Mix Examples-Moving Charges and Magnetism Questions in English

(D) The permeability $\mu$ is related to magnetic field $B$,magnetic field intensity $H$,and magnetic flux $\Phi$.
From the relation $B = \mu H$,the unit of $\mu$ is $\text{Tesla} / (\text{Ampere/metre}) = \text{Tesla metre per ampere}$ $(T \cdot m \cdot A^{-1})$.
Since $1 \text{ Tesla} = 1 \text{ Weber/metre}^2$,the unit becomes $(\text{Weber/metre}^2) \cdot \text{metre} / \text{Ampere} = \text{Weber per ampere metre}$ $(Wb \cdot A^{-1} \cdot m^{-1})$.
Also,the inductance $L$ of a solenoid is given by $L = \mu n^2 A l$,where $\mu$ has units of $\text{Henry/metre}$ $(H \cdot m^{-1})$.
Therefore,all the given units are equivalent and correct.

(C) An electric charge at rest produces only an electric field.
When an electric charge is in uniform motion,it constitutes a steady current.
$A$ steady current produces a magnetic field in the space surrounding it,in addition to the electric field that is always associated with the charge.
Therefore,a charge in uniform motion produces both an electric field and a magnetic field.

(C) stationary electron produces an electric field in its rest frame. When an observer moves with respect to the electron,the electron appears to be in motion relative to the observer. $A$ moving charge constitutes an electric current,which generates a magnetic field. Therefore,the observer detects both an electric field (due to the charge) and a magnetic field (due to the relative motion of the charge).

(C) The relationship between the thermo $e.m.f.$ $(E)$ and the temperature of the hot junction $(T_h)$ is given by the parabolic equation $E = a(T_h - T_c) + \frac{1}{2}b(T_h - T_c)^2$, where $T_c$ is the temperature of the cold junction.
Since the graph is parabolic, the $e.m.f.$ increases with temperature until it reaches a maximum value at the neutral temperature $(T_n)$.
Beyond the neutral temperature, the $e.m.f.$ starts to decrease as the temperature of the hot junction increases further.
Therefore, the thermo $e.m.f.$ may increase or decrease depending on the range of temperature relative to the neutral temperature.

(B) The thermo-electric power $P$ is defined as the rate of change of thermo-electric $EMF$ $(E)$ with respect to temperature $(\theta)$,given by $P = \frac{dE}{d\theta}$.
At the neutral temperature $(t_n)$,the thermo-electric $EMF$ $(E)$ reaches its maximum value.
Since the derivative of a function at its maximum point is zero,the thermo-electric power $P$ becomes zero at the neutral temperature.
Therefore,the correct option is $B$.

(D) The relationship between neutral temperature $(t_n)$,cold junction temperature $(t_c)$,and inversion temperature $(t_i)$ is given by the formula: $t_n = \frac{t_i + t_c}{2}$.
Given the initial condition: $t_n = 350\,^{\circ}C$ when $t_c = 0\,^{\circ}C$.
Using the formula $t_n = \frac{t_i + t_c}{2}$,we find the constant property of the thermocouple: $350 = \frac{t_i + 0}{2}$,which implies $t_i = 700\,^{\circ}C$ for $t_c = 0\,^{\circ}C$.
However,the neutral temperature $t_n$ is a constant for a given thermocouple pair. The formula can be rearranged as $t_i = 2t_n - t_c$.
Substituting the new cold junction temperature $t_c = 30\,^{\circ}C$ and the constant $t_n = 350\,^{\circ}C$:
$t_i = 2(350) - 30 = 700 - 30 = 670\,^{\circ}C$.

(C) The unit of magnetic field in the $SI$ system is Tesla $(T)$.
In the $CGS$ system,the unit of magnetic field is Gauss $(G)$.
The relationship between these two units is defined as $1 \, T = 10^4 \, G$.
Therefore,one Tesla is equal to $10^4 \, \text{gauss}$.

(D) The force on a proton moving with velocity $\overrightarrow{v}$ in the presence of electric field $\overrightarrow{E}$ and magnetic field $\overrightarrow{B}$ is given by the Lorentz force law: $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})$.
Since the proton moves with a constant velocity,the net force $\overrightarrow{F}$ must be zero.
Case $1$: If $\overrightarrow{E} = 0$ and $\overrightarrow{B} = 0$,then $\overrightarrow{F} = 0$. The proton continues in a straight line.
Case $2$: If $\overrightarrow{E} = 0$ and $\overrightarrow{B} \neq 0$,the force $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$. If $\overrightarrow{B}$ is parallel or anti-parallel to $\overrightarrow{v}$,then $\overrightarrow{v} \times \overrightarrow{B} = 0$,so $\overrightarrow{F} = 0$.
Case $3$: If $\overrightarrow{E} \neq 0$ and $\overrightarrow{B} \neq 0$,the forces can cancel each other if $q\overrightarrow{E} = -q(\overrightarrow{v} \times \overrightarrow{B})$,i.e.,$\overrightarrow{E} = -(\overrightarrow{v} \times \overrightarrow{B})$. This is the principle of a velocity selector.
Thus,all the given conditions are possible.

(B) When two beams of electrons move parallel to each other in the same direction,each beam acts as a current carrying wire. Since the electrons are moving in the same direction,the conventional current in both beams is also in the same direction. According to the magnetic force law between two parallel current-carrying conductors,currents in the same direction attract each other. However,for electron beams,we must consider the electrostatic force as well. The electrostatic force of repulsion between the negatively charged electrons is significantly stronger than the magnetic force of attraction. Therefore,the net force between two parallel electron beams moving in the same direction is a force of repulsion in the plane of the paper.

(A) $1$. The moving beams of electrons and protons constitute electric currents. Since they are moving in the same direction,the magnetic force between them is attractive.
$2$. However,electrons and protons are charged particles. The electrostatic force between an electron and a proton is attractive because they have opposite charges.
$3$. In this specific scenario,the electrostatic force is significantly stronger than the magnetic force.
$4$. Therefore,the net force between the beam of electrons and the beam of protons is attractive.

