Magnetic Flux and Gauss law for Magnetism Questions in English

(C) The $Weber$ $(Wb)$ is defined as the $SI$ unit of magnetic flux. It is the amount of magnetic flux which,when linked with a circuit of one turn,would produce an electromotive force of $1 \ V$ if it were reduced to zero at a uniform rate in $1 \ s$.

(B) The $SI$ unit of Electric field is $Newton/Coulomb$ $(N/C)$ or $Volt/meter$ $(V/m)$.
The $SI$ unit of Magnetic flux is $Weber$ $(Wb)$.
The $SI$ unit of Power is $Watt$ $(W)$.
The $SI$ unit of Capacitance is $Farad$ $(F)$.
Comparing the given options,Magnetic flux is correctly matched with $Weber$.

(A) Magnetic flux $\phi$ is defined as the product of magnetic field $B$ and area $A$,i.e.,$\phi = B \cdot A$.
From the Lorentz force formula,$F = B I L$,we can write $B = \frac{F}{I L}$.
Substituting the dimensions of force $[F] = [M L T^{-2}]$,current $[I] = [A]$,and length $[L] = [L]$,we get the dimensions of $B$ as $[B] = \frac{[M L T^{-2}]}{[A] [L]} = [M T^{-2} A^{-1}]$.
Now,the dimensions of magnetic flux $\phi$ are $[B] \times [A] = [M T^{-2} A^{-1}] \times [L^2] = [M L^2 T^{-2} A^{-1}]$.

(A) The magnetic flux $\Phi$ is defined as the product of magnetic induction $B$ and the area $A$ perpendicular to the magnetic field,given by the formula: $\Phi = B \times A$.
From this,we can express magnetic induction $B$ as: $B = \frac{\Phi}{A}$.
Since the unit of magnetic flux $\Phi$ is $Weber$ $(Wb)$ and the unit of area $A$ is $m^2$,the unit of magnetic induction $B$ is $Weber/m^2$.

(A) The magnetic flux $\phi$ through a surface is given by the formula $\phi = B \cdot A \cdot \cos(\theta)$,where $B$ is the magnetic field intensity,$A$ is the area of the coil,and $\theta$ is the angle between the magnetic field vector and the area vector.
Since the coil is placed perpendicular to the magnetic field,the area vector (which is normal to the surface) is parallel to the magnetic field vector. Thus,$\theta = 0^\circ$ and $\cos(0^\circ) = 1$.
Given: $B = 10^3 \ Wb/m^2$ and $A = 10^{-2} \ m^2$.
Calculating the flux: $\phi = (10^3) \times (10^{-2}) \times \cos(0^\circ) = 10^1 = 10 \ Wb$.
Therefore,the correct option is $A$.

(B) The magnetic flux $\Phi_B$ through a surface is defined as the product of the magnetic field $B$ and the area $A$ perpendicular to it,given by $\Phi_B = B \cdot A$.
The $SI$ unit of magnetic field is Tesla $(T)$ and the $SI$ unit of area is square meter $(m^2)$.
Thus,the unit of magnetic flux is $Tesla \cdot m^2$,which is defined as the Weber $(Wb)$.
Therefore,the correct option is $(b)$.

(D) Magnetic flux is defined as $\phi = B \cdot A$.
From the Lorentz force law,$F = B I l$,we can write $B = \frac{F}{I l}$.
The dimensions of force $F$ are $[M L T^{-2}]$,current $I$ are $[A]$,and length $l$ is $[L]$.
Thus,the dimensions of magnetic field $B$ are $\frac{[M L T^{-2}]}{[A][L]} = [M T^{-2} A^{-1}]$.
Now,substituting this into the formula for magnetic flux:
$\phi = [M T^{-2} A^{-1}] \cdot [L^2] = [M L^2 T^{-2} A^{-1}]$.

(A) The magnetic flux $\phi$ linked with a coil of $N$ turns is given by the formula: $\phi = N B A \cos \theta$,where $\theta$ is the angle between the normal to the coil and the magnetic field direction.
Given:
$N = 100$
$B = 0.2 \text{ T}$
$A = 5 \text{ cm}^2 = 5 \times 10^{-4} \text{ m}^2$
$\theta = 60^o$
Substituting the values:
$\phi = 100 \times 0.2 \times (5 \times 10^{-4}) \times \cos(60^o)$
$\phi = 20 \times 5 \times 10^{-4} \times 0.5$
$\phi = 100 \times 10^{-4} \times 0.5$
$\phi = 0.5 \times 10^{-2} \text{ Wb} = 5 \times 10^{-3} \text{ Wb}$.

