Ampere’s circuital law and its application (Solenoid and Toroid) Questions in English

(C) According to Ampere's circuital law,the magnetic field $B$ at a distance $r$ from the axis of a cylindrical conductor carrying a current $I$ is given by:
$1$. For $r < R$ (inside the rod): $B = \frac{\mu_0 I r}{2 \pi R^2}$,where $R$ is the radius of the rod.
$2$. For $r > R$ (outside the rod): $B = \frac{\mu_0 I}{2 \pi r}$.
Since the magnetic field is non-zero in both regions,the magnetic field exists both inside and outside the rod.

(B) According to Ampere's circuital law,$\oint B \cdot dl = \mu_0 I_{enclosed}$.
For a long hollow pipe,the current $I$ flows only through the surface of the pipe.
Inside the pipe,any Amperian loop will enclose zero current $(I_{enclosed} = 0)$,so the magnetic field $B = 0$.
Outside the pipe,at a distance $r$ from the axis,the Amperian loop encloses the total current $I$,resulting in a magnetic field $B = \frac{\mu_0 I}{2\pi r}$.
Therefore,the magnetic field is produced only outside the pipe.

(B) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around a closed loop is equal to $\mu_0$ times the net current enclosed by the loop,i.e.,$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$.
For a point outside the coaxial cable at a distance $r$ from the axis,we consider an Amperian loop of radius $r$.
The net current enclosed by this loop is the sum of the current flowing through the inner conductor $(+i)$ and the current returning through the outer conductor $(-i)$.
Therefore,$I_{\text{enclosed}} = i + (-i) = 0$.
Since the net current enclosed is zero,the magnetic field $\vec{B}$ at any point outside the cable must be zero.

(A) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n I$.
Here,the number of turns per unit length $n = 10 \ turns/cm = 1000 \ turns/m$.
The current $I = 5 \ A$.
The permeability of free space $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$.
Substituting these values into the formula:
$B = (4\pi \times 10^{-7}) \times 1000 \times 5$
$B = 4\pi \times 10^{-7} \times 5 \times 10^3$
$B = 20\pi \times 10^{-4} \ T$
$B = 2\pi \times 10^{-3} \ T$.

(B) The magnetic field $B$ inside a long solenoid is derived using Ampere's circuital law.
For a solenoid with $n$ turns per unit length carrying a current $i$,the magnetic field at any point well inside the solenoid is given by the formula:
$B = \mu_0 ni$
where $\mu_0$ is the permeability of free space.

(B) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n I$,where $\mu_0$ is the permeability of free space,$n$ is the number of turns per unit length,and $I$ is the current flowing through the solenoid.
From this relation,it is clear that the magnetic field $B$ is directly proportional to the current $I$ flowing through the solenoid.
Therefore,option $B$ is correct.

(B) The magnetic field $B$ inside a long solenoid at any point on its axis is given by the formula $B = \mu_0 ni$,where $\mu_0$ is the permeability of free space,$n$ is the number of turns per unit length,and $i$ is the current flowing through the solenoid.
From this expression,it is clear that the magnetic field $B$ is directly proportional to the product of the number of turns per unit length $n$ and the current $i$.
Therefore,$B \propto ni$.
Thus,the correct option is $B$.

(A) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.
Given:
$B = 20 \text{ mT} = 20 \times 10^{-3} \text{ T}$
$n = 20 \text{ turns/cm} = 20 \times 100 \text{ turns/m} = 2000 \text{ turns/m}$
$\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A} \Rightarrow \mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$
Rearranging the formula for current $i$:
$i = \frac{B}{\mu_0 n}$
Substituting the values:
$i = \frac{20 \times 10^{-3}}{4\pi \times 10^{-7} \times 2000}$
$i = \frac{20 \times 10^{-3}}{8\pi \times 10^{-4}}$
$i = \frac{200}{8\pi} = \frac{25}{\pi} \approx \frac{25}{3.14} \approx 7.96 \text{ A}$
Rounding to the nearest integer,we get $i \approx 8 \text{ A}$.