(A) Let the magnetic field due to a long straight wire at distance $r$ be $B_0 = \frac{\mu_0 i}{2\pi r}$. The field due to a full circular loop of radius $r$ is $B_{loop} = \frac{\mu_0 i}{2r} = \pi B_0$.
Case $1$: The field is due to a semi-circular arc and two semi-infinite wires. The wires produce fields in opposite directions. The net field is $B_1 = \frac{1}{2} \left( \frac{\mu_0 i}{2r} \right) - \frac{\mu_0 i}{4\pi r} - \frac{\mu_0 i}{4\pi r} = \frac{\mu_0 i}{4r} - \frac{\mu_0 i}{2\pi r} = \frac{\mu_0 i}{4\pi r} (\pi - 2)$.
Case $2$: The field is due to a semi-circular arc and two semi-infinite wires. The fields add up: $B_2 = \frac{1}{2} \left( \frac{\mu_0 i}{2r} \right) + \frac{\mu_0 i}{4\pi r} + \frac{\mu_0 i}{4\pi r} = \frac{\mu_0 i}{4r} + \frac{\mu_0 i}{2\pi r} = \frac{\mu_0 i}{4\pi r} (\pi + 2)$.
Case $3$: The field is due to a $3/4$ circular arc and two semi-infinite wires. $B_3 = \frac{3}{4} \left( \frac{\mu_0 i}{2r} \right) + \frac{\mu_0 i}{4\pi r} + \frac{\mu_0 i}{4\pi r} = \frac{3\mu_0 i}{8r} + \frac{\mu_0 i}{2\pi r} = \frac{\mu_0 i}{4\pi r} (\frac{3\pi}{2} + 2)$.
Comparing the magnitudes and directions,the ratio is $\left( -\frac{\pi}{2} \right) : \left( \frac{\pi}{2} \right) : \left( \frac{3\pi}{4} - \frac{1}{2} \right)$.

(C) The magnetic field due to a long straight wire at a distance $a$ is given by $B = \frac{\mu_0 i}{2\pi a}$.
For the conductor $AOB$ carrying current $i_1$,the magnetic field at point $P$ is $B_1 = \frac{\mu_0 i_1}{2\pi a}$.
For the conductor $COD$ carrying current $i_2$,the magnetic field at point $P$ is $B_2 = \frac{\mu_0 i_2}{2\pi a}$.
Since the conductors are perpendicular,the magnetic field vectors $B_1$ and $B_2$ are also perpendicular to each other.
The net magnetic field at point $P$ is $B_{net} = \sqrt{B_1^2 + B_2^2}$.
Substituting the values,$B_{net} = \sqrt{\left(\frac{\mu_0 i_1}{2\pi a}\right)^2 + \left(\frac{\mu_0 i_2}{2\pi a}\right)^2}$.
$B_{net} = \frac{\mu_0}{2\pi a} \sqrt{i_1^2 + i_2^2} = \frac{\mu_0}{2\pi a}(i_1^2 + i_2^2)^{1/2}$.

(D) The kinetic energy of the particle at point $P$ is $K_P = \frac{1}{2}mv^2$.
The kinetic energy of the particle at point $Q$ is $K_Q = \frac{1}{2}m(2v)^2 = 2mv^2$.
The increase in kinetic energy is $\Delta K = K_Q - K_P = 2mv^2 - \frac{1}{2}mv^2 = \frac{3}{2}mv^2$.
This increase is due to the work done by the electric force $F_e = qE$ as the particle moves a distance $2a$ along the $x$-axis.
Thus,$W = qE(2a) = \frac{3}{2}mv^2$,which gives $E = \frac{3}{4}\frac{mv^2}{qa}$. So,option $A$ is correct.
The rate of work done by the electric field is $P_e = \vec{F_e} \cdot \vec{v}$. At point $P$,$\vec{F_e} = qE\hat{i}$ and $\vec{v} = v\hat{i}$,so $P_e = (qE)(v) = q(\frac{3}{4}\frac{mv^2}{qa})v = \frac{3}{4}\frac{mv^3}{a}$. So,option $B$ is correct.
At point $Q$,the velocity is $\vec{v} = -2v\hat{j}$. The electric force is $\vec{F_e} = qE\hat{i}$. Since $\vec{F_e} \perp \vec{v}$,the rate of work done by the electric field is $P_e = \vec{F_e} \cdot \vec{v} = 0$. The magnetic force $\vec{F_m} = q(\vec{v} \times \vec{B})$ is always perpendicular to the velocity $\vec{v}$,so the rate of work done by the magnetic field is $P_m = \vec{F_m} \cdot \vec{v} = 0$. Thus,the total rate of work done at $Q$ is $0+0=0$. So,option $C$ is correct.
Therefore,all statements are correct.

(B) $1$. For the conductors $A$ and $B$: When two parallel conductors carry current in the same direction, they produce magnetic fields that result in an attractive force between them. The force per unit length is given by $F/L = \frac{\mu_0 I_1 I_2}{2 \pi d}$. Since the currents are in the same direction, the force is attractive.
$2$. For the electron beams $x$ and $y$: Electrons are negatively charged particles. When they move in the same direction, they constitute a current in the opposite direction to their motion. However, because they are both negatively charged, they experience a strong electrostatic repulsive force. While there is a magnetic attractive force between the parallel currents formed by the moving electrons, the electrostatic repulsion between the like charges dominates. Therefore, the net force between two parallel electron beams is repulsive.

(A) Each current-carrying ring acts as a magnetic dipole with a magnetic moment $\vec{\mu}$ perpendicular to its plane.
When currents are set up in the rings,they experience a magnetic torque $\vec{\tau} = \vec{\mu} \times \vec{B}$ due to the magnetic field of the other ring.
This torque acts to align the magnetic moments of the two rings in the same direction.
Since the magnetic moment is perpendicular to the plane of the ring,aligning the magnetic moments means the planes of the two rings will rotate until they become coplanar,with the currents flowing in the same direction to achieve a state of minimum potential energy.
Therefore,both rings rotate to reach a common plane.

(C) Let the proton beam be at $x = 0$ and the electron beam be at $x = d$. The current $I$ in both beams is equal in magnitude.
For a point $P$ at distance $x$ from the proton beam,the magnetic field due to the proton beam is $B_p = \frac{\mu_0 I}{2 \pi x}$ (directed inwards,$\otimes$).
The magnetic field due to the electron beam at distance $(d - x)$ is $B_e = \frac{\mu_0 I}{2 \pi (d - x)}$ (also directed inwards,$\otimes$,because the electron current is in the opposite direction).
The net magnetic field is $B = B_p + B_e = \frac{\mu_0 I}{2 \pi} \left( \frac{1}{x} + \frac{1}{d - x} \right)$.
As $x \to 0$,$B \to \infty$. As $x \to d$,$B \to \infty$.
At $x = d/2$,$B$ is minimum.
This corresponds to a $U$-shaped curve where $B$ is positive and approaches infinity at both boundaries $x=0$ and $x=d$. Among the given options,graph $(c)$ represents this behavior.