(C) Magnetic flux is defined as the measure of the total magnetic field passing through a given surface area.
Mathematically,for a uniform magnetic field $\overrightarrow{B}$ passing through a surface with area vector $\overrightarrow{A}$,the magnetic flux $\Phi$ is defined as the scalar product (dot product) of the magnetic field vector and the area vector.
Therefore,$\Phi = \overrightarrow{B} \cdot \overrightarrow{A} = BA \cos \theta$,where $\theta$ is the angle between the magnetic field vector and the normal to the surface area.
Thus,the correct expression is $\overrightarrow{B} \cdot \overrightarrow{A}$.

(D) circular loop carrying current $I$ behaves as a magnetic dipole.
Magnetic field lines emerge from the North pole and enter the South pole.
Since the loop lies in the $x-y$ plane,the magnetic field lines pass through the loop and then loop back around to form closed paths.
For every magnetic field line that passes through the $x-y$ plane in one direction (e.g.,upwards),an equal number of field lines must pass back through the $x-y$ plane in the opposite direction (e.g.,downwards) to complete the closed loop.
Therefore,the net magnetic flux through the entire $x-y$ plane is zero.

(C) The magnetic flux $\phi$ is given by the formula $\phi = \vec{B} \cdot \vec{A} = BA \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (which is normal to the surface).
Given,radius $R = 0.2 \; m$,so the area $A = \pi R^2 = \pi (0.2)^2 = 0.04\pi \; m^2$.
The magnetic field $B = \frac{1}{\pi} \; Wb/m^2$.
The axis of the disc makes an angle of $60^{\circ}$ with $\vec{B}$. Since the area vector $\vec{A}$ is along the axis of the disc,the angle $\theta$ between $\vec{B}$ and $\vec{A}$ is $60^{\circ}$.
Therefore,$\phi = B A \cos 60^{\circ} = \left(\frac{1}{\pi}\right) \times (0.04\pi) \times \cos 60^{\circ}$.
$\phi = 0.04 \times \frac{1}{2} = 0.02 \; Wb$.

(A) Magnetic flux $\Phi$ through a surface area $A$ in a uniform magnetic field $B$ is given by $\Phi = B \cdot A \cdot \cos(\theta)$. Since the magnetic field is perpendicular to the plane of the loops,$\theta = 0^\circ$ and $\cos(0^\circ) = 1$,so $\Phi = B \cdot A$.
In Case $I$,the total area enclosed by the wire frame is the sum of the areas of the two squares: $A_I = L^2 + l^2$. Thus,$\Phi_I = (L^2 + l^2)B$.
In Case $II$,the smaller square is inside the larger square. The effective area enclosed by the wire frame is the area of the larger square minus the area of the smaller square: $A_{II} = L^2 - l^2$. Thus,$\Phi_{II} = (L^2 - l^2)B$.
Therefore,the correct option is $A$.

(A) As the coil moves from above the magnet towards the center,the magnetic flux increases,reaches a maximum at the center,and then decreases as it moves away to the bottom. When moving back from bottom to top,the flux changes in the opposite direction (or follows a symmetric path depending on the orientation). The magnetic flux $\Phi$ is defined as $\int \vec{B} \cdot d\vec{A}$. As the coil passes through the magnet,the field lines first point in one direction and then in the opposite direction relative to the coil's area vector. Graph $A$ shows the flux increasing to a positive maximum,passing through zero (at the center of the magnet where field lines are parallel to the coil plane),and then reaching a negative minimum as it moves to the other side. This represents the change in flux correctly for a coil moving through the magnetic field of a bar magnet.

(A) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $i$ is given by $B = \frac{\mu_0 i}{2 \pi r}$.
The area $A$ under the $B-r$ curve between $r_1$ and $r_2$ is defined as $A = \int_{r_1}^{r_2} B \, dr = \int_{r_1}^{r_2} \frac{\mu_0 i}{2 \pi r} \, dr = \frac{\mu_0 i}{2 \pi} \ln\left(\frac{r_2}{r_1}\right)$.
The magnetic flux $\phi$ through the rectangular loop of length $l$ is given by $\phi = \int B \, dA_{loop}$.
Considering a strip of width $dr$ at a distance $r$ from the wire,the area of the strip is $dA_{loop} = l \, dr$.
Thus,the flux is $\phi = \int_{r_1}^{r_2} B \, l \, dr = l \int_{r_1}^{r_2} B \, dr$.
Since $A = \int_{r_1}^{r_2} B \, dr$,we substitute this into the flux equation:
$\phi = l \cdot A = Al$.

(B) The magnetic flux $\phi$ is given by the formula $\phi = \vec{B} \cdot \vec{A} = BA \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (which is normal to the surface).
From the figure,the angle between the surface and the vertical is $60^\circ$. The area vector $\vec{A}$ is perpendicular to the surface,so the angle between the horizontal magnetic field $\vec{B}$ and the area vector $\vec{A}$ is $\theta = 60^\circ$.
Given: $\phi = 2 \times 10^{-3} \ Wb$,$A = 10 \ cm^2 = 10 \times 10^{-4} \ m^2 = 10^{-3} \ m^2$,and $\theta = 60^\circ$.
Using $\phi = BA \cos \theta$,we have $B = \frac{\phi}{A \cos \theta}$.
Substituting the values: $B = \frac{2 \times 10^{-3}}{(10^{-3}) \cos 60^\circ} = \frac{2 \times 10^{-3}}{(10^{-3}) \times (1/2)} = \frac{2}{1/2} = 4 \ Wb/m^2$.