(D) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed path is equal to $\mu_0$ times the net current $i_{enclosed}$ passing through the surface bounded by the path,i.e.,$\oint \vec{B} \cdot d\vec{l} = \mu_0 i_{enclosed}$.
For a long hollow copper tube,if we consider an Amperian loop inside the tube (at a radius $r < R$),the current enclosed by this loop is zero because the current flows only through the material of the tube.
Since $i_{enclosed} = 0$,the magnetic field $B$ inside the tube must be zero.

(C) The number of turns per unit length $n = 50 \, turns/cm = 5000 \, turns/m$. The current $I = 4 \, A$. The permeability of free space $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$.
$(i)$ Magnetic field at an internal point: $B_{in} = \mu_0 n I = (4\pi \times 10^{-7}) \times 5000 \times 4 = 25.12 \times 10^{-3} \, Wb/m^2 \approx 25.1 \times 10^{-3} \, Wb/m^2$.
(ii) Magnetic field at one end: $B_{end} = \frac{1}{2} B_{in} = \frac{25.1 \times 10^{-3}}{2} = 12.55 \times 10^{-3} \, Wb/m^2 \approx 12.6 \times 10^{-3} \, Wb/m^2$.

(B) The magnetic field $B$ at the centre of a long solenoid is given by the formula: $B = \mu_0 n I$.
Here, $n$ is the number of turns per unit length, which is calculated as $n = N/L$.
Given: $N = 4250$, $L = 1.0 \ m$, $I = 5.0 \ A$, and $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$.
First, calculate $n$: $n = 4250 / 1.0 = 4250 \ turns/m$.
Now, substitute the values into the formula: $B = (4\pi \times 10^{-7}) \times 4250 \times 5.0$.
$B = (4 \times 3.14159 \times 10^{-7}) \times 21250$.
$B \approx 12.566 \times 10^{-7} \times 21250 \approx 0.0267 \ T$.
Thus, $B \approx 2.7 \times 10^{-2} \ Wb/m^2$.

(D) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n I$,where $n$ is the number of turns per unit length.
Given that the total number of turns is $N$ and the length of the solenoid is $L$,the number of turns per unit length is $n = \frac{N}{L}$.
Substituting this into the formula,we get $B = \mu_0 \left( \frac{N}{L} \right) I$.
Therefore,the correct expression is $\mu_0 \frac{N}{L} I$.

(C) The magnetic field $B$ at the center of a long solenoid is given by the formula:
$B = \mu_0 n i$
where $n$ is the number of turns per unit length $(n = N/l)$,
$i$ is the current flowing through the solenoid,
and $\mu_0$ is the permeability of free space.
From this formula,it is evident that the magnetic field $B$ depends only on the number of turns per unit length $(n)$ and the current $(i)$.
It does not depend on the radius of the solenoid.
Therefore,the correct option is $C$.

(B) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the total current $I_{\text{enclosed}}$ passing through the surface bounded by the loop: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$.
For an infinitely long,thin-walled pipe carrying current $I$ along its length,any closed loop chosen inside the pipe encloses zero current $(I_{\text{enclosed}} = 0)$.
Therefore,$\oint \vec{B} \cdot d\vec{l} = 0$,which implies that the magnetic field $\vec{B}$ at any point inside the pipe is zero.

(A) toroid is essentially a solenoid bent into a circular shape.
For a toroid with $n$ turns per unit length and carrying a current $i$,the magnetic field $B$ inside the core is derived using Ampere's Circuital Law.
According to Ampere's Law,$\oint B \cdot dl = \mu_0 I_{\text{enclosed}}$.
For a circular path of radius $r$ inside the toroid,the length is $2\pi r$ and the total number of turns is $N = n(2\pi r)$.
Thus,$B(2\pi r) = \mu_0 (n \cdot 2\pi r) i$.
Simplifying this,we get $B = \mu_0 ni$.