(B) The net force on a charged particle moving in a region of electric field $\overrightarrow{E}$ and magnetic field $\overrightarrow{B}$ is given by the Lorentz force equation: $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})$.
For the net force to be zero,we must have $\overrightarrow{F} = 0$,which implies $\overrightarrow{E} = -(\overrightarrow{v} \times \overrightarrow{B})$.
This means the electric force $\overrightarrow{F_e} = q\overrightarrow{E}$ and the magnetic force $\overrightarrow{F_m} = q(\overrightarrow{v} \times \overrightarrow{B})$ must be equal in magnitude and opposite in direction.
In Figure $B$,the velocity $\overrightarrow{v}$ is along the $x$-axis,the magnetic field $\overrightarrow{B}$ is along the $z$-axis,and the electric field $\overrightarrow{E}$ is along the $y$-axis.
The magnetic force $\overrightarrow{F_m} = q(\overrightarrow{v} \times \overrightarrow{B})$ will be in the direction of $(\hat{i} \times \hat{k}) = -\hat{j}$,which is along the negative $y$-axis.
The electric force $\overrightarrow{F_e} = q\overrightarrow{E}$ is along the positive $y$-axis.
Since the forces are opposite in direction,they can cancel each other out if their magnitudes are equal $(|qE| = |qvB|)$. Thus,the net force can be zero in Figure $B$.

(A) The energy density $U$ of a magnetic field $B$ is given by the formula $U = \frac{B^2}{2\mu_0}$.
Here,$U$ is directly proportional to the square of the magnetic field induction $(U \propto B^2)$.
This relationship represents a parabola that passes through the origin $(0,0)$ and is symmetric about the $U$-axis.
Therefore,the graph of $U$ versus $B$ is an upward-opening parabola starting from the origin,which corresponds to the graph provided in option $A$.

(B) When a current $i$ flows through a thin cylindrical shell of radius $R$,according to Ampere's circuital law,the magnetic field $B$ inside the shell $(r < R)$ is zero.
Since the magnetic energy density $U$ is given by $U = \frac{B^2}{2\mu_0}$,it follows that $U = 0$ for $r < R$.
For the region outside the shell $(r > R)$,the magnetic field is $B = \frac{\mu_0 i}{2\pi r}$.
Substituting this into the energy density formula,we get $U = \frac{1}{2\mu_0} \left( \frac{\mu_0 i}{2\pi r} \right)^2 = \frac{\mu_0 i^2}{8\pi^2 r^2}$.
Thus,$U \propto \frac{1}{r^2}$ for $r > R$,and $U = 0$ for $r < R$.
The graph representing this behavior shows $U = 0$ from $r = 0$ to $r = R$,and a curve following $U \propto r^{-2}$ for $r > R$. This corresponds to the graph shown in option $B$.

(D) Magnetic flux density $(B)$ is defined as the magnetic flux per unit area. The $SI$ unit of magnetic flux is Weber $(Wb)$. Therefore,the unit of magnetic flux density is $Wb/m^2$,which is also known as Tesla $(T)$.
From the Lorentz force law,$F = qvB \sin(\theta)$ or $F = IlB \sin(\theta)$,we can express $B$ as $B = F / (Il)$. Thus,the unit is also Newton per ampere-metre $(N/(A \cdot m))$.
Since all three options represent the same physical quantity,the correct answer is $D$.

(D) In a Thomson mass spectrograph,the electric field $\overrightarrow{E}$ and the magnetic field $\overrightarrow{B}$ are applied parallel to each other. This configuration allows for the separation of positive ions based on their charge-to-mass ratio $(q/m)$ by observing the deflection on a fluorescent screen. Therefore,the correct option is $D$.

(C) In Thomson's mass spectrograph,the equation of the parabola is given by $y = \frac{k q}{m} x^2$,where $k = \frac{B^2 L D}{E}$.
For the parabolas to coincide,the value of $\frac{k q}{m}$ must be the same for both cases.
Let the singly ionized ion have charge $q_1 = e$ and mass $m_1$,and the doubly ionized ion have charge $q_2 = 2e$ and mass $m_2$.
Given the ratios of electric fields $\frac{E_1}{E_2} = \frac{1}{2}$ and magnetic fields $\frac{B_1}{B_2} = \frac{3}{2}$.
Equating the constants: $\frac{B_1^2}{E_1 m_1} q_1 = \frac{B_2^2}{E_2 m_2} q_2$.
Substituting the values: $\frac{(3/2)^2}{1/2 \cdot m_1} \cdot e = \frac{1^2}{1 \cdot m_2} \cdot 2e$.
$\frac{9/4}{1/2 \cdot m_1} = \frac{2}{m_2} \implies \frac{9}{2 m_1} = \frac{2}{m_2}$.
Therefore,$\frac{m_1}{m_2} = \frac{9}{4}$.

(B) Let the length of the wire be $L$. For a circle of radius $r$,$L = 2\pi r$,so $r = \frac{L}{2\pi}$. The magnetic field at the center is $B_c = \frac{\mu_0 i}{2r} = \frac{\mu_0 i}{2(L/2\pi)} = \frac{\mu_0 i \pi}{L}$.
For a square of side $a$,$L = 4a$,so $a = \frac{L}{4}$. The magnetic field at the center due to one side is $B_s = \frac{\mu_0 i}{4\pi (a/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2\pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{\pi a \sqrt{2}}$.
Since there are $4$ sides,the total magnetic field is $B_{square} = 4 \times \frac{\mu_0 i}{\pi a \sqrt{2}} = \frac{4\mu_0 i}{\pi (L/4) \sqrt{2}} = \frac{16\mu_0 i}{\pi L \sqrt{2}}$.
The ratio is $\frac{B_c}{B_{square}} = \frac{\mu_0 i \pi / L}{16\mu_0 i / (\pi L \sqrt{2})} = \frac{\pi^2 \sqrt{2}}{16} = \frac{\pi^2}{8\sqrt{2}}$.

(A) For a closed loop carrying a current in a uniform magnetic field,the net magnetic force on the loop is zero.
Let the force on segment $QP$ be $F_4$.
Since the loop is in equilibrium under the influence of these forces,the vector sum of all forces must be zero:
$\vec{F_1} + \vec{F_2} + \vec{F_3} + \vec{F_4} = 0$
$\vec{F_4} = -(\vec{F_1} + \vec{F_2} + \vec{F_3})$
Resolving the forces into horizontal and vertical components:
Let the horizontal direction be along the $x$-axis (positive towards the right) and the vertical direction be along the $y$-axis (positive upwards).
$F_{4x} = -(F_{3x} + F_{2x} + F_{1x}) = -(F_3 + 0 - F_1) = F_1 - F_3$
$F_{4y} = -(F_{3y} + F_{2y} + F_{1y}) = -(0 - F_2 + 0) = F_2$
The magnitude of the force $F_4$ is given by:
$F_4 = \sqrt{F_{4x}^2 + F_{4y}^2} = \sqrt{(F_1 - F_3)^2 + F_2^2} = \sqrt{(F_3 - F_1)^2 + F_2^2}$