(D) The loop lies in the $xy$-plane with corners at $(L, L, 0)$,$(-L, L, 0)$,$(-L, -L, 0)$,and $(L, -L, 0)$.
The side length of the square loop is $2L$.
Therefore,the area of the loop is $A = (2L) \times (2L) = 4L^2$.
The area vector $\vec A$ is perpendicular to the $xy$-plane,so $\vec A = 4L^2 \hat k$.
The magnetic field is given by $\vec B = B_0(\hat i + \hat k)$.
The magnetic flux $\phi$ is given by the dot product of the magnetic field and the area vector:
$\phi = \vec B \cdot \vec A$
$\phi = B_0(\hat i + \hat k) \cdot (4L^2 \hat k)$
$\phi = 4B_0L^2 (\hat i \cdot \hat k + \hat k \cdot \hat k)$
Since $\hat i \cdot \hat k = 0$ and $\hat k \cdot \hat k = 1$,we get:
$\phi = 4B_0L^2 (0 + 1) = 4B_0L^2 \text{ Wb}$.

(D) The magnetic flux $\phi$ linked with a coil is given by the formula $\phi = NBA \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (which is normal to the plane of the coil).
Given that the plane of the coil is parallel to the magnetic field $B$,the angle between the normal to the coil and the magnetic field is $\theta = 90^{\circ}$.
Substituting this into the formula: $\phi = NBA \cos(90^{\circ})$.
Since $\cos(90^{\circ}) = 0$,the magnetic flux $\phi = 0$.

(A) The area vector $\vec A$ of a square of side $x$ lying in the $x-y$ plane is given by $\vec A = x^2 \hat k \, m^2$.
The magnetic field is given as $\vec B = B_0 (3\hat i + 4\hat j + 5\hat k) \, T$.
The magnetic flux $\phi$ is defined as the dot product of the magnetic field vector and the area vector: $\phi = \vec B \cdot \vec A$.
Substituting the values: $\phi = [B_0 (3\hat i + 4\hat j + 5\hat k)] \cdot [x^2 \hat k]$.
Using the dot product properties $(\hat i \cdot \hat k = 0, \hat j \cdot \hat k = 0, \hat k \cdot \hat k = 1)$:
$\phi = B_0 x^2 (3 \cdot 0 + 4 \cdot 0 + 5 \cdot 1) = 5 B_0 x^2 \, Wb$.

(C) The magnetic flux $\phi$ is given by $\phi = \vec{B} \cdot \vec{A} = B A \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (normal to the surface).
Given: Magnetic field $B = 0.2 \, T$ along the positive $x-$axis.
The top surface is a rectangle with dimensions $10 \, cm \times 10 \, cm = 0.1 \, m \times 0.1 \, m = 0.01 \, m^2$.
The normal to the top surface makes an angle of $60^{\circ}$ with the vertical,which means it makes an angle of $90^{\circ} - 60^{\circ} = 30^{\circ}$ with the $x-$axis (or based on the geometry,the angle $\theta$ between the normal and the $x-$axis is $60^{\circ}$ as given in the diagram).
Using $\theta = 60^{\circ}$:
$\phi = B A \cos 60^{\circ} = 0.2 \times 0.01 \times \cos 60^{\circ} = 0.2 \times 0.01 \times 0.5 = 0.001 \, Wb$.
Converting to $m-Wb$: $0.001 \, Wb = 1 \, m-Wb$.

(B) The magnetic field lines originate from the North pole and terminate at the South pole of a bar magnet.
In the given figure,the South pole of the bar magnet is facing the open end of the cylinder.
Therefore,the magnetic field lines are directed towards the South pole,meaning they enter the cylinder through the open end.
By convention,the area vector $d\vec A$ for a closed surface is taken to be pointing outwards.
Since the magnetic field $\vec B$ is entering the surface and the area vector $d\vec A$ is pointing outwards,the angle between them is obtuse (greater than $90^{\circ}$).
Consequently,the dot product $\vec B \cdot d\vec A = B \, dA \cos \theta$ will be negative because $\cos \theta < 0$ for $\theta > 90^{\circ}$.

(B) The equation $\oint \vec{B} \cdot d\vec{A} = 0$ represents Gauss's Law for magnetism.
It states that the net magnetic flux through any closed surface is always zero.
This implies that there are no isolated magnetic charges (magnetic monopoles) in nature.
Magnetic field lines always form closed loops,meaning every north pole must be accompanied by a south pole.