(B) The magnetic field $B$ at the centre of a long solenoid is given by the formula $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current.
Given:
Turns per cm, $n' = 200 \, \text{turns/cm} = 200 \times 10^2 \, \text{turns/m} = 2 \times 10^4 \, \text{turns/m}$.
Current, $I = 2.5 \, A$.
Permeability, $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$.
Substituting these values into the formula:
$B = (4\pi \times 10^{-7}) \times (2 \times 10^4) \times 2.5$
$B = 4\pi \times 10^{-7} \times 5 \times 10^4$
$B = 20\pi \times 10^{-3} \, T$
$B = 2 \times 3.14 \times 10^{-2} \, T = 6.28 \times 10^{-2} \, Wb/m^2$.
Thus, the correct option is $B$.

(D) According to Ampere's circuital law,the magnetic field $B$ at a distance $r$ from the axis of a long straight cylindrical wire carrying a current $I$ is given by $\oint B \cdot dl = \mu_0 I_{enclosed}$.
For a point on the axis of the wire,the distance $r = 0$.
Since the current is distributed over the cross-section of the wire,the current enclosed by a loop of radius $r = 0$ is zero $(I_{enclosed} = 0)$.
Therefore,the magnetic field $B$ at the axis of the wire is $0\, T$.

(A) The magnetic field $B$ inside a long solenoid at its center is given by $B_{center} = \mu_0 nI$.
At the ends of a long solenoid,the magnetic field is exactly half of the field at the center.
Therefore,the magnetic field at the ends is $B_{ends} = \frac{\mu_0 nI}{2}$.

(B) The statement "On flowing current in a conducting wire, a magnetic field is produced around it" describes the fundamental observation that led to the development of Ampere's Law.
Ampere's Law relates the integrated magnetic field around a closed loop to the electric current passing through the loop.
Therefore, the correct option is $B$.

(A) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.
Let the initial magnetic field be $B = \mu_0 n i$.
When the current is doubled,the new current $i' = 2i$.
When the number of turns per cm is halved,the new number of turns per unit length $n' = n/2$.
The new magnetic field $B'$ is given by $B' = \mu_0 n' i' = \mu_0 (n/2) (2i) = \mu_0 n i$.
Therefore,$B' = B$.

(D) The magnetic field inside a long solenoid is given by $B = \mu_0 ni$.
Since the two solenoids are coaxial and the currents flow in opposite directions,the net magnetic field $B_{net}$ inside the inner coil is the difference between the individual magnetic fields: $B_{net} = |B_1 - B_2|$.
For the magnetic field to be zero,we must have $B_1 = B_2$.
Substituting the formula,we get $\mu_0 n_1 i_1 = \mu_0 n_2 i_2$,which simplifies to $n_1 i_1 = n_2 i_2$.
This condition is satisfied if $n_1 i_1 = n_2 i_2$ (Option $b$).
Additionally,if $n_1 = n_2$ and $i_1 = i_2$,the condition $n_1 i_1 = n_2 i_2$ is also satisfied (Option $c$).
Therefore,both conditions $(b)$ and $(c)$ are correct.

(A) For a long solenoid,the magnetic field $B$ at the center is given by $B_{center} = \mu_0 n i$.
At the ends of the solenoid,the magnetic field is $B_{end} = \frac{1}{2} \mu_0 n i = \frac{1}{2} B_{center}$.
As we move from one end $(x=0)$ towards the center,the magnetic field increases from $\frac{1}{2} B_{center}$ to $B_{center}$ and then remains approximately constant near the center before decreasing again as we approach the other end.
Graph $A$ correctly represents this variation,showing a gradual increase from the end,a plateau at the center,and a symmetric decrease towards the other end.