(B) The net magnetic force on a current-carrying closed loop in a uniform magnetic field is always zero.
Let the forces on the arms $AB,$ $BC,$ and $AC$ be $\vec{F}_{AB},$ $\vec{F}_{BC},$ and $\vec{F}_{AC}$ respectively.
According to the principle of superposition,$\vec{F}_{AB} + \vec{F}_{BC} + \vec{F}_{AC} = 0.$
Since the magnetic field is acting along the arm $AB,$ the angle between the current element $I\vec{dl}$ of arm $AB$ and the magnetic field $\vec{B}$ is $0^\circ$ or $180^\circ.$
Therefore,the magnetic force on arm $AB$ is $\vec{F}_{AB} = I(\vec{L}_{AB} \times \vec{B}) = 0.$
Substituting this into the net force equation: $0 + \vec{F}_{BC} + \vec{F}_{AC} = 0.$
Given that $\vec{F}_{BC} = \vec{F},$ we get $\vec{F} + \vec{F}_{AC} = 0.$
Thus,$\vec{F}_{AC} = -\vec{F}.$

(B) $1$. When the proton is at rest,the only force acting on it is the electric force $F_E = qE = eE$. Given the acceleration is $a_0$ towards west,we have $ma_0 = eE$,which implies $E = \frac{ma_0}{e}$ towards west.
$2$. When the proton is projected towards north with speed $v_0$,the total force is the vector sum of the electric force and the magnetic force: $F_{net} = F_E + F_B = ma_{net}$.
$3$. The net acceleration is $3a_0$ towards west. Since $F_E$ is $ma_0$ towards west,the magnetic force $F_B$ must be $2ma_0$ towards west to result in a total force of $3ma_0$ towards west $(F_B = F_{net} - F_E = 3ma_0 - ma_0 = 2ma_0)$.
$4$. The magnetic force is given by $F_B = q(v \times B) = ev_0B \sin(\theta)$. For the force to be towards west,with velocity towards north,the magnetic field $B$ must be directed downwards (using Fleming's Left-Hand Rule or the cross product $v \times B$).
$5$. Setting the magnitudes equal: $ev_0B = 2ma_0$,which gives $B = \frac{2ma_0}{ev_0}$ downwards.

(D) For the electron to move in a straight line,the electric force must be balanced by the magnetic force: $F_E = F_B$.
$eE = evB$,where $E$ is the electric field and $v$ is the velocity.
Since $E = \frac{\sigma}{\varepsilon_0}$,we have $\frac{\sigma}{\varepsilon_0} = vB$,which implies $v = \frac{\sigma}{\varepsilon_0 B}$.
The time $t$ taken to cover the length $l$ is $t = \frac{l}{v}$.
Substituting $v$,we get $t = \frac{l}{\frac{\sigma}{\varepsilon_0 B}} = \frac{\varepsilon_0 l B}{\sigma}$.

(D) Initially,when the wires carry currents in the same direction as shown in figure $(A)$,the magnetic fields at the midpoint $O$ due to wires $1$ and $2$ are in opposite directions.
$B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2i_1}{x}$ (inward)
$B_2 = \frac{\mu_0}{4\pi} \cdot \frac{2i_2}{x}$ (outward)
The net magnetic field at $O$ is $B_{net} = B_1 - B_2 = \frac{\mu_0}{4\pi} \cdot \frac{2}{x} (i_1 - i_2)$.
Given $B_{net} = 10 \, \mu T$,so $10 \times 10^{-6} = \frac{\mu_0}{4\pi} \cdot \frac{2}{x} (i_1 - i_2)$ ..... $(i)$
When the direction of $i_2$ is reversed as shown in figure $(B)$,the magnetic fields at $O$ due to both wires are in the same direction (inward).
$B_{net}' = B_1 + B_2 = \frac{\mu_0}{4\pi} \cdot \frac{2}{x} (i_1 + i_2)$.
Given $B_{net}' = 40 \, \mu T$,so $40 \times 10^{-6} = \frac{\mu_0}{4\pi} \cdot \frac{2}{x} (i_1 + i_2)$ ..... $(ii)$
Dividing equation $(ii)$ by $(i)$:
$\frac{i_1 + i_2}{i_1 - i_2} = \frac{40}{10} = 4$
$i_1 + i_2 = 4i_1 - 4i_2$
$5i_2 = 3i_1$
$\frac{i_1}{i_2} = \frac{5}{3}$

(C) The magnetic field on the axis of a circular coil of radius $R$ at a distance $x$ from its center is given by $B = \frac{\mu_0 N i R^2}{2(x^2 + R^2)^{3/2}}$.
For the smaller coil $X$,radius is $r$ and distance from $O$ is $x = d/2$. Thus,$B_X = \frac{\mu_0 N i r^2}{2((d/2)^2 + r^2)^{3/2}} = \frac{\mu_0 N i r^2}{2(\frac{d^2+4r^2}{4})^{3/2}} = \frac{4\mu_0 N i r^2}{2(d^2+4r^2)^{3/2}}$.
For the bigger coil $Y$,radius is $2r$ and distance from $O$ is $x = d$. Thus,$B_Y = \frac{\mu_0 N i (2r)^2}{2(d^2 + (2r)^2)^{3/2}} = \frac{4\mu_0 N i r^2}{2(d^2+4r^2)^{3/2}}$.
Wait,re-evaluating the geometry: The coils subtend the same solid angle,meaning the ratio of radius to distance is constant. For $X$,$r/(d/2) = 2r/d$. For $Y$,$2r/d$. This is consistent.
Recalculating $B_X$: $B_X = \frac{\mu_0 N i r^2}{2(d^2/4 + r^2)^{3/2}} = \frac{\mu_0 N i r^2}{2(\frac{d^2+4r^2}{4})^{3/2}} = \frac{8\mu_0 N i r^2}{2(d^2+4r^2)^{3/2}}$.
Recalculating $B_Y$: $B_Y = \frac{\mu_0 N i (2r)^2}{2(d^2 + 4r^2)^{3/2}} = \frac{4\mu_0 N i r^2}{2(d^2+4r^2)^{3/2}}$.
Therefore,$\frac{B_Y}{B_X} = \frac{4}{8} = \frac{1}{2}$.