(C) The magnetic flux $\phi$ is given by the formula $\phi = B A \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{n}$ (normal to the surface).
Given that the area makes an angle of $60^o$ with the magnetic field,the angle $\theta$ between the normal to the area and the magnetic field is $\theta = 90^o - 60^o = 30^o$.
Substituting the values: $\phi = 2.0 \times 0.5 \times \cos(30^o)$.
$\phi = 1.0 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \, Wb$.

(A) For a current-carrying circular coil,the magnetic field lines originate from the coil and form closed loops.
By Gauss's law for magnetism,the net magnetic flux through any closed surface is zero,i.e.,$\oint \vec{B} \cdot d\vec{A} = 0$.
Consider the infinite plane containing the coil. The magnetic flux passing through the area of the coil $(\phi_{0})$ is in one direction (e.g.,upward).
The magnetic flux passing through the rest of the infinite plane $(\phi_{i})$ must be in the opposite direction (downward) to ensure that the total flux through the entire infinite plane is zero.
Thus,$\phi_{0} + \phi_{i} = 0$,which implies $\phi_{i} = -\phi_{0}$.

(D) The loop $ABCDEFA$ consists of two planar surfaces: the rectangle $ABCD$ in the $xy$-plane and the rectangle $ADEF$ in the $yz$-plane.
The area vector for the rectangle $ABCD$ (in the $xy$-plane) is $\overrightarrow{A}_{ABCD} = (5 \times 5) \hat{k} = 25 \hat{k} \; \text{m}^2$.
The area vector for the rectangle $ADEF$ (in the $yz$-plane) is $\overrightarrow{A}_{ADEF} = (5 \times 5) \hat{i} = 25 \hat{i} \; \text{m}^2$.
The net area vector is $\overrightarrow{A}_{net} = \overrightarrow{A}_{ABCD} + \overrightarrow{A}_{ADEF} = 25 \hat{i} + 25 \hat{k} \; \text{m}^2$.
The magnetic field is given as $\overrightarrow{B} = 3 \hat{i} + 4 \hat{k} \; \text{T}$.
The magnetic flux $\phi$ is calculated using the dot product: $\phi = \overrightarrow{B} \cdot \overrightarrow{A}_{net}$.
$\phi = (3 \hat{i} + 4 \hat{k}) \cdot (25 \hat{i} + 25 \hat{k})$
$\phi = (3 \times 25) + (4 \times 25) = 75 + 100 = 175 \; \text{Wb}$.
Thus,the magnetic flux through the loop is $175 \; \text{Wb}$.

(N/A) No. The magnetic force is always normal to $B$ (remember magnetic force $= q(v \times B)$). It is misleading to call magnetic field lines as lines of force.
$(b)$ If field lines were entirely confined between two ends of a straight solenoid,the flux through the cross-section at each end would be non-zero. But the flux of field $B$ through any closed surface must always be zero. For a toroid,this difficulty is absent because it has no 'ends'.
$(c)$ Gauss's law of magnetism states that the flux of $B$ through any closed surface is always zero,$\int_{S} B \cdot dS = 0$. If monopoles existed,the right-hand side would be equal to the monopole (magnetic charge) $q_m$ enclosed by $S$. Analogous to Gauss's law of electrostatics,$\int_{S} B \cdot dS = \mu_0 q_m$,where $q_m$ is the magnetic charge enclosed by $S$.
$(d)$ No. There is no force or torque on an element due to the field produced by that element itself. However,there is a force (or torque) on one element of a wire due to another element of the same wire (though for a straight wire,this net force is zero).
$(e)$ Yes. The net charge in a system may be zero,yet the magnetic moments due to various current loops within the system may not cancel out. We encounter such examples in paramagnetic materials where atoms have a net dipole moment even though their net charge is zero.

(N/A) Magnetic flux through a plane of area $A$ placed in a uniform magnetic field $B$ as shown in the figure can be written as $\Phi_{B} = \vec{B} \cdot \vec{A} = BA \cos \theta$.
Where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$. The equation can be extended to curved surfaces and non-uniform fields.
If the magnetic field has different magnitudes and directions at various parts of a surface,then the magnetic flux through the surface is given by the summation of flux through small area elements $d\vec{A}_i$:
$\Phi_{B} = \sum_{i} \vec{B}_{i} \cdot d \vec{A}_{i}$
In the limit where the area elements become infinitesimally small,this summation becomes an integral:
$\Phi_{B} = \int_{S} \vec{B} \cdot d\vec{A}$
Magnetic flux is a scalar quantity. Its $SI$ unit is weber $(Wb)$,which is equivalent to tesla meter squared $(T \cdot m^2)$ or volt-second $(V \cdot s)$. Its dimensional formula is $[M^1 L^2 T^{-2} A^{-1}]$.
Definition of $1 \ Wb$: If a magnetic field of $1 \ T$ is applied perpendicularly to a surface of area $1 \ m^2$,then the magnetic flux linked with the surface is $1 \ Wb$. Thus,$1 \ Wb = 1 \ T \cdot m^2$.