(B) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.
Total number of turns $N = 3 \times 1000 = 3000$.
Length $L = 1.5 \, m$.
Number of turns per unit length $n = N/L = 3000 / 1.5 = 2000 \, turns/m$.
Current $I = 2.0 \, A$.
Radius $r = 2.0 \, cm = 0.02 \, m$.
Area of cross-section $A = \pi r^2 = \pi (0.02)^2 = 4\pi \times 10^{-4} \, m^2$.
Magnetic flux $\phi = B \cdot A = (\mu_0 n I) A$.
$\phi = (4\pi \times 10^{-7}) \times 2000 \times 2.0 \times (4\pi \times 10^{-4})$.
$\phi = 16 \pi^2 \times 10^{-7} \times 4000 \times 10^{-4} = 64 \pi^2 \times 10^{-7} \times 10^{-1} \approx 64 \times 9.87 \times 10^{-8} \approx 6.31 \times 10^{-6} \, Wb$.

(A) The magnetic field inside a long solenoid is given by the formula: $B = \mu_0 n i$, where $n$ is the number of turns per unit length and $i$ is the current.
Given: $n = 20 \text{ turns/cm} = 2000 \text{ turns/m}$.
Given: $B = 20 \text{ mT} = 20 \times 10^{-3} \text{ T}$.
We know $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
Substituting the values into the formula:
$20 \times 10^{-3} = (4\pi \times 10^{-7}) \times 2000 \times i$
$20 \times 10^{-3} = 8\pi \times 10^{-4} \times i$
$i = \frac{20 \times 10^{-3}}{8\pi \times 10^{-4}} = \frac{200}{8\pi} = \frac{25}{\pi} \approx 7.96 \text{ A}$.
Rounding to the nearest integer, the current is $8 \text{ A}$.

(D) The magnetic field $B$ inside a toroid is given by the formula $B = \mu_0 n i$,where $n$ is the number of turns per unit length.
Given: Radius $R = 0.1 \ m$,Number of turns $N = 500$,Current $i = 0.5 \ A$.
The number of turns per unit length is $n = \frac{N}{2\pi R}$.
Substituting the values: $n = \frac{500}{2\pi \times 0.1} = \frac{500}{0.2\pi} = \frac{2500}{\pi} \ m^{-1}$.
Now,$B = (4\pi \times 10^{-7}) \times \left(\frac{2500}{\pi}\right) \times 0.5$.
$B = 4 \times 10^{-7} \times 2500 \times 0.5$.
$B = 4 \times 10^{-7} \times 1250$.
$B = 5000 \times 10^{-7} \ T = 5 \times 10^{-4} \ T$.

(A) The magnetic field on the axis of a finite solenoid at a point $P$ is given by the formula:
$B = \frac{{{\mu _0}ni}}{2}(\sin \theta_1 + \sin \theta_2)$
where $\theta_1$ and $\theta_2$ are the angles subtended by the ends of the solenoid at point $P$.
From the figure,$\theta_1 = 60^\circ$ and $\theta_2 = 30^\circ$.
Substituting these values into the formula:
$B = \frac{{{\mu _0}ni}}{2}(\sin 60^\circ + \sin 30^\circ)$
$B = \frac{{{\mu _0}ni}}{2}(\frac{\sqrt{3}}{2} + \frac{1}{2})$
$B = \frac{{{\mu _0}ni}}{4}(\sqrt{3} + 1)$

(D) For a cylindrical conductor with inner radius $a = R$ and outer radius $b = 2R$,the magnetic field $B$ at a distance $r$ from the center (where $a < r < b$) is given by Ampere's Law:
$B(2\pi r) = \mu_0 I_{enclosed}$
$I_{enclosed} = i \left( \frac{r^2 - a^2}{b^2 - a^2} \right)$
Substituting $a = R$,$b = 2R$,and $r = \frac{3R}{2}$:
$I_{enclosed} = i \left( \frac{(\frac{3R}{2})^2 - R^2}{(2R)^2 - R^2} \right) = i \left( \frac{\frac{9R^2}{4} - R^2}{4R^2 - R^2} \right) = i \left( \frac{\frac{5R^2}{4}}{3R^2} \right) = i \left( \frac{5}{12} \right)$
Now,$B = \frac{\mu_0 I_{enclosed}}{2\pi r} = \frac{\mu_0 (i \cdot \frac{5}{12})}{2\pi (\frac{3R}{2})}$
$B = \frac{5\mu_0 i}{12 \cdot 2\pi \cdot \frac{3R}{2}} = \frac{5\mu_0 i}{36\pi R}$

