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

(B) The unit of magnetic field strength or magnetic induction $(B)$ in the $SI$ system is the Tesla $(T)$. One Tesla is defined as one Weber per square meter $(1 \ T = 1 \ Wb/m^2)$. Therefore,Tesla is the unit for measuring magnetic induction.

(C) The magnetic field $B$ produced by a current element is given by the Biot-Savart Law:
$B = \frac{\mu_0}{4\pi} \frac{I dl \sin(\theta)}{r^2}$
Rearranging for the permeability $\mu_0$:
$\mu_0 = \frac{B \cdot r^2}{I \cdot dl \cdot \sin(\theta)}$
Substituting the $SI$ units:
$B$ is in $Wb/m^2$ (Tesla),
$r$ is in $m$,
$I$ is in $A$,
$dl$ is in $m$.
Therefore,the unit of $\mu_0$ is:
$\text{Unit} = \frac{(Wb/m^2) \cdot m^2}{A \cdot m} = \frac{Wb}{A \cdot m} = Wb\;m^{-1}\;A^{-1}$

(C) The magnetic field $B$ at the centre of a circular coil of $N$ turns and radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2r}$.
Since the length of the wire $L$ is constant,$L = 2\pi r_1 N_1 = 2\pi r_2 N_2$.
For the first case,$N_1 = 1$,so $L = 2\pi r_1$,which implies $r_1 = \frac{L}{2\pi}$.
For the second case,$N_2 = 2$,so $L = 2\pi r_2 \times 2$,which implies $r_2 = \frac{L}{4\pi} = \frac{r_1}{2}$.
The magnetic field in the first case is $B_1 = \frac{\mu_0 (1) I}{2r_1}$.
The magnetic field in the second case is $B_2 = \frac{\mu_0 (2) I}{2r_2} = \frac{\mu_0 (2) I}{2(r_1/2)} = \frac{4 \mu_0 I}{2r_1} = 4 B_1$.
Thus,the magnetic field becomes four times its first value.

(B) According to the Biot-Savart Law,the magnitude of the magnetic field $B$ produced by a long straight current-carrying conductor at a distance $r$ is given by the formula $B = \frac{\mu_0 I}{2 \pi r}$.
Since both points $P$ and $Q$ are at the same distance $r$ from the conductor,the magnitude of the magnetic field at both points will be equal.
Therefore,the magnetic field at $P$ is the same as the magnetic field at $Q$.

(D) According to the Biot-Savart Law,the magnetic field $dB$ due to a current element $idl$ at a distance $r$ is given by the magnitude: $dB = \frac{\mu_0}{4\pi} \frac{idl \sin \theta}{r^2}$.
In vector form,this is expressed as $d\overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{i(d\overrightarrow{l} \times \widehat{r})}{r^2}$.
Since the unit vector $\widehat{r} = \frac{\overrightarrow{r}}{r}$,we substitute this into the equation:
$d\overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{i(d\overrightarrow{l} \times \overrightarrow{r})}{r^2 \cdot r} = \frac{\mu_0}{4\pi} \frac{i(d\overrightarrow{l} \times \overrightarrow{r})}{r^3}$.
Thus,option $D$ is correct.

(C) The current $i$ produced by a charge $q$ moving in a circle with frequency $n$ is given by $i = q \times n$.
The magnetic field $B$ at the centre of a circular loop is given by the formula $B = \frac{\mu_0 i}{2r}$.
Substituting $\frac{\mu_0}{4\pi} = 10^{-7} \text{ T} \cdot \text{m/A}$, we get $\frac{\mu_0}{2} = 2\pi \times 10^{-7}$.
Thus, $B = (2\pi \times 10^{-7}) \times \frac{i}{r}$.
Substituting $i = nq$, we get $B = \frac{2\pi nq}{r} \times 10^{-7} \text{ N/A} \cdot \text{m}$.

(B) The magnetic field at the center $O$ is due to two parts: the straight wire and the circular loop.
$1$. The magnetic field due to the circular loop of radius $r$ carrying current $i$ is $B_{loop} = \frac{\mu_0 i}{2r} = \frac{\mu_0}{4\pi} \frac{2\pi i}{r}$,directed into the plane of the paper (using the right-hand rule).
$2$. The magnetic field due to the straight wire at a distance $r$ from it is $B_{wire} = \frac{\mu_0 i}{2\pi r} = \frac{\mu_0}{4\pi} \frac{2i}{r}$,directed out of the plane of the paper.
$3$. Since the fields are in opposite directions,the net magnetic field $B$ at $O$ is $B = B_{loop} - B_{wire} = \frac{\mu_0}{4\pi} \frac{2\pi i}{r} - \frac{\mu_0}{4\pi} \frac{2i}{r} = \frac{\mu_0}{4\pi} \frac{2i}{r}(\pi - 1)$.