(N/A) Magnetic flux $\Phi_B$ is defined as the product of magnetic field $B$ and the area $A$ perpendicular to it,given by $\Phi_B = B \cdot A$.
From the Lorentz force law,$F = qvB$,we can express the magnetic field as $B = \frac{F}{qv}$.
The dimensional formula for force $F$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula for charge $q$ is $[I^1 T^1]$.
The dimensional formula for velocity $v$ is $[L^1 T^{-1}]$.
Thus,the dimensions of $B$ are $[B] = \frac{[M^1 L^1 T^{-2}]}{[I^1 T^1][L^1 T^{-1}]} = [M^1 T^{-2} I^{-1}]$.
Since magnetic flux $\Phi_B = B \cdot A$,and the area $A$ has dimensions $[L^2]$,we have:
$[\Phi_B] = [M^1 T^{-2} I^{-1}] \cdot [L^2] = [M^1 L^2 T^{-2} I^{-1}]$.
Therefore,the dimensional formula for magnetic flux is $[M^1 L^2 T^{-2} I^{-1}]$.

(C) The magnetic flux $\Phi$ linked with a surface is given by the formula $\Phi = B A \cos(\theta)$,where $B$ is the magnetic field,$A$ is the area of the surface,and $\theta$ is the angle between the magnetic field vector and the area vector (normal to the surface).
When the surface of the sheet is parallel to the magnetic field,the normal to the surface makes an angle of $90^{\circ}$ with the magnetic field lines.
Therefore,$\theta = 90^{\circ}$.
Substituting this into the formula: $\Phi = B A \cos(90^{\circ})$.
Since $\cos(90^{\circ}) = 0$,the magnetic flux $\Phi = 0$.

(B) The magnetic flux $\Phi_B$ through a surface is given by the formula $\Phi_B = \vec{B} \cdot \vec{A} = BA \cos \theta$,where $\vec{B}$ is the magnetic field vector,$\vec{A}$ is the area vector,and $\theta$ is the angle between them.
For the magnetic flux to be zero,we must have $\cos \theta = 0$.
This occurs when $\theta = 90^{\circ}$.
An angle of $90^{\circ}$ between the magnetic field vector and the area vector means the magnetic field is perpendicular to the area vector (or parallel to the plane of the surface).
Therefore,the correct option is $B$.

(A) The unit $VS$ stands for Volt-second $(V \cdot s)$.
According to Faraday's law of electromagnetic induction,the induced electromotive force $(e)$ is given by $e = -\frac{d\Phi_B}{dt}$,where $\Phi_B$ is the magnetic flux.
Rearranging this,we get $d\Phi_B = e \cdot dt$.
The unit of magnetic flux $(\Phi_B)$ is the Weber $(Wb)$.
Since the unit of electromotive force $(e)$ is the Volt $(V)$ and the unit of time $(t)$ is the second $(s)$,we have $1 \ Wb = 1 \ V \cdot s$.
Therefore,$VS$ is the unit of magnetic flux.

(N/A) Consider a closed surface $S$ placed in a magnetic field $\vec{B}$. We need to determine the magnetic flux associated with this surface.
Imagine the surface $S$ is divided into small area elements. For one such element $\Delta \vec{S}$, the magnetic field is $\vec{B}$. The magnetic flux $\Delta \phi_{B}$ through this element is defined as:
$\Delta \phi_{B} = \vec{B} \cdot \Delta \vec{S}$
The total magnetic flux $\phi_{B}$ through the closed surface is the sum of the fluxes through all such elements:
$\phi_{B} = \sum_{\text{all}} \Delta \phi_{B} = \sum_{\text{all}} \vec{B} \cdot \Delta \vec{S} = 0 \quad \dots (1)$
Since the number of magnetic field lines leaving the closed surface is equal to the number of lines entering it, the net magnetic flux through any closed surface is always zero.
Gauss's law for magnetism states:
"The net magnetic flux through any closed surface is zero."
In the limit where $\Delta S \rightarrow 0$, the summation becomes an integral:
$\phi_{B} = \oint_{S} \vec{B} \cdot d\vec{S} = 0$
This integral form is the mathematical representation of Gauss's law for magnetism.

(N/A) Gauss's law for electrostatics is given by:
$\oint \vec{E} \cdot d\vec{S} = \frac{q_{enclosed}}{\varepsilon_{0}}$
Where $\vec{E}$ is the electric field and $q_{enclosed}$ is the net charge enclosed by the Gaussian surface.
Gauss's law for magnetism is given by:
$\oint \vec{B} \cdot d\vec{S} = 0$
Where $\vec{B}$ is the magnetic field.
The fundamental difference is that for electrostatics,the net electric flux through a closed surface is proportional to the enclosed charge,implying the existence of electric monopoles (charges). For magnetism,the net magnetic flux through any closed surface is always zero,which implies that isolated magnetic monopoles do not exist; magnetic field lines always form continuous closed loops.