(C) The magnetic field inside a solenoid is given by $B = \frac{\mu_0 N i}{l}$,where $N$ is the total number of turns,$i$ is the current,and $l$ is the length of the solenoid.
Given: $B = 0.2 \, T$,$l = 0.8 \, m$,$i = 10 \, A$,$\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$.
Substituting the values: $0.2 = \frac{4\pi \times 10^{-7} \times N \times 10}{0.8}$.
Solving for $N$: $N = \frac{0.2 \times 0.8}{4\pi \times 10^{-6}} = \frac{0.16}{4\pi \times 10^{-6}} = \frac{4 \times 10^4}{\pi}$.
The total length of the wire $L$ is given by $L = (2\pi r) \times N$,where $r = 3 \, cm = 3 \times 10^{-2} \, m$.
$L = 2\pi \times (3 \times 10^{-2}) \times \frac{4 \times 10^4}{\pi}$.
$L = 6 \times 10^{-2} \times 4 \times 10^4 = 24 \times 10^2 = 2.4 \times 10^3 \, m$.

(C) The magnetic field inside a solenoid is given by $B = \frac{\mu_0 N i}{l}$,where $N$ is the total number of turns,$i$ is the current,and $l$ is the length of the solenoid.
Given: $B = 0.2\, T$,$i = 10\, A$,$l = 0.8\, m$,and $\mu_0 = 4\pi \times 10^{-7}\, T\cdot m/A$.
Substituting the values: $0.2 = \frac{4\pi \times 10^{-7} \times N \times 10}{0.8}$.
Solving for $N$: $N = \frac{0.2 \times 0.8}{4\pi \times 10^{-6}} = \frac{0.16}{4\pi \times 10^{-6}} = \frac{4 \times 10^4}{\pi}$.
The length of the winding wire $L$ is equal to the circumference of one turn multiplied by the total number of turns: $L = (2\pi r) \times N$.
Given $r = 3\, cm = 3 \times 10^{-2}\, m$.
$L = 2\pi \times (3 \times 10^{-2}) \times \frac{4 \times 10^4}{\pi} = 6 \times 10^{-2} \times 4 \times 10^4 = 24 \times 10^2 = 2.4 \times 10^3\, m$.

(D) According to Ampere's Circuital Law,the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the total current $I_{enclosed}$ passing through the surface bounded by the loop.
For a hollow cylinder of infinite length with a surface current density $\lambda$ (current per unit length along the axis),the magnetic field inside the cylinder is zero. This is because the current flows along the circumference,creating a solenoid-like configuration where the magnetic field is confined to the interior of the solenoid (if it were a coil) or,in this specific case of a surface current flowing around the cylinder,the field inside is zero due to the symmetry and the nature of the current distribution.
Alternatively,using Ampere's law for a loop inside the cylinder,the enclosed current is zero,hence $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed} = 0$,which implies $B = 0$ inside the hollow cylinder.

(D) For a long cylindrical conductor with uniform current density $J$,the magnetic field at a distance $x$ from the center (where $x < r$) is given by Ampere's circuital law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$.
$B(2\pi x) = \mu_0 (J \cdot \pi x^2) \implies B = \frac{\mu_0 J x}{2}$.
Let the center of the left cylinder be at $x = -d/2$ and the right cylinder at $x = d/2$. Point $A$ is at the origin $(0,0)$.
The field due to the left cylinder (current out of the page) at $A$ is $B_1 = \frac{\mu_0 J (d/2)}{2} = \frac{\mu_0 J d}{4}$ in the $+y$-direction.
The field due to the right cylinder (current into the page) at $A$ is $B_2 = \frac{\mu_0 J (d/2)}{2} = \frac{\mu_0 J d}{4}$ in the $+y$-direction.
The total magnetic field is $B_{net} = B_1 + B_2 = \frac{\mu_0 J d}{2}$ in the $+y$-direction.