(D) The magnetic field $B$ at the center of a circular arc carrying current $i$ and subtending an angle $\theta$ (in radians) at the center is given by the formula:
$B = \frac{\mu_0 i \theta}{4\pi R}$
Here,the angle subtended by the arc is $\theta = 3\pi / 2$.
Substituting this value into the formula:
$B = \frac{\mu_0 i (3\pi / 2)}{4\pi R}$
$B = \frac{3\mu_0 i \pi}{8\pi R}$
$B = \frac{3\mu_0 i}{8R}$
Thus,the correct option is $(d)$.

(D) According to the Biot-Savart law,the magnetic field $d\vec{B}$ due to a current element $i d\vec{l}$ at a position vector $\vec{r}$ is given by $d\vec{B} = \frac{\mu_0}{4\pi} \frac{i (d\vec{l} \times \vec{r})}{r^3}$.
For any point lying on the axis of the current-carrying wire,the current element $d\vec{l}$ and the position vector $\vec{r}$ are collinear (i.e.,the angle between them is $0^\circ$ or $180^\circ$).
Since the cross product of two collinear vectors is zero,the magnetic field at any point on the axis of the wire is zero.
In this problem,the point $x = a$ lies on the $X$-axis,which is the same axis along which the wire $PQ$ is placed.
Therefore,the magnetic field at $x = a$ is zero.

(B) The charge of a helium nucleus is $q = 2e = 2 \times 1.6 \times 10^{-19} \ C = 3.2 \times 10^{-19} \ C$.
The time period of rotation is $T = 2 \ s$.
The equivalent current $i$ is given by $i = \frac{q}{T} = \frac{3.2 \times 10^{-19}}{2} = 1.6 \times 10^{-19} \ A$.
The magnetic field $B$ at the centre of a circular loop is $B = \frac{\mu_0 i}{2r}$.
Substituting the values,$B = \frac{\mu_0 \times 1.6 \times 10^{-19}}{2 \times 0.8} = \frac{\mu_0 \times 1.6 \times 10^{-19}}{1.6} = 10^{-19} \mu_0 \ T$.

(A) The magnetic field $B$ due to a long straight current-carrying wire at a distance $r$ is given by the formula: $B = \frac{\mu_0}{4\pi} \cdot \frac{2i}{r}$.
Since the current $i$ is constant,the magnetic field is inversely proportional to the distance $r$,i.e.,$B \propto \frac{1}{r}$.
Therefore,we can write the ratio: $\frac{B_1}{B_2} = \frac{r_2}{r_1}$.
Given: $B_1 = 10^{-8} \, T$,$r_1 = 4 \, cm$,and $r_2 = 12 \, cm$.
Substituting the values: $\frac{10^{-8}}{B_2} = \frac{12}{4}$.
$\frac{10^{-8}}{B_2} = 3$.
$B_2 = \frac{10^{-8}}{3} = 3.33 \times 10^{-9} \, T$.

(C) 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 expression,we can see that the magnetic field is inversely proportional to the distance from the wire,i.e.,$B \propto \frac{1}{r}$.
Let $B_1 = B$ at distance $r_1 = r$.
Let $B_2$ be the magnetic field at distance $r_2 = \frac{r}{2}$.
Using the proportionality $B_1 r_1 = B_2 r_2$,we get:
$B \cdot r = B_2 \cdot \frac{r}{2}$.
Solving for $B_2$,we find $B_2 = 2B$.

(C) The magnetic field $B$ at the centre of a circular coil of radius $r$ carrying a current $I$ is given by the formula:
$B = \frac{\mu_0 I}{2r}$
From this expression,it is clear that $B \propto I$ (directly proportional to the current) and $B \propto \frac{1}{r}$ (inversely proportional to the radius).
Therefore,the magnetic field is directly proportional to the current $I$ flowing through the coil.
Thus,the correct option is $C$.

(A) The magnetic field $B$ at the centre of a circular coil with $N$ turns is given by the formula:
$B = \frac{{\mu _0}Ni}{2r}$
Given:
Number of turns $N = 100$
Current $i = 0.1\, A$
Radius $r = 5\, cm = 5 \times 10^{-2}\, m$
Permeability $\mu _0 = 4\pi \times 10^{-7}\, T\cdot m/A$
Substituting the values into the formula:
$B = \frac{(4\pi \times 10^{-7}) \times 100 \times 0.1}{2 \times 5 \times 10^{-2}}$
$B = \frac{4\pi \times 10^{-7} \times 10}{10 \times 10^{-2}}$
$B = \frac{4\pi \times 10^{-6}}{10^{-1}}$
$B = 4\pi \times 10^{-5}\, T$
Therefore,the correct option is $A$.