(N/A) Gauss's law for magnetism states that the net magnetic flux through any closed surface is always zero.
Mathematically,it is expressed as: $\oint_{S} \vec{B} \cdot d\vec{A} = 0$
Here,$\vec{B}$ is the magnetic field and $d\vec{A}$ is the area vector of the closed surface $S$.
This law implies that magnetic monopoles do not exist; magnetic field lines always form closed loops,meaning every north pole is accompanied by a south pole.

(N/A) Gauss's law in electrostatics states that the net electric flux through any closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space: $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}$. This implies that electric charges exist as isolated monopoles.
Gauss's law in magnetism states that the net magnetic flux through any closed surface is always zero: $\oint \vec{B} \cdot d\vec{A} = 0$. This implies that magnetic monopoles do not exist and magnetic field lines always form continuous closed loops.

(N/A) Gauss's law of magnetism states that the net magnetic flux through any closed surface is zero: $\oint \vec{B} \cdot d\vec{S} = 0$.
Consider a magnetic dipole of moment $\vec{m} = m \hat{k}$ placed at the origin $O$. The magnetic field $\vec{B}$ at a point $P$ on the surface of a sphere of radius $R$ (where $OP$ makes an angle $\theta$ with the $z$-axis) is given by:
$\vec{B} = \frac{\mu_0}{4\pi} \frac{m}{r^3} (2 \cos \theta \hat{r} + \sin \theta \hat{\theta})$.
The area element $d\vec{S}$ on the spherical surface is $d\vec{S} = R^2 \sin \theta d\theta d\phi \hat{r}$.
Now,calculate the flux $\Phi_B = \oint \vec{B} \cdot d\vec{S}$:
$\Phi_B = \int_0^{2\pi} \int_0^{\pi} \left( \frac{\mu_0}{4\pi} \frac{m}{R^3} (2 \cos \theta \hat{r} + \sin \theta \hat{\theta}) \right) \cdot (R^2 \sin \theta d\theta d\phi \hat{r})$.
Since $\hat{r} \cdot \hat{r} = 1$ and $\hat{\theta} \cdot \hat{r} = 0$:
$\Phi_B = \frac{\mu_0 m}{4\pi R} \int_0^{2\pi} d\phi \int_0^{\pi} 2 \cos \theta \sin \theta d\theta$.
$\Phi_B = \frac{\mu_0 m}{4\pi R} (2\pi) \int_0^{\pi} \sin(2\theta) d\theta$.
$\Phi_B = \frac{\mu_0 m}{2R} \left[ -\frac{\cos(2\theta)}{2} \right]_0^{\pi} = \frac{\mu_0 m}{4R} [-\cos(2\pi) + \cos(0)] = \frac{\mu_0 m}{4R} [-1 + 1] = 0$.
Thus,Gauss's law for magnetism is verified.

(N/A) Yes,we would get the same answer for the magnetic flux.
According to Gauss's Law for magnetism,the net magnetic flux through any closed surface is zero $(\oint \vec{B} \cdot d\vec{A} = 0)$.
If we consider the closed surface formed by the union of surfaces $S_1$ and $S_2$,the total flux through this closed surface is zero.
Let the flux through $S_1$ be $\phi_1$ and through $S_2$ be $\phi_2$. Since the magnetic field lines enter through one surface and exit through the other,the flux through $S_1$ must equal the flux through $S_2$ in magnitude.
Therefore,the magnetic flux linked with any surface bounded by the same loop $C$ is the same,as it depends only on the number of magnetic field lines passing through the loop.

(C) The magnetic flux $\phi$ is given by $\phi = \vec{B} \cdot \vec{A} = B A \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (normal to the surface).
The magnetic field is directed along the positive $X$-axis. The top surface is inclined at an angle of $30^{\circ}$ with the horizontal. The normal to this surface makes an angle of $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$ with the $X$-axis.
The dimensions of the top surface are $10 \, cm \times 10 \, cm = 0.1 \, m \times 0.1 \, m = 0.01 \, m^2$.
Substituting the values:
$\phi = (0.2 \, T) \times (0.01 \, m^2) \times \cos(60^{\circ})$
$\phi = 0.002 \times 0.5 = 0.001 \, Wb$
Since $1 \, Wb = 1000 \, m-Wb$,we have:
$\phi = 0.001 \times 1000 = 1 \, m-Wb$.

(C) Magnetic flux $(\Phi_B)$ is defined as the product of the magnetic field $(B)$ and the area $(A)$ through which it passes, given by $\Phi_B = B \cdot A \cdot \cos(\theta)$.
The $SI$ unit of magnetic field is Tesla $(T)$ and the unit of area is square meters $(m^2)$.
Therefore, $1 \text{ Tesla} \cdot m^2 = 1 \text{ Weber}$ $(Wb)$.
Thus, the $SI$ unit of magnetic flux is Weber.