(C) The loop consists of two parts: a circular arc and a straight wire segment.
According to the Biot-Savart law,the magnetic field due to a circular arc of radius $R$ subtending an angle $\theta$ at the center is $B_{arc} = \frac{\mu_0 i \theta}{4 \pi R}$. Using the right-hand rule,for the given counter-clockwise current,the field is directed perpendicular to the paper outwards.
For the straight wire segment,the magnetic field at the center $O$ is given by $B_{wire} = \frac{\mu_0 i}{4 \pi d} (\sin \alpha + \sin \beta)$,where $d$ is the perpendicular distance from $O$ to the wire. Using the right-hand rule,the field due to this straight wire is directed perpendicular to the paper inwards.
The net magnetic field is $B_{net} = B_{arc} - B_{wire}$. Since the arc length is greater than the chord length for any $\theta < 180^\circ$,the magnetic field contribution from the arc is always greater than that of the straight wire. Thus,the net magnetic field is directed perpendicular to the paper outwards.

(A) Let the positively charged beam be at $x = 0$ and the negatively charged beam be at $x = d$.
Since the positive charges move in one direction and the negative charges move in the opposite direction,the currents $I_1$ and $I_2$ produced by these beams are in the same direction.
Let the distance from the first beam be $x$. The magnetic field due to the first beam is $B_1 = \frac{\mu_0 I}{2\pi x}$ (directed into the page).
The magnetic field due to the second beam at distance $(d - x)$ is $B_2 = \frac{\mu_0 I}{2\pi (d - x)}$ (directed out of the page).
The net magnetic field is $B = B_1 - B_2 = \frac{\mu_0 I}{2\pi} \left( \frac{1}{x} - \frac{1}{d - x} \right)$.
At the midpoint $x = d/2$,$B = \frac{\mu_0 I}{2\pi} \left( \frac{1}{d/2} - \frac{1}{d/2} \right) = 0$.
As $x \to 0$,$B \to \infty$,and as $x \to d$,$B \to -\infty$.
This behavior is correctly represented by the graph in option $A$.

(B) Let the current in the thin conductor be $I$ and the current in the hollow conductor be $I$. Both currents are in the same direction.
For $r < R$,the magnetic field $B$ is due only to the thin conductor at the axis. Using Ampere's circuital law,$\oint B \cdot dl = \mu_0 I_{enclosed}$,we get $B(2\pi r) = \mu_0 I$,so $B = \frac{\mu_0 I}{2\pi r}$. Thus,$B \propto \frac{1}{r}$.
For $r > R$,the total current enclosed by an Amperian loop of radius $r$ is $I + I = 2I$. Using Ampere's circuital law,$B(2\pi r) = \mu_0 (2I)$,so $B = \frac{\mu_0 (2I)}{2\pi r} = \frac{\mu_0 I}{\pi r}$. Thus,$B \propto \frac{1}{r}$.
At $r = R$,the magnetic field due to the hollow conductor is zero inside $(r < R)$ and the field due to the thin conductor is $\frac{\mu_0 I}{2\pi R}$. Just outside $(r > R)$,the field is the sum of the fields from both conductors,which is $\frac{\mu_0 I}{2\pi R} + \frac{\mu_0 I}{2\pi R} = \frac{\mu_0 I}{\pi R}$.
Therefore,the magnetic field $B$ jumps from $\frac{\mu_0 I}{2\pi R}$ to $\frac{\mu_0 I}{\pi R}$ at $r = R$,and follows a $1/r$ dependence in both regions. This corresponds to the graph where there is a discontinuous jump at $r = R$ and both segments follow a $1/r$ curve.

(A) Consider a small element of the pipe subtending an angle $d\theta$ at the center. The current in this element is $dI = \frac{I}{2\pi} d\theta$.
The magnetic field $dB$ at the center due to this long current-carrying strip (acting as an infinite wire) is given by $dB = \frac{\mu_0 (dI)}{2\pi R} = \frac{\mu_0 I}{4\pi^2 R} d\theta$.
The direction of $dB$ is perpendicular to the radius vector of the element. For the quarter portion,the angle $\theta$ ranges from $0$ to $\pi/2$. Resolving the field into components:
$dB_x = dB \sin\theta$ and $dB_y = -dB \cos\theta$.
Integrating these components from $0$ to $\pi/2$:
$B_x = \int_0^{\pi/2} \frac{\mu_0 I}{4\pi^2 R} \sin\theta d\theta = \frac{\mu_0 I}{4\pi^2 R} [-\cos\theta]_0^{\pi/2} = \frac{\mu_0 I}{4\pi^2 R}$.
$B_y = \int_0^{\pi/2} \frac{\mu_0 I}{4\pi^2 R} \cos\theta d\theta = \frac{\mu_0 I}{4\pi^2 R} [\sin\theta]_0^{\pi/2} = \frac{\mu_0 I}{4\pi^2 R}$.
The magnitude of the resultant magnetic field is $B = \sqrt{B_x^2 + B_y^2} = \sqrt{(\frac{\mu_0 I}{4\pi^2 R})^2 + (\frac{\mu_0 I}{4\pi^2 R})^2} = \frac{\mu_0 I \sqrt{2}}{4\pi^2 R}$.

(C) Let a point $P$ be at a distance $x'$ from the wire at $(-d, 0, 0)$. The distance from the wire at $(d, 0, 0)$ is $(2d - x')$.
Using the right-hand rule,the magnetic field $B_z$ at point $P$ due to the wire at $(-d, 0, 0)$ carrying current $I$ in $-y$ direction is $B_1 = \frac{\mu_0 I}{2 \pi x'}$ (directed into the page,i.e.,$-z$ direction).
The magnetic field $B_z$ due to the wire at $(d, 0, 0)$ carrying current $4I$ in $+y$ direction is $B_2 = \frac{\mu_0 (4I)}{2 \pi (2d - x')}$ (directed into the page,i.e.,$-z$ direction).
Since both fields are in the same direction,the total field is $B_z = -\frac{\mu_0}{2 \pi} \left( \frac{I}{x'} + \frac{4I}{2d - x'} \right)$.
As $x' \to 0$ or $x' \to 2d$,$B_z \to -\infty$. The magnitude $|B_z|$ has a minimum where $\frac{d|B_z|}{dx'} = 0$.
$\frac{d}{dx'} \left( \frac{1}{x'} + \frac{4}{2d - x'} \right) = -\frac{1}{(x')^2} + \frac{4}{(2d - x')^2} = 0 \Rightarrow (2d - x')^2 = 4(x')^2 \Rightarrow 2d - x' = 2x' \Rightarrow x' = \frac{2d}{3}$.
At $x' = \frac{2d}{3}$,the field magnitude is minimum. The graph shows a $U$-shaped curve with vertical asymptotes at the wire positions,which matches option $C$.