(B) The magnetic field at the centre of a long solenoid is given by $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.
For the first solenoid: $B_1 = \mu_0 n_1 i_1 = 6.28 \times 10^{-2} \ Wb/m^2$,where $n_1 = 200 \ turns/cm$ and $i_1 = i$.
For the second solenoid: $n_2 = 100 \ turns/cm$ and $i_2 = i/3$.
Taking the ratio: $\frac{B_2}{B_1} = \frac{\mu_0 n_2 i_2}{\mu_0 n_1 i_1} = \frac{n_2 i_2}{n_1 i_1}$.
Substituting the values: $\frac{B_2}{6.28 \times 10^{-2}} = \frac{100 \times (i/3)}{200 \times i} = \frac{100}{200 \times 3} = \frac{1}{6}$.
Therefore,$B_2 = \frac{6.28 \times 10^{-2}}{6} \approx 1.05 \times 10^{-2} \ Wb/m^2$.

(B) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the net current $I_{\text{enclosed}}$ passing through the surface bounded by the loop.
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$
For a thin-walled pipe carrying current $I$ along its length,any closed loop drawn inside the pipe encloses zero net current $(I_{\text{enclosed}} = 0)$.
Since the current is distributed only on the surface of the pipe,there is no current flowing through the interior region.
Therefore,$\oint \vec{B} \cdot d\vec{l} = 0$,which implies that the magnetic field $\vec{B}$ at any point inside the pipe is zero.

(D) For two coaxial solenoids carrying current in the same direction,the magnetic field produced by the outer solenoid is uniform inside it and zero outside it. The inner solenoid is placed in this uniform magnetic field. Since the current in the inner solenoid is distributed symmetrically,the net magnetic force on the inner solenoid is zero because the forces on opposite sides cancel each other out.
Similarly,the magnetic field produced by the inner solenoid is confined within its own volume and is zero outside it. Therefore,the outer solenoid is placed in a region where the magnetic field due to the inner solenoid is zero. Thus,the net magnetic force on the outer solenoid is also zero.
Hence,$\overrightarrow{F_1} = 0$ and $\overrightarrow{F_2} = 0$.

(C) For a long cylindrical pipe (hollow cylinder) carrying current $i$ uniformly distributed over its cross-section:
$1$. Inside the hollow region $(x < r)$: By Ampere's circuital law,$\oint B \cdot dl = \mu_0 I_{enclosed}$. Since no current is enclosed,$B = 0$.
$2$. Within the material of the pipe $(r < x < R)$: The current enclosed by a loop of radius $x$ is $I_{enclosed} = i \cdot \frac{\pi x^2 - \pi r^2}{\pi R^2 - \pi r^2}$. Thus,$B(2\pi x) = \mu_0 i \frac{x^2 - r^2}{R^2 - r^2}$,which implies $B \propto \frac{x^2 - r^2}{x}$. This shows a non-linear increase.
$3$. Outside the pipe $(x > R)$: The total current enclosed is $i$. Thus,$B(2\pi x) = \mu_0 i$,which implies $B = \frac{\mu_0 i}{2\pi x}$,so $B \propto 1/x$.
Comparing these characteristics,the graph shows $B=0$ for $x < r$,a non-linear increase for $r < x < R$,and a $1/x$ decay for $x > R$. Option $C$ correctly depicts this behavior.

(D) Ampere's circuital law is most effective for calculating the magnetic field when there is high degree of symmetry,such as in an infinitely long straight wire,a solenoid,or a toroid.
For a finite length wire,the magnetic field does not possess the required symmetry to apply Ampere's law easily,as the field lines are not simple circles centered on the wire axis throughout the length.
Therefore,Statement-$1$ is false.
Regarding Statement-$2$,the magnetic field of a finite wire is indeed symmetric about the wire (cylindrical symmetry),but this symmetry is not sufficient to apply Ampere's law to find the field magnitude.
Thus,Statement-$2$ is true.