(A) Given:
Side length of the square loop,$l = 1\,m$.
Area of the loop,$A = l^2 = (1\,m)^2 = 1\,m^2$.
Magnetic field,$B = 0.5\,T$.
The plane of the loop is perpendicular to the magnetic field,which means the area vector $\overrightarrow{A}$ is parallel to the magnetic field vector $\overrightarrow{B}$.
Therefore,the angle $\theta$ between $\overrightarrow{B}$ and $\overrightarrow{A}$ is $0^\circ$.
The magnetic flux $\phi$ is given by the formula:
$\phi = B A \cos\theta$
Substituting the values:
$\phi = 0.5 \times 1 \times \cos(0^\circ)$
Since $\cos(0^\circ) = 1$,
$\phi = 0.5 \times 1 \times 1 = 0.5\,Wb$.

(A) The correct option is $A$.
$1$. According to Gauss's Law for magnetism,the net magnetic flux through any closed surface is always zero because magnetic monopoles do not exist. This means the number of magnetic field lines entering a closed surface is equal to the number of lines exiting it. Thus,statement $A$ is correct.
$2$. Magnetic field lines are always perpendicular to the surface of a magnetic material (like a pole) at the points where they originate or terminate. Therefore,statement $B$ is incorrect.
$3$. Magnetic field lines always pass through a conductor (unless it is a superconductor exhibiting the Meissner effect). Therefore,statement $C$ is incorrect.
$4$. Since only statement $A$ is correct,the correct choice is $A$.

(D) The magnetic flux $\phi$ through a coil is given by the formula $\phi = \vec{B} \cdot \vec{A} = BA \cos \theta$,where $B$ is the magnetic field magnitude,$A$ is the area of the coil,and $\theta$ is the angle between the magnetic field vector and the area vector.
$1$. By changing the magnitude of the magnetic field $(B)$,the flux $\phi$ changes. (Statement $A$ is correct).
$2$. By changing the area of the coil $(A)$ within the magnetic field,the flux $\phi$ changes. (Statement $B$ is correct).
$3$. By changing the angle $(\theta)$ between the direction of the magnetic field and the plane of the coil (which changes the angle between the magnetic field and the area vector),the flux $\phi$ changes. (Statement $C$ is correct).
$4$. By reversing the magnetic field direction abruptly,the angle $\theta$ changes (e.g.,from $0^\circ$ to $180^\circ$),which changes the flux $\phi$. (Statement $D$ is correct).
Therefore,all four methods can change the magnetic flux through the coil.

(B) According to Gauss's Law for magnetism,the net magnetic flux through any closed surface is given by the surface integral of the magnetic field $\vec{B}$ over that surface.
$\oint \vec{B} \cdot d\vec{s} = 0$
This result arises because magnetic monopoles do not exist; magnetic field lines always form closed loops,meaning every field line entering a closed surface must also exit it.
Therefore,the net magnetic flux through any closed surface is always $0$.

(C) Assertion $(A)$ is correct because, according to Gauss's Law for magnetism, the net magnetic flux through any closed surface is zero $( \oint \vec{B} \cdot d\vec{A} = 0)$. This implies that there are no isolated magnetic charges (monopoles).
Reason $(R)$ is also correct because magnetic field lines always form continuous closed loops, starting from the North pole and ending at the South pole outside the magnet, and continuing from South to North inside the magnet.
Since magnetic field lines form closed loops, they do not have a starting or ending point (source or sink), which directly explains why magnetic monopoles cannot exist. Therefore, $(R)$ is the correct explanation of $(A)$.

(B) The magnetic flux $\Phi$ is given by the formula $\Phi = B A \cos(\theta)$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$.
Given that the plane of the strip makes an angle of $60^{\circ}$ with the magnetic field,the angle $\theta$ between the area vector (which is perpendicular to the plane) and the magnetic field is $\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Given values: $\Phi = 1 \ mWb = 10^{-3} \ Wb$,$A = 0.02 \ m^2$,and $\theta = 30^{\circ}$.
Substituting these values into the formula: $10^{-3} = B \times 0.02 \times \cos(30^{\circ})$.
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866$,we have $10^{-3} = B \times 0.02 \times 0.866$.
$B = \frac{10^{-3}}{0.02 \times 0.866} = \frac{0.001}{0.01732} \approx 0.0577 \ T$.
Rounding this value gives $B \approx 0.058 \ T$.