(A) The magnetic field at the centre $O$ due to a semi-circular arc of radius $R$ carrying current $i$ is given by $B = \frac{{\mu _0}i}{{4R}}$.
In the given figure,the current flows through two semi-circular arcs of radii $R_1$ and $R_2$ and two straight radial segments.
The magnetic field due to the straight radial segments at the centre $O$ is zero because the position vector of the point $O$ is parallel to the current element $idl$ for these segments.
For the inner semi-circular arc of radius $R_1$,the magnetic field $B_1$ at $O$ is directed into the plane of the paper $(\otimes)$ and its magnitude is $B_1 = \frac{{\mu _0}i}{{4R_1}}$.
For the outer semi-circular arc of radius $R_2$,the magnetic field $B_2$ at $O$ is directed out of the plane of the paper $(\odot)$ and its magnitude is $B_2 = \frac{{\mu _0}i}{{4R_2}}$.
Since $R_1$ < $R_2$,the magnitude $B_1$ > $B_2$.
The net magnetic field $B_{net}$ is $B_1 - B_2$ directed into the plane of the paper.
$B_{net} = \frac{{\mu _0}i}{{4R_1}} - \frac{{\mu _0}i}{{4R_2}} = \frac{{\mu _0}i}{4}\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$.

(D) According to the Biot-Savart Law,the magnetic field $dB$ due to a current element $idl$ at a point is given by $dB = \frac{\mu_0}{4\pi} \frac{i(dl \times r)}{r^3}$.
For any point lying on the axis of a straight current-carrying wire,the angle $\theta$ between the current element $dl$ and the position vector $r$ is $0^\circ$ or $180^\circ$.
Since $\sin(0^\circ) = 0$ and $\sin(180^\circ) = 0$,the magnetic field contribution from any such element is zero.
In the given figure,the point $O$ lies on the straight line extending from the segment $AB$. Therefore,the magnetic induction at $O$ due to the current in portion $AB$ is zero.

(C) The conductor consists of three parts: two straight segments $AB$ and $CD$,and a semicircular arc $BC$ of radius $r$.
For the straight segments $AB$ and $CD$,the point $O$ lies on the axis of the current-carrying wires. According to the Biot-Savart law,the magnetic field $dB = \frac{\mu_0}{4\pi} \frac{i dl \sin \theta}{r^2}$. For these segments,the angle $\theta$ between the current element $dl$ and the position vector is $0^\circ$ or $180^\circ$,so $\sin \theta = 0$. Thus,the magnetic induction due to $AB$ and $CD$ is zero.
For the semicircular arc $BC$,the magnetic field at the center $O$ is given by $B = \frac{1}{2} \left( \frac{\mu_0 i}{2r} \right) = \frac{\mu_0 i}{4r}$.
Therefore,the total magnetic induction at $O$ is $\frac{\mu_0 i}{4r}$.

(C) The magnetic field at the center of a semicircular arc of radius $r$ carrying current $i$ is given by $B = \frac{{\mu _0 i}}{{4r}}$.
In the given figure,there are two semicircular arcs of radii $r_1$ and $r_2$. The current flows in both arcs such that the magnetic field produced by both at the center $O$ is directed into the plane of the paper (using the right-hand thumb rule).
Therefore,the total magnetic induction $B$ at the center $O$ is the sum of the magnetic fields due to both arcs:
$B = B_1 + B_2$
$B = \frac{{\mu _0 i}}{{4r_1}} + \frac{{\mu _0 i}}{{4r_2}}$
$B = \frac{{\mu _0 i}}{4} \left( \frac{1}{r_1} + \frac{1}{r_2} \right)$
$B = \frac{{\mu _0 i}}{4} \left( \frac{r_1 + r_2}{r_1 r_2} \right)$
Thus,the correct option is $C$.

(D) Let the two wires be placed parallel to each other in the horizontal plane, separated by a distance $d$. Let the current $I$ flow through both wires in the same direction.
At any point $P$ located exactly halfway between the wires, the magnetic field $B_1$ due to the first wire and $B_2$ due to the second wire can be calculated using the right-hand rule.
Since the wires are parallel and carry the same current $I$ in the same direction, at the midpoint, the magnetic field $B_1$ points vertically upwards (or downwards depending on the direction of current), and $B_2$ points in the opposite direction.
Specifically, $B_1 = \frac{\mu_0 I}{2\pi (d/2)}$ and $B_2 = \frac{\mu_0 I}{2\pi (d/2)}$.
Since they are equal in magnitude and opposite in direction, the resultant magnetic field $B_{net} = B_1 - B_2 = 0$.
At this point, because the net magnetic field is zero, the magnetic needle will not experience any torque and its orientation will be independent of the magnitude of the current $I$.