(D) For an electric field:
The force on the electron is $F = qE$.
This force acts as the centripetal force,so $F = \frac{mv_0^2}{r_1}$.
Equating the two,$qE = \frac{mv_0^2}{r_1}$,which gives $r_1 = \frac{mv_0^2}{qE}$.
For a magnetic field:
The magnetic force on the electron is $F = qv_0B$ (since velocity is perpendicular to the field).
This force acts as the centripetal force,so $F = \frac{mv_0^2}{r_2}$.
Equating the two,$qv_0B = \frac{mv_0^2}{r_2}$,which gives $r_2 = \frac{mv_0}{qB}$.
Calculating the ratio:
$\frac{r_1}{r_2} = \frac{\frac{mv_0^2}{qE}}{\frac{mv_0}{qB}} = \frac{mv_0^2}{qE} \times \frac{qB}{mv_0} = \frac{Bv_0}{E}$.
Thus,the ratio $r_1:r_2$ is $\frac{Bv_0}{E}$.

(A) The electric force $F_e$ between two protons separated by distance $r$ is given by Coulomb's Law: $F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$.
The magnetic force $F_m$ between two parallel moving charges is given by $F_m = qvB$,where $B$ is the magnetic field produced by one proton at the location of the other: $B = \frac{\mu_0}{4\pi} \frac{qv}{r^2}$.
Thus,$F_m = qv \left( \frac{\mu_0}{4\pi} \frac{qv}{r^2} \right) = \frac{\mu_0}{4\pi} \frac{q^2 v^2}{r^2}$.
Taking the ratio of electric force to magnetic force:
$\frac{F_e}{F_m} = \frac{\frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}}{\frac{\mu_0}{4\pi} \frac{q^2 v^2}{r^2}} = \frac{1}{\epsilon_0 \mu_0 v^2}$.
Since the speed of light $c$ is defined as $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,we have $c^2 = \frac{1}{\mu_0 \epsilon_0}$.
Substituting this into the ratio,we get $\frac{F_e}{F_m} = \frac{c^2}{v^2}$.

(C) The speed of the charged particle just before entering the magnetic field is $v_0$.
By the work-energy theorem,the kinetic energy gained is equal to the work done by the electric field:
$qV = \frac{1}{2}mv_0^2$
Solving for $v_0$:
$v_0 = \sqrt{\frac{2qV}{m}}$
The radius $r$ of the circular path in the magnetic field is given by:
$r = \frac{mv_0}{qB}$
From the figure,the distance $d$ between the holes is the diameter of the circular path,so $r = \frac{d}{2}$.
Substituting this into the radius formula:
$\frac{d}{2} = \frac{m}{qB} \sqrt{\frac{2qV}{m}}$
Squaring both sides:
$\frac{d^2}{4} = \frac{m^2}{q^2B^2} \cdot \frac{2qV}{m}$
Simplifying the expression:
$\frac{d^2}{4} = \frac{2mV}{qB^2}$
Solving for $m$:
$m = \frac{qB^2d^2}{8V}$

(C) The electric field $\vec{E}$ is along the $+y$ direction,so the acceleration of the particle is $a_y = (qE/m)$.
For the particle to return to the origin,its displacement along the $y$-axis must be zero at some time $t > 0$.
Using the equation of motion along the $y$-axis: $y = u_y t + \frac{1}{2} a_y t^2$.
Setting $y = 0$ and $u_y = -v$,we get: $0 = -vt + \frac{1}{2} (qE/m) t^2$.
Solving for $t$,we find: $t = \frac{2mv}{qE}$.
In this time $t$,the particle must complete an integer number of revolutions in the $x$-$z$ plane due to the magnetic field $\vec{B}$ along the $y$-axis.
The time period of revolution in the magnetic field is $T = \frac{2\pi m}{qB}$.
For the particle to return to the origin,the time $t$ must be an integer multiple of the time period $T$,i.e.,$t = nT$ where $n = 1, 2, 3, \dots$.
Substituting the expressions for $t$ and $T$: $\frac{2mv}{qE} = n \left( \frac{2\pi m}{qB} \right)$.
Simplifying this,we get: $\frac{v}{E} = n \frac{\pi}{B}$,which implies $\frac{vB}{\pi E} = n$.
Thus,$[vB / \pi E]$ must be an integer.

(B) The radius of the circular path of a charged particle in a uniform magnetic field is given by $r = \frac{mv}{qB}$.
From the figure,the radius $r$ of the path is decreasing as the particle moves.
Statement $[1]$: If the magnetic field strength $B$ increases,the radius $r$ decreases $(r \propto \frac{1}{B})$. Thus,this is a possible explanation.
Statement $[2]$: As the particle moves through air,it loses kinetic energy $(K = \frac{1}{2}mv^2)$ due to collisions and ionization of air molecules. Since $r = \frac{\sqrt{2mK}}{qB}$,a decrease in kinetic energy $K$ leads to a decrease in radius $r$. Thus,this is a possible explanation.
Statement $[3]$: If the particle loses charge $q$,the radius $r$ would increase $(r \propto \frac{1}{q})$,which contradicts the observed path. Therefore,this statement is incorrect.
Thus,statements $[1]$ and $[2]$ are correct.

(D) The magnetic field at the centre of a closed loop carrying current is zero if the current distribution is symmetric such that the magnetic field produced by each segment is cancelled by an equal and opposite field from another segment.
For a hexagon,the magnetic field at the centre is zero if the current in each segment is such that the net magnetic field vectors sum to zero.
If the hexagon is made of two different materials $P$ and $Q$ with different resistances,the current will divide unequally between the two paths connecting $X$ and $Y$.
If the arrangement of materials $P$ and $Q$ is not symmetric with respect to the entry and exit points $X$ and $Y$,the currents in the segments will not be such that their magnetic fields cancel each other out.
In option $A$,the arrangement is symmetric. In option $B$,the arrangement is symmetric. In option $C$,the arrangement is symmetric. In option $D$,the arrangement is not symmetric with respect to the points $X$ and $Y$ because the distribution of materials $P$ and $Q$ along the two paths is different,leading to unequal currents and a non-zero magnetic field at the centre.

(C) For a square frame where current enters and leaves at opposite corners (as in option $A$ and $B$),the symmetry of the current distribution ensures that the magnetic field produced by each segment of the wire at the centre cancels out,resulting in a net magnetic field of $0$.
In option $C$,the current enters at one corner and leaves at the adjacent corner. The current divides into two paths of different lengths. Let the total current be $I$. The current in the shorter path is $i_1 = \frac{3}{4}I$ and in the longer path is $i_2 = \frac{1}{4}I$.
The magnetic field at the centre due to a straight wire segment of length $L$ carrying current $i$ at a perpendicular distance $d = L/2$ is given by $B = \frac{\mu_0 i}{4\pi d}(\sin \theta_1 + \sin \theta_2)$. For each side of the square,$\theta_1 = \theta_2 = 45^\circ$.
Because the currents $i_1$ and $i_2$ are unequal,the magnetic fields produced by the segments do not cancel each other out. Specifically,the contributions from the segments do not sum to zero because the current distribution is asymmetric with respect to the centre. Therefore,the magnetic field at the centre is not zero.