(A) The magnetic flux $\Phi$ through a surface is given by the dot product of the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$.
Given,$\vec{B} = B_0(2 \hat{i} + 3 \hat{j} + 4 \hat{k})$.
The square lies in the $x-y$ plane,so its area vector $\vec{A}$ is perpendicular to the $x-y$ plane,which is along the $z$-axis.
Thus,$\vec{A} = L^2 \hat{k}$.
The magnetic flux $\Phi$ is calculated as:
$\Phi = \vec{B} \cdot \vec{A}$
$\Phi = [B_0(2 \hat{i} + 3 \hat{j} + 4 \hat{k})] \cdot [L^2 \hat{k}]$
$\Phi = B_0 L^2 (2 \hat{i} \cdot \hat{k} + 3 \hat{j} \cdot \hat{k} + 4 \hat{k} \cdot \hat{k})$
Since $\hat{i} \cdot \hat{k} = 0$,$\hat{j} \cdot \hat{k} = 0$,and $\hat{k} \cdot \hat{k} = 1$,we get:
$\Phi = B_0 L^2 (0 + 0 + 4(1)) = 4 B_0 L^2$.
Therefore,the magnitude of the flux is $4 B_0 L^2$.

(B) The magnetic flux $\phi$ through a surface is given by the dot product of the magnetic field vector $\overrightarrow{B}$ and the area vector $\overrightarrow{A}$.
Since the square of side $L$ lies in the $x-y$ plane,its area vector $\overrightarrow{A}$ is directed along the $z$-axis.
Therefore,$\overrightarrow{A} = L^2 \hat{k}$.
The magnetic field is $\overrightarrow{B} = B_0(2 \hat{i} + 4 \hat{j} + 3 \hat{k})$.
The magnetic flux is calculated as:
$\phi = \overrightarrow{B} \cdot \overrightarrow{A}$
$\phi = [B_0(2 \hat{i} + 4 \hat{j} + 3 \hat{k})] \cdot [L^2 \hat{k}]$
$\phi = B_0 L^2 (2 \hat{i} \cdot \hat{k} + 4 \hat{j} \cdot \hat{k} + 3 \hat{k} \cdot \hat{k})$
Since $\hat{i} \cdot \hat{k} = 0$,$\hat{j} \cdot \hat{k} = 0$,and $\hat{k} \cdot \hat{k} = 1$,we get:
$\phi = B_0 L^2 (0 + 0 + 3(1)) = 3 B_0 L^2 \text{ Wb}$.

(C) The correct option is $C$.
Magnetic flux $\phi$ is defined as the product of magnetic field intensity $B$ and the area $A$ perpendicular to the field,given by the formula $\phi = B \times A$.
To find the unit of magnetic field intensity $B$,we rearrange the formula: $B = \frac{\phi}{A}$.
Since the unit of magnetic flux $\phi$ is weber $(Wb)$ and the unit of area $A$ is square meters $(m^2)$,the unit of magnetic field intensity $B$ is $\frac{Wb}{m^2}$ (also known as Tesla).

(A) Given,$B = 2 \text{ mT} = 2 \times 10^{-3} \text{ T}$.
Radius $R = \frac{10}{\sqrt{\pi}} \text{ cm} = \frac{10}{\sqrt{\pi}} \times 10^{-2} \text{ m} = \frac{10^{-1}}{\sqrt{\pi}} \text{ m}$.
The area of the circular cross-section $S_3$ is $A = \pi R^2 = \pi \times \left(\frac{10^{-1}}{\sqrt{\pi}}\right)^2 = \pi \times \frac{10^{-2}}{\pi} = 10^{-2} \text{ m}^2$.
For the hemisphere $S_1$,the magnetic field lines enter the surface. The angle between the area vector (outward normal) and the magnetic field vector is $180^{\circ}$.
Flux $\Phi_{S_1} = B A \cos(180^{\circ}) = (2 \times 10^{-3} \text{ T}) \times (10^{-2} \text{ m}^2) \times (-1) = -2 \times 10^{-5} \text{ Wb} = -20 \times 10^{-6} \text{ Wb} = -20 \mu \text{ Wb}$.
Since the total magnetic flux through a closed surface is zero (Gauss's Law for magnetism),the flux leaving the cone $S_2$ must be equal in magnitude to the flux entering the hemisphere $S_1$.
Therefore,$\Phi_{S_1} + \Phi_{S_2} = 0 \implies \Phi_{S_2} = -\Phi_{S_1} = +20 \mu \text{ Wb}$.

(A) The area vector $\vec{A}$ of a square loop of side $2 \ m$ lying in the $Y-Z$ plane is directed along the $X$-axis.
$\vec{A} = (2 \times 2) \hat{i} = 4 \hat{i} \ m^2$.
The magnetic field is given by $\vec{B} = (5 \hat{i} + 3 \hat{j} - 4 \hat{k}) \ T$.
The magnetic flux $\phi$ is given by the dot product of the magnetic field and the area vector:
$\phi = \vec{B} \cdot \vec{A}$
$\phi = (5 \hat{i} + 3 \hat{j} - 4 \hat{k}) \cdot (4 \hat{i})$
$\phi = (5 \times 4) + (3 \times 0) + (-4 \times 0)$
$\phi = 20 \ Wb$.
Thus,the magnitude of the magnetic flux is $20 \ Wb$.