(A) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around a closed path is equal to $\mu_0$ times the total current $i_{\text{enclosed}}$ passing through the surface bounded by the path.
$\oint \vec{B} \cdot d\vec{\ell} = \mu_0 i_{\text{enclosed}}$
For a point $P$ outside the cable at a distance $r > c$ from the axis,we consider a circular Amperian loop of radius $r$.
The total current enclosed by this loop is the sum of the current in the inner conductor $(+i)$ and the current in the outer conductor $(-i)$.
$i_{\text{enclosed}} = i + (-i) = 0$
Therefore,the magnetic induction $B$ at point $P$ is:
$B(2 \pi r) = \mu_0 (0)$
$B = 0$

(D) Consider a thin cylindrical shell of radius $r$ and thickness $dr$ within the thick cylinder.
The charge on this thin shell is $dQ = \rho (2\pi r dr) L$.
The current $dI$ produced by this rotating shell is $dI = \frac{dQ}{T} = \frac{dQ \cdot \omega}{2\pi} = \rho \omega r L dr$.
This acts as a solenoid with $n = \frac{1}{L}$ turns per unit length (effectively). The magnetic field produced by this shell at the center $P$ is $dB = \mu_0 n dI = \mu_0 \left(\frac{1}{L}\right) (\rho \omega r L dr) = \mu_0 \omega \rho r dr$.
Integrating from $r = a$ to $r = b$,the total magnetic field at $P$ is $B_P = \int_a^b \mu_0 \omega \rho r dr = \frac{\mu_0 \omega \rho (b^2 - a^2)}{2}$.
For a long solenoid,the magnetic field at the ends ($A$ or $B$) is exactly half of the field at the center: $B_A = B_B = \frac{B_P}{2} = \frac{\mu_0 \omega \rho (b^2 - a^2)}{4}$.
The direction of the magnetic field is determined by the right-hand rule and is the same along the axis. The graph of the magnetic field along the axis shows a constant value in the middle and a decrease towards the ends,which is consistent with the properties of a long solenoid.

(B) Consider two large parallel sheets carrying surface current density $K$ in the same direction.
Using the right-hand rule for a single sheet,the magnetic field $B$ is parallel to the sheet and perpendicular to the current direction.
For a sheet with current $K$ flowing to the right,the magnetic field points 'out' (towards us) above the sheet and 'in' (away from us) below the sheet.
Let the top sheet be $S_1$ and the bottom sheet be $S_2$.
Above $S_1$: Both sheets produce a field pointing 'out' (towards us).
Between $S_1$ and $S_2$: $S_1$ produces a field pointing 'in' (away from us),and $S_2$ produces a field pointing 'out' (towards us). Since the magnitudes are equal,they cancel out,resulting in zero field.
Below $S_2$: Both sheets produce a field pointing 'in' (away from us).
Thus,the magnetic field is zero between the plates and points towards us above the plates and away from us below the plates.

(D) For a long hollow cylindrical wire with inner radius $R_1$ and outer radius $R_2$ carrying a total current $i$ uniformly distributed:
$1$. For $r < R_1$: The enclosed current is zero,so by Ampere's Law,$\oint B \cdot dl = \mu_0 I_{enclosed} = 0$,which implies $B = 0$.
$2$. For $R_1 \le r \le R_2$: The enclosed current $I(r)$ is $i \cdot \frac{\pi(r^2 - R_1^2)}{\pi(R_2^2 - R_1^2)}$. Applying Ampere's Law: $B(2\pi r) = \mu_0 i \frac{r^2 - R_1^2}{R_2^2 - R_1^2}$,so $B = \frac{\mu_0 i}{2\pi r} \frac{r^2 - R_1^2}{R_2^2 - R_1^2}$. This shows $B$ increases from $0$ at $r = R_1$ to a maximum value at $r = R_2$.
$3$. For $r > R_2$: The enclosed current is the total current $i$. Applying Ampere's Law: $B(2\pi r) = \mu_0 i$,so $B = \frac{\mu_0 i}{2\pi r}$. This shows $B$ decreases as $1/r$.
The correct graph shows $B=0$ for $r < R_1$,a smooth increase for $R_1 \le r \le R_2$,and a $1/r$ decay for $r > R_2$. Option $D$ correctly represents this behavior.