(D) Option $(A)$ is incorrect because a charged particle entering a magnetic field at an angle other than $90^o$ follows a helical path,not a circular one.
Option $(B)$ is correct. When a charged particle enters a uniform magnetic field perpendicular to its velocity,it experiences a magnetic force $F = q(v \times B)$ which acts as a centripetal force,causing the particle to move in a circular path. Both the electron and proton will follow circular paths.
Option $(C)$ is correct. The magnetic force acting on a moving charged particle is always perpendicular to its velocity $(F = q(v \times B))$. Since the force is perpendicular to the displacement,the work done by the magnetic force is zero $(W = F \cdot ds = 0)$. According to the work-energy theorem,the kinetic energy of the particle remains constant.
Since both $(B)$ and $(C)$ are correct,option $(D)$ is the correct answer.

(C) In a velocity selector,the electric force and magnetic force must balance for the particle to pass undeflected: $qE = qvB$,so $E = vB$.
Given $v = 1.5 \times 10^6 \, ms^{-1}$ and $B = 1.0 \, T$,we have $E = (1.5 \times 10^6)(1.0) = 1.5 \times 10^6 \, Vm^{-1}$.
Since the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ acts in the $+x$ direction (using right-hand rule for $\vec{v} = v\hat{j}$ and $\vec{B} = -B\hat{k}$),the electric force must act in the $-x$ direction. Thus,$\vec{E} = -1.5 \times 10^6 \hat{i} \, NC^{-1}$.
For the circular motion in chamber $X$,the radius is given by $r = \frac{mv}{qB}$.
Since $v, q,$ and $B$ are the same for both particles,$r \propto m$.
Therefore,$\frac{r_A}{r_B} = \frac{m_A}{m_B} = \frac{12}{13}$.
Thus,option $C$ is correct.

(A) Consider a small elemental ring of radius $x$ and thickness $dx$ on the disc.
The surface charge density $\sigma = \frac{Q}{\pi R^2}$.
The charge on the elemental ring is $dq = \sigma (2 \pi x dx) = \frac{Q}{\pi R^2} (2 \pi x dx) = \frac{2Qx dx}{R^2}$.
The current $di$ due to the rotation of this ring with angular velocity $\omega$ is $di = \frac{dq}{T} = \frac{dq \omega}{2 \pi} = \frac{2Qx dx}{R^2} \cdot \frac{\omega}{2 \pi} = \frac{Q \omega x dx}{\pi R^2}$.
The magnetic field $dB$ at the centre due to this ring is $dB = \frac{\mu_0 di}{2x} = \frac{\mu_0}{2x} \cdot \frac{Q \omega x dx}{\pi R^2} = \frac{\mu_0 Q \omega}{2 \pi R^2} dx$.
Integrating from $x = 0$ to $x = R$:
$B = \int_0^R \frac{\mu_0 Q \omega}{2 \pi R^2} dx = \frac{\mu_0 Q \omega}{2 \pi R^2} [x]_0^R = \frac{\mu_0 Q \omega}{2 \pi R^2} \cdot R = \frac{\mu_0 Q \omega}{2 \pi R}$.
Thus,$B \propto \frac{1}{R}$.
This relationship is represented by a rectangular hyperbola,which corresponds to Figure $A$.

(A) Let us consider a length $l$ of the current-carrying wire.
At equilibrium,the forces acting on the wire are tension $T$,gravitational force $(\lambda l)g$,and magnetic force $F_B$.
The distance between the two wires is $r = 2L \sin \theta$.
The magnetic force per unit length is $\frac{F_B}{l} = \frac{\mu_0 I^2}{2 \pi r} = \frac{\mu_0 I^2}{2 \pi (2L \sin \theta)} = \frac{\mu_0 I^2}{4 \pi L \sin \theta}$.
Resolving forces in the vertical and horizontal directions:
$T \cos \theta = \lambda l g$
$T \sin \theta = F_B = \frac{\mu_0 I^2 l}{4 \pi L \sin \theta}$
Dividing the two equations:
$\tan \theta = \frac{F_B}{\lambda l g} = \frac{\mu_0 I^2 l}{4 \pi L \sin \theta \cdot \lambda l g} = \frac{\mu_0 I^2}{4 \pi \lambda g L \sin \theta}$
Solving for $I$:
$I^2 = \frac{4 \pi \lambda g L \sin \theta \tan \theta}{\mu_0} = \frac{4 \pi \lambda g L \sin^2 \theta}{\mu_0 \cos \theta}$
$I = 2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}$.

(A) The force on a current-carrying conductor in a magnetic field is $\vec{F} = I(\vec{L} \times \vec{B})$.
Here,$\vec{L} = 3 \hat{a}_z \text{ m}$ (length of the conductor from $-1.5$ to $1.5$) and $I = 10 \text{ A}$ in the $-\hat{a}_z$ direction,so $\vec{I} = -10 \hat{a}_z \text{ A}$.
The magnetic field is $\vec{B} = 3.0 \times 10^{-4} e^{-0.2x} \hat{a}_y \text{ T}$.
The magnetic force is $\vec{F} = I \vec{L} \times \vec{B} = (-10 \hat{a}_z) \times (3.0 \times 10^{-4} e^{-0.2x} \hat{a}_y) = -30 \times 10^{-4} e^{-0.2x} (\hat{a}_z \times \hat{a}_y) = 3.0 \times 10^{-3} e^{-0.2x} \hat{a}_x \text{ N}$.
To move the conductor at a constant speed,an external force $\vec{F}_{ext} = -\vec{F} = -3.0 \times 10^{-3} e^{-0.2x} \hat{a}_x \text{ N}$ must be applied.
The work done is $W = \int_0^2 F_{ext} dx = \int_0^2 3.0 \times 10^{-3} e^{-0.2x} dx$.
$W = 3.0 \times 10^{-3} \left[ \frac{e^{-0.2x}}{-0.2} \right]_0^2 = \frac{3.0 \times 10^{-3}}{-0.2} (e^{-0.4} - 1) = 0.015 (1 - e^{-0.4}) \approx 0.015 (1 - 0.6703) = 0.0049455 \text{ J}$.
The power is $P = \frac{W}{t} = \frac{0.0049455}{5 \times 10^{-3}} = 0.989 \text{ W}$.
Re-evaluating the integral based on the provided options: If the force magnitude is taken as $F = I L B = 10 \times 3 \times 3 \times 10^{-4} e^{-0.2x} = 9 \times 10^{-3} e^{-0.2x}$,then $W = \int_0^2 9 \times 10^{-3} e^{-0.2x} dx = 9 \times 10^{-3} [\frac{e^{-0.2x}}{-0.2}]_0^2 = 0.045(1 - e^{-0.4}) \approx 0.01485 \text{ J}$.
$P = \frac{0.01485}{5 \times 10^{-3}} = 2.97 \text{ W}$.