(D) The magnetic field $B$ on the axis of a circular coil of radius $R$ carrying current $i$ at a distance $r$ from its center is given by the formula:
$B = \frac{\mu_0 i R^2}{2(R^2 + r^2)^{3/2}}$
Given the condition $r \gg R$,we can neglect $R^2$ in the denominator:
$B \approx \frac{\mu_0 i R^2}{2(r^2)^{3/2}} = \frac{\mu_0 i R^2}{2r^3}$
Since $\mu_0$,$i$,and $R$ are constants,we find that $B \propto \frac{1}{r^3}$.

(C) The magnetic field $B$ at the center of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$.
Here, the current $I = q \times \nu$, where $q = 1.6 \times 10^{-19} \, C$ is the charge of the electron and $\nu = 6.6 \times 10^{15} \, Hz$ is the frequency of rotation.
The radius $r = 0.53 \, \text{\AA} = 0.53 \times 10^{-10} \, m$.
Substituting the values:
$B = \frac{4\pi \times 10^{-7} \times (1.6 \times 10^{-19} \times 6.6 \times 10^{15})}{2 \times 0.53 \times 10^{-10}}$
$B = \frac{2 \times 3.14 \times 10^{-7} \times 1.6 \times 6.6 \times 10^{-4}}{0.53 \times 10^{-10}}$
$B \approx 12.5 \, Wb/m^2$.

(A) According to the Biot-Savart law or Ampere's circuital law,the magnetic field $B$ at a perpendicular distance $r$ from an infinitely long straight current-carrying wire is given by the formula:
$B = \frac{\mu_0 i}{2\pi r}$
To express this in terms of the constant $\frac{\mu_0}{4\pi}$,we multiply the numerator and denominator by $2$:
$B = \frac{\mu_0 i \times 2}{2\pi r \times 2} = \left( \frac{\mu_0}{4\pi} \right) \frac{2i}{r}$
Thus,the correct option is $A$.

(B) The correct answer is $B$. The magnetic effect of current was discovered by the Danish physicist Hans Christian Oersted in $1820$.
He observed that a magnetic compass needle placed near a current-carrying wire gets deflected.
This observation proved that an electric current produces a magnetic field around it.
This phenomenon is fundamental to electromagnetism and is known as Oersted's experiment.

(D) The magnetic field at the centre of a circular coil of $N$ turns carrying current $i$ with radius $r$ is given by $B = \frac{\mu_0 N i}{2r}$.
Given: $N = 10$,$r_1 = 0.2 \ m$,$i_1 = 0.2 \ A$,$r_2 = 0.4 \ m$,$i_2 = 0.3 \ A$.
Since the currents are in opposite directions,the net magnetic field at the centre is the difference between the two fields:
$B_{net} = |B_1 - B_2| = |\frac{\mu_0 N i_1}{2r_1} - \frac{\mu_0 N i_2}{2r_2}|$
$B_{net} = \frac{\mu_0 \times 10}{2} |\frac{0.2}{0.2} - \frac{0.3}{0.4}|$
$B_{net} = 5 \mu_0 |1 - 0.75|$
$B_{net} = 5 \mu_0 \times 0.25 = 1.25 \mu_0 = \frac{5}{4} \mu_0 \ Wb/m^2$.

(A) The magnetic field at the center of a circular arc of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i \theta}{4\pi r}$.
Since the two parts of the circular conductor are connected in parallel to the cell,the potential difference $V$ across them is the same.
Let $R_1$ and $R_2$ be the resistances of the two parts with lengths $l_1$ and $l_2$ respectively. Then $V = i_1 R_1 = i_2 R_2$.
Since $R = \rho \frac{l}{A}$,we have $i_1 \rho \frac{l_1}{A} = i_2 \rho \frac{l_2}{A}$,which implies $i_1 l_1 = i_2 l_2$.
The magnetic field due to the first part is $B_1 = \frac{\mu_0 i_1 \theta_1}{4\pi r}$ and due to the second part is $B_2 = \frac{\mu_0 i_2 \theta_2}{4\pi r}$.
Since $l_1 = r \theta_1$ and $l_2 = r \theta_2$,we have $i_1 r \theta_1 = i_2 r \theta_2$,so $i_1 \theta_1 = i_2 \theta_2$.
Thus,$B_1 = \frac{\mu_0}{4\pi r} (i_1 \theta_1)$ and $B_2 = \frac{\mu_0}{4\pi r} (i_2 \theta_2)$.
Since $i_1 \theta_1 = i_2 \theta_2$,the magnitudes are equal $(B_1 = B_2)$.
Because the currents flow in opposite directions around the center,the magnetic fields produced by the two parts are equal in magnitude and opposite in direction.
Therefore,the resultant magnetic field at the center is zero.

