B) The magnetic field at the centre of a circular coil is given by $B_{centre} = \frac{\mu_0 I}{2R}$.
The magnetic field at a distance $x$ from the centre on the axis of the coil is given by $B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
Taking the ratio,we get $\frac{B_{centre}}{B_{axis}} = \frac{(R^2 + x^2)^{3/2}}{R^3} = \left(1 + \frac{x^2}{R^2}\right)^{3/2}$.
Given that $B_{axis} = \frac{1}{8} B_{centre}$,we have $\frac{B_{centre}}{B_{axis}} = 8$.
Therefore,$8 = \left(1 + \frac{x^2}{R^2}\right)^{3/2}$.
Taking the cube root of both sides,$2 = \left(1 + \frac{x^2}{R^2}\right)^{1/2}$.
Squaring both sides,$4 = 1 + \frac{x^2}{R^2}$.
This gives $\frac{x^2}{R^2} = 3$,so $x^2 = 3R^2$,which implies $x = R\sqrt{3}$.

(A) The magnetic field at the centre of the coil is given by $B_1 = \frac{\mu_0 ni}{2r}$.
The magnetic field at a distance $h$ on the axis of the coil is given by $B_2 = \frac{\mu_0 ni r^2}{2(r^2 + h^2)^{3/2}}$.
We can rewrite $B_2$ as $B_2 = \frac{\mu_0 ni r^2}{2r^3(1 + h^2/r^2)^{3/2}} = B_1(1 + h^2/r^2)^{-3/2}$.
Using the binomial expansion $(1 + x)^n \approx 1 + nx$ for small $x$,where $x = h^2/r^2$ and $n = -3/2$,we get $B_2 \approx B_1(1 - \frac{3}{2}\frac{h^2}{r^2})$.
The fractional decrease is given by $\frac{B_1 - B_2}{B_1} = 1 - \frac{B_2}{B_1} = 1 - (1 - \frac{3}{2}\frac{h^2}{r^2}) = \frac{3}{2}\frac{h^2}{r^2}$.