(D) Given: $B = 1\, T$,$E = 1\, V/m$,$m = 1\, kg$,$q = 1\, C$,$\vec{v}_0 = 1\hat{i}\, m/s$. The electric field is along the $y-$ axis,so the acceleration in the $y-$ direction is $a_y = \frac{qE}{m} = \frac{1 \times 1}{1} = 1\, m/s^2$. The position in the $y-$ direction at time $t$ is $y(t) = v_{0y}t + \frac{1}{2}a_yt^2 = 0 + \frac{1}{2}(1)t^2 = \frac{t^2}{2}$. At $t = \pi$,$y = \frac{\pi^2}{2}$.
The magnetic field is along the $y-$ axis. The Lorentz force $\vec{F} = q(\vec{v} \times \vec{B})$ causes circular motion in the $x-z$ plane. The radius of this circular path is $R = \frac{mv_{\perp}}{qB} = \frac{1 \times 1}{1 \times 1} = 1\, m$. The angular frequency is $\omega = \frac{qB}{m} = 1\, rad/s$. The position in the $x-z$ plane is given by $x(t) = R \sin(\omega t)$ and $z(t) = R(1 - \cos(\omega t))$ (since it starts at origin with velocity along $x$). At $t = \pi$,$x(\pi) = 1 \sin(\pi) = 0$ and $z(\pi) = 1(1 - \cos(\pi)) = 1(1 - (-1)) = 2$. Thus,the coordinates are $(0, \pi^2/2, 2)$.

(A) The net force on a moving charge in the presence of both electric and magnetic fields is given by the Lorentz force equation: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
For the net force to be zero,we must have $\vec{F} = 0$,which implies $\vec{E} = -(\vec{v} \times \vec{B})$.
This means the electric force $\vec{F}_e = q\vec{E}$ and the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ must be equal in magnitude and opposite in direction.
In Figure $A$,the velocity $\vec{v}$ is along the $+x$ direction,and the magnetic field $\vec{B}$ is along the $+y$ direction. The magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ will be in the $+z$ direction (using the right-hand rule).
The electric field $\vec{E}$ is shown along the $-z$ direction. Thus,the electric force $\vec{F}_e = q\vec{E}$ will be in the $-z$ direction.
Since the electric force and magnetic force are in opposite directions,they can cancel each other out if their magnitudes are equal. Therefore,the net force can be zero in Figure $A$.

(C) The magnetic field due to an infinite wire at a distance $r$ is $B = \frac{\mu_0 i}{2\pi r}$.
$1$. Wire $(1)$ along the $x$-axis carries current $i$ in the $+x$ direction. At point $A(r, -r, 0)$,the position vector is $r\hat{i} - r\hat{j}$. The magnetic field is given by $\vec{B}_1 = \frac{\mu_0 i}{2\pi r} (\hat{i} \times \hat{n})$,where $\hat{n}$ is the unit vector from the wire to $A$. The direction is $-\hat{k}$. Thus,$\vec{B}_1 = \frac{\mu_0 i}{2\pi r} (-\hat{k})$.
$2$. Wire $(2)$ along the $y$-axis carries current $i$ in the $+y$ direction. At point $A(r, -r, 0)$,the magnetic field is $\vec{B}_2 = \frac{\mu_0 i}{2\pi r} (\hat{k})$.
$3$. Wire $(3)$ along the $z$-axis carries current $i$ out of the plane ($+\hat{k}$ direction). The distance from the $z$-axis to $A$ is $d = \sqrt{r^2 + (-r)^2} = r\sqrt{2}$. The magnetic field magnitude is $B_3 = \frac{\mu_0 i}{2\pi (r\sqrt{2})}$. The direction is perpendicular to the position vector $\vec{r} = r\hat{i} - r\hat{j}$,which is $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$. Thus,$\vec{B}_3 = \frac{\mu_0 i}{2\pi r\sqrt{2}} \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \right) = \frac{\mu_0 i}{4\pi r} \hat{i} + \frac{\mu_0 i}{4\pi r} \hat{j}$.
Summing the fields: $\vec{B}_{net} = \vec{B}_1 + \vec{B}_2 + \vec{B}_3 = 0 + \frac{\mu_0 i}{4\pi r} \hat{i} + \frac{\mu_0 i}{4\pi r} \hat{j} = \frac{\mu_0 i}{4\pi r} \hat{i} + \frac{\mu_0 i}{4\pi r} \hat{j}$.
Wait,re-evaluating the directions: Wire $(1)$ is at $y=0$,$z=0$. Point $A$ is at $(r, -r, 0)$. The vector from wire to $A$ is $-r\hat{j}$. $\vec{B}_1 = \frac{\mu_0 i}{2\pi r} (\hat{i} \times (-\hat{j})) = -\frac{\mu_0 i}{2\pi r} \hat{k}$.
Wire $(2)$ is at $x=0$,$z=0$. Point $A$ is at $(r, -r, 0)$. The vector from wire to $A$ is $r\hat{i}$. $\vec{B}_2 = \frac{\mu_0 i}{2\pi r} (\hat{j} \times \hat{i}) = -\frac{\mu_0 i}{2\pi r} \hat{k}$.
Wire $(3)$ is at $x=0$,$y=0$. $\vec{B}_3 = \frac{\mu_0 i}{2\pi (r\sqrt{2})} \frac{\hat{i} + \hat{j}}{\sqrt{2}} = \frac{\mu_0 i}{4\pi r} \hat{i} + \frac{\mu_0 i}{4\pi r} \hat{j}$.
Total $\vec{B} = \frac{\mu_0 i}{4\pi r} \hat{i} + \frac{\mu_0 i}{4\pi r} \hat{j} - \frac{\mu_0 i}{\pi r} \hat{k}$.

(D) $1$. According to the Biot-Savart Law,the magnetic field $d\vec{B}$ due to a current element $I d\vec{l}$ is given by $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \vec{r}}{r^3}$. Since it involves a cross product,$d\vec{B}$ is perpendicular to both $d\vec{l}$ and the position vector $\vec{r}$. Thus,statement $A$ is correct.
$2$. The electric field $\vec{E}$ of a point charge $q$ is given by $\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}$,where $\hat{r}$ is the unit vector along the position vector. Thus,the electric field is directed along the position vector. Statement $B$ is correct.
$3$. Magnetic monopoles have never been observed in nature; magnetic fields are always produced by dipoles or current loops. Thus,statement $C$ is correct.
Since all statements are correct,the correct option is $D$.