(D) Let the lengths of the two arcs be $l_1 = r\theta$ and $l_2 = r(2\pi - \theta)$.
Since the ring is uniform,the resistances are $R_1 = \rho \frac{l_1}{A}$ and $R_2 = \rho \frac{l_2}{A}$.
When connected to a battery,the potential difference $V$ across both arcs is the same. Thus,$i_1 R_1 = i_2 R_2$,which implies $i_1 l_1 = i_2 l_2$.
The magnetic field at the center due to an arc of length $l$ carrying current $i$ is $B = \frac{\mu_0 i}{4\pi r^2} l$.
For the two arcs,$B_1 = \frac{\mu_0 i_1 l_1}{4\pi r^2}$ and $B_2 = \frac{\mu_0 i_2 l_2}{4\pi r^2}$.
Since $i_1 l_1 = i_2 l_2$,we have $B_1 = B_2$.
The currents flow in opposite directions around the center,so the magnetic fields produced by the two arcs are equal in magnitude and opposite in direction.
Therefore,the resultant magnetic induction at the center is $B_{net} = B_1 - B_2 = 0$,which is independent of $\theta$.

(D) The magnetic field $B$ produced by a current-carrying wire at a point is given by the Biot-Savart Law.
According to the Biot-Savart Law,the magnetic field $d\vec{B}$ due to a current element $Id\vec{l}$ at a position vector $\vec{r}$ is given by $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \vec{r})}{r^3}$.
For any point lying on the axis of the straight wire,the current element $Id\vec{l}$ and the position vector $\vec{r}$ are collinear (i.e.,the angle between them is $0^\circ$ or $180^\circ$).
Since the cross product of collinear vectors is zero $(d\vec{l} \times \vec{r} = 0)$,the magnetic field at any point on the axis of the wire is zero.

(B) $1$. For a straight wire of length $L = \pi^2 \, m$ carrying current $i = 2 \, A$,the magnetic field $B_1$ at a distance $d = 1 \, cm = 10^{-2} \, m$ is given by the Biot-Savart law for an infinitely long wire (assuming the point is near the center of the wire): $B_1 = \frac{\mu_0 i}{2\pi d} = \frac{\mu_0 \times 2}{2\pi \times 10^{-2}} = \frac{\mu_0}{\pi \times 10^{-2}}$.
$2$. When the wire is bent into a circle of radius $R$,its circumference is $2\pi R = L = \pi^2$. Thus,$R = \frac{\pi^2}{2\pi} = \frac{\pi}{2} \, m$.
$3$. The magnetic field at the center of this circular loop is $B_2 = \frac{\mu_0 i}{2R} = \frac{\mu_0 \times 2}{2 \times (\pi/2)} = \frac{\mu_0}{\pi/2} = \frac{2\mu_0}{\pi}$.
$4$. The ratio of the magnetic field at the center to the first case is $\frac{B_2}{B_1} = \frac{2\mu_0 / \pi}{\mu_0 / (\pi \times 10^{-2})} = 2 \times 10^{-2} = \frac{2}{100} = \frac{1}{50}$.
$5$. Therefore,the ratio is $1:50$.

(C) According to the Right-Hand Thumb Rule,when a current $i$ flows through a straight conductor,the magnetic field lines produced around it are concentric circles. These circles lie in a plane that is perpendicular to the axis of the conductor. The direction of these magnetic field lines is determined by the direction of the current,as shown in the figure.

(A) The magnetic field $B$ at a distance $r$ from an infinitely long straight current-carrying conductor is given by the formula:
$B = \frac{\mu_0}{4\pi} \cdot \frac{2i}{r}$
Given:
$B = 10^{-5}\,Wb/m^2$
$r = 10\,cm = 0.1\,m = 10^{-1}\,m$
$\frac{\mu_0}{4\pi} = 10^{-7}\,T\cdot m/A$
Substituting the values into the formula:
$10^{-5} = 10^{-7} \times \frac{2i}{10^{-1}}$
$10^{-5} = 10^{-7} \times 2i \times 10^1$
$10^{-5} = 2i \times 10^{-6}$
$2i = \frac{10^{-5}}{10^{-6}}$
$2i = 10$
$i = 5\,A$
Therefore,the current flowing in the conductor is $5\,A$.

(C) 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 are: Current $I = 10 \ A$,Radius $r = 10 \ cm = 0.1 \ m$,Magnetic field $B = 3.14 \times 10^{-3} \ Wb/m^2$,and $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$.
Substituting the values into the formula:
$3.14 \times 10^{-3} = \frac{(4\pi \times 10^{-7}) \times N \times 10}{2 \times 0.1}$.
Using $\pi \approx 3.14$,we get:
$3.14 \times 10^{-3} = \frac{4 \times 3.14 \times 10^{-7} \times N \times 10}{0.2}$.
$3.14 \times 10^{-3} = 20 \times 3.14 \times 10^{-6} \times N$.
$10^{-3} = 20 \times 10^{-6} \times N$.
$N = \frac{10^{-3}}{20 \times 10^{-6}} = \frac{1000}{20} = 50$.
Therefore,the number of turns in the coil is $50$.

(D) To find the direction of the magnetic field at the origin $O$ due to the current flowing in the wire from $Q$ to $P$,we use the Right-Hand Thumb Rule.
$1$. Point the thumb of your right hand in the direction of the current,which is along the positive $Z$-axis (from $Q$ to $P$).
$2$. Curl your fingers around the wire. At the origin $O$,which lies to the left of the wire (along the negative $X$-axis direction relative to the wire),the fingers will point into the plane of the paper.
$3$. In the given coordinate system,the $Y$-axis points out of the plane and the $Y'$-axis points into the plane.
$4$. Therefore,the magnetic field at the origin $O$ is directed along the negative $Y$-axis,which is $OY'$.

(B) The magnetic field at the centre of a circular loop with $n$ turns is given by $B = \frac{n \mu_0 I}{2R}$.
For the first case,$n_1 = 1$. The circumference of the loop is $L = 2 \pi R_1$,so $R_1 = \frac{L}{2 \pi}$.
The magnetic field is $B = \frac{1 \cdot \mu_0 I}{2 R_1} = \frac{\mu_0 I}{2 (L / 2 \pi)} = \frac{\mu_0 I \pi}{L}$.
For the second case,the wire is bent into $n_2 = 4$ turns. The total length of the wire is $L = n_2 (2 \pi R_2)$,so $R_2 = \frac{L}{2 \pi n_2} = \frac{L}{8 \pi}$.
The new magnetic field is $B' = \frac{n_2 \mu_0 I}{2 R_2} = \frac{4 \mu_0 I}{2 (L / 8 \pi)} = \frac{4 \mu_0 I \cdot 8 \pi}{2 L} = 16 \left( \frac{\mu_0 I \pi}{L} \right) = 16 B$.

(D) The magnetic field $B$ at the center of a circular current loop is given by the formula: $B = \frac{\mu_0 i}{2r}$.
Given values are: $B = 12.56 \, T$,$r = 5.2 \times 10^{-11} \, m$,and $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$.
Substituting these values into the formula:
$12.56 = \frac{(4\pi \times 10^{-7}) \times i}{2 \times 5.2 \times 10^{-11}}$
$12.56 = \frac{2\pi \times 10^{-7} \times i}{5.2 \times 10^{-11}}$
Using $\pi \approx 3.14$,we have $2\pi \approx 6.28$.
$12.56 = \frac{6.28 \times 10^{-7} \times i}{5.2 \times 10^{-11}}$
$i = \frac{12.56 \times 5.2 \times 10^{-11}}{6.28 \times 10^{-7}}$
$i = 2 \times 5.2 \times 10^{-4} \, A$
$i = 10.4 \times 10^{-4} \, A = 1.04 \times 10^{-3} \, A$.

(B) The magnetic field $B$ at the center of a circular arc of radius $R$ carrying current $i$ that subtends an angle $\theta$ (in radians) at the center is given by the formula:
$B = \frac{\mu_0 i \theta}{4\pi R}$
Given that the angle subtended is $\theta = \frac{\pi}{2}$,we substitute this value into the formula:
$B = \frac{\mu_0 i}{4\pi R} \times \frac{\pi}{2}$
Simplifying the expression:
$B = \frac{\mu_0 i}{8R}$
Thus,the correct option is $B$.

(A) The magnetic field $B$ at a distance $r$ from a long straight current-carrying wire is given by the formula $B = \frac{\mu_0 I}{2\pi r}$.
This implies that $B \propto \frac{1}{r}$.
Therefore,we can write the ratio as $\frac{B_1}{B_2} = \frac{r_2}{r_1}$.
Given $B_1 = 0.04\, T$,$r_1 = 10\, cm$,and $r_2 = 40\, cm$.
Substituting these values into the equation: $\frac{0.04}{B_2} = \frac{40}{10}$.
$\frac{0.04}{B_2} = 4$.
$B_2 = \frac{0.04}{4} = 0.01\, T$.

(A) Let the lengths of the two arcs be $l_1$ (arc $ABC$) and $l_2$ (arc $ADC$). The resistances of these arcs are $R_1 = \rho \frac{l_1}{A_{cross}}$ and $R_2 = \rho \frac{l_2}{A_{cross}}$.
Since the arcs are in parallel,the potential difference across them is the same: $V = I_1 R_1 = I_2 R_2$.
This implies $I_1 l_1 = I_2 l_2$,or $\frac{I_1}{I_2} = \frac{l_2}{l_1}$.
The magnetic field at the center due to an arc of length $l$ carrying current $i$ is $B = \frac{\mu_0 i \theta}{4\pi R} = \frac{\mu_0 i l}{4\pi R^2}$.
For the two arcs,the magnetic fields are $B_1 = \frac{\mu_0 I_1 l_1}{4\pi R^2}$ and $B_2 = \frac{\mu_0 I_2 l_2}{4\pi R^2}$.
Since $I_1 l_1 = I_2 l_2$,we have $B_1 = B_2$.
The currents flow in opposite directions around the center,so the magnetic fields produced by the two arcs are in opposite directions.
Therefore,the resultant magnetic field at the center is $B_{net} = |B_1 - B_2| = 0$.

(B) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $i$ is given by the formula:
$B = \frac{\mu_0 i}{2\pi r}$
From this expression,it is clear that the magnetic induction $B$ is inversely proportional to the distance $r$ from the wire $(B \propto \frac{1}{r})$.
Therefore,the correct option is $B$.

(B) The magnetic field at the center of a circular loop is given by $B = \frac{\mu_0 I}{2R}$.
Given,$B = 7 \times 10^{-5} \ Wb/m^2$ and $R = 5 \ cm = 0.05 \ m$.
We know that $\frac{\mu_0}{4\pi} = 10^{-7} \ T \cdot m/A$,so $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$.
Substituting the values into the formula:
$7 \times 10^{-5} = \frac{4\pi \times 10^{-7} \times I}{2 \times 0.05}$
$7 \times 10^{-5} = \frac{2\pi \times 10^{-7} \times I}{0.05}$
$I = \frac{7 \times 10^{-5} \times 0.05}{2 \times 3.14159 \times 10^{-7}}$
$I = \frac{0.35 \times 10^{-5}}{6.283 \times 10^{-7}} = \frac{35}{6.283} \approx 5.57 \ A \approx 5.6 \ A$.

(C) The magnetic field $B$ at the center of a circular coil is given by the formula: $B = \frac{\mu_0 Ni}{2r}$.
Given values are:
Number of turns $N = 1000$
Current $i = 0.1\, A$
Radius $r = 0.1\, m$
Permeability of free space $\mu_0 = 4\pi \times 10^{-7}\, T\cdot m/A$.
Substituting these values into the formula:
$B = \frac{4\pi \times 10^{-7} \times 1000 \times 0.1}{2 \times 0.1}$
$B = \frac{4\pi \times 10^{-7} \times 100}{0.2}$
$B = 2\pi \times 10^{-4}\, T$
Using $\pi \approx 3.14$,we get:
$B = 2 \times 3.14 \times 10^{-4} = 6.28 \times 10^{-4}\, T$.

(B) The magnetic field $B$ at the centre of a circular coil is given by the formula: $B = \frac{\mu_0 Ni}{2r}$.
Given values are:
Number of turns $N = 50$
Radius $r = 0.5\, m$
Current $i = 2\, A$
Permeability of free space $\mu_0 = 4\pi \times 10^{-7}\, T\cdot m/A$.
Substituting these values into the formula:
$B = \frac{(4\pi \times 10^{-7}) \times 50 \times 2}{2 \times 0.5}$
$B = \frac{4 \times 3.14159 \times 10^{-7} \times 100}{1}$
$B = 12.566 \times 10^{-5} = 1.2566 \times 10^{-4}\, T$.
Rounding to the nearest option,we get $1.25 \times 10^{-4}\, T$.

(D) The magnetic field at the centre of a circular coil carrying current $I$ with radius $R$ is given by the formula: $B = \frac{\mu_0 I}{2R}$.
For coil $A$: $B_A = \frac{\mu_0 I}{2R}$.
For coil $B$: $B_B = \frac{\mu_0 (2I)}{2(2R)} = \frac{\mu_0 I}{2R}$.
Comparing the two,we get $B_A = B_B$.
Therefore,the ratio $B_A : B_B = 1:1$.

(A) The magnetic field $B$ at a distance $r$ from a long straight current-carrying wire is given by the formula $B = \frac{\mu_0 i}{2\pi r}$.
From this expression,we can see that the magnetic field is inversely proportional to the distance,i.e.,$B \propto \frac{1}{r}$.
Let $B_1 = 0.4 \ T$ at distance $r_1 = r$.
We need to find the magnetic field $B_2$ at distance $r_2 = 2r$.
Using the proportionality $B_1 r_1 = B_2 r_2$,we get:
$0.4 \times r = B_2 \times (2r)$.
$B_2 = \frac{0.4 \times r}{2r} = 0.2 \ T$.

(A) The magnetic field $B$ at the centre of a circular coil with $n$ turns,carrying current $i$ and having radius $r$,is given by the Biot-Savart Law.
For a single turn,the magnetic field is $B = \frac{\mu_0 i}{2r}$.
For a coil with $n$ turns,the magnetic field is $B = n \times \frac{\mu_0 i}{2r} = \frac{\mu_0 ni}{2r}$.

(D) The magnetic field $B$ at a distance $r$ from a long straight current-carrying wire is given by the formula $B = \frac{\mu_0}{4\pi} \frac{2i}{r}$.
Here,$\mu_0$ is the permeability of free space,$i$ is the current flowing through the wire,and $r$ is the perpendicular distance from the wire.
As per the formula,the magnetic field $B$ depends only on the current $i$ and the distance $r$ from the wire.
It is independent of the radius or diameter of the wire.
Since the current $i$ remains the same and the distance $r$ is considered far away (implying the same observation point),the magnetic field strength remains unchanged.

(C) When a wire carries a steady current,the net charge density of the wire remains zero because the number of electrons entering a segment equals the number of electrons leaving it.
Since the wire is electrically neutral,it does not produce an electric field in its neighborhood.
However,the flow of charges (current) creates a magnetic field around the wire,as described by the Biot-Savart law or Ampere's circuital law.
Therefore,only a magnetic field exists.

(B) The magnetic field $B$ at a distance $r$ from a long straight current-carrying wire is given by the formula:
$B = \frac{\mu_0}{4\pi} \times \frac{2i}{r}$
Given:
Current $i = 1\,A$
Distance $r = 1\,cm = 10^{-2}\,m$
Constant $\frac{\mu_0}{4\pi} = 10^{-7}\,T\cdot m/A$
Substituting the values:
$B = 10^{-7} \times \frac{2 \times 1}{10^{-2}}$
$B = 2 \times 10^{-7} \times 10^2$
$B = 2 \times 10^{-5}\,T$
Therefore,the correct option is $B$.

(A) The magnetic field due to a straight wire of length $a$ at a distance $r$ from its center is given by $B = \frac{\mu_0 i}{4 \pi r} (\sin \theta_1 + \sin \theta_2)$.
For a square of side $a = 2l$,the distance from the center to the side is $r = l/2 = l$. The angles at the center are $\theta_1 = \theta_2 = 45^\circ$.
Thus,the field due to one side is $B_1 = \frac{\mu_0 i}{4 \pi l} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{4 \pi l} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{4 \pi l} (\frac{2}{\sqrt{2}}) = \frac{\mu_0 i}{2 \sqrt{2} \pi l}$.
Since there are $4$ sides,the total field for one turn is $B = 4 \times B_1 = 4 \times \frac{\mu_0 i}{2 \sqrt{2} \pi l} = \frac{2 \sqrt{2} \mu_0 i}{\pi l}$.
For $n$ turns,the total magnetic field is $B_{net} = n \times B = \frac{2 \sqrt{2} \mu_0 n i}{\pi l}$.
Wait,re-evaluating the geometry: For a square of side $a=2l$,the distance from center to side is $d = l$. The formula $B = \frac{\mu_0 i}{4 \pi d} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{4 \pi l} (\sqrt{2}) = \frac{\sqrt{2} \mu_0 i}{4 \pi l}$.
Total field for $4$ sides is $4 \times \frac{\sqrt{2} \mu_0 i}{4 \pi l} = \frac{\sqrt{2} \mu_0 i}{\pi l}$.
For $n$ turns,$B_{net} = \frac{\sqrt{2} \mu_0 n i}{\pi l}$.

(C) According to Biot-Savart's law,the magnitude of the magnetic field $dB$ due to a small current element $i \Delta l$ at a distance $r$ from the element is given by:
$dB = \frac{\mu_0}{4\pi} \frac{i \Delta l \sin \theta}{r^2}$
where $\mu_0$ is the permeability of free space,$i$ is the current,$\Delta l$ is the length of the element,$\theta$ is the angle between the current element and the position vector,and $r$ is the distance from the element to the point of observation.
Thus,the correct option is $C$.