(A) The energy stored in a solenoid is given by $U = \frac{1}{2} LI^2$.
The self-inductance $L$ of a solenoid is $L = \frac{\mu_0 \mu_r N^2 A}{\ell}$, where $N$ is the total number of turns and $\ell$ is the length.
Energy per unit length $u = \frac{U}{\ell} = \frac{1}{\ell} \left( \frac{1}{2} \frac{\mu_0 \mu_r N^2 A}{\ell} I^2 \right) = \frac{1}{2} \mu_0 \mu_r \left( \frac{N}{\ell} \right)^2 A I^2$.
Given: turn density $n = \frac{N}{\ell} = 5000 \, \text{turns/m}$, area $A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2$, current $I = 1 \, \text{A}$, relative permeability $\mu_r = 1000$, and $\mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}$.
Substituting the values:
$u = \frac{1}{2} \times (4\pi \times 10^{-7}) \times 1000 \times (5000)^2 \times (10 \times 10^{-4}) \times (1)^2$
$u = \frac{1}{2} \times 4\pi \times 10^{-7} \times 10^3 \times 25 \times 10^6 \times 10^{-3} \times 1$
$u = 2\pi \times 10^{-7} \times 25 \times 10^9 \times 10^{-3} = 50\pi \approx 157.08 \times 0.1 = 15.7 \, \text{J/m}$.

(B) According to Ampere's circuital law,$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$.
Let the current flowing out of the plane be positive and into the plane be negative.
For path $a$: It encloses $2\, A$ (out) and $2\, A$ (in). Net current $I_a = 2 - 2 = 0\, A$.
For path $b$: It encloses $2\, A$ (out) and $3\, A$ (out). Net current $I_b = 2 + 3 = 5\, A$.
For path $c$: It encloses $2\, A$ (in) and $3\, A$ (out). Net current $I_c = 3 - 2 = 1\, A$.
For path $d$: It encloses $2\, A$ (out),$2\, A$ (in),and $3\, A$ (out). Net current $I_d = 2 - 2 + 3 = 3\, A$.
Comparing the net currents: $I_a = 0\, A$,$I_c = 1\, A$,$I_d = 3\, A$,$I_b = 5\, A$.
Therefore,the order from smallest to largest is $a, c, d, b$.

(B) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the net current $I_{\text{en}}$ enclosed by the loop: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{en}}$.
For any point inside an infinitely long thin-walled tube,we can consider a circular Amperian loop of radius $r$ (where $r < R$,the radius of the tube).
Since the current $i$ flows only along the surface of the tube,the current enclosed by this Amperian loop is $I_{\text{en}} = 0$.
Therefore,$\oint \vec{B} \cdot d\vec{l} = \mu_0(0) = 0$.
This implies that the magnetic induction $B$ at any point inside the tube is $0$.

(C) The magnetic moment $m$ of a solenoid is given by $m = N i A$,where $N$ is the number of turns,$i$ is the current,and $A$ is the cross-sectional area.
Magnetization $\vec M$ is defined as the magnetic moment per unit volume $V$.
$V = A \ell$,where $\ell$ is the length of the solenoid.
Therefore,$|\vec M| = \frac{m}{V} = \frac{N i A}{A \ell} = \frac{N i}{\ell}$.
Given $N = 500$,$i = 15\,A$,and $\ell = 25\,cm = 0.25\,m$.
$|\vec M| = \frac{500 \times 15}{0.25} = \frac{7500}{0.25} = 30000\,A m^{-1}$.

(C) The total number of turns $N$ for $6$ layers, each with $450$ turns, is given by:
$N = 6 \times 450 = 2700$
The number of turns per unit length $n$ is calculated as:
$n = \frac{N}{l} = \frac{2700}{90 \times 10^{-2}\, m} = \frac{2700}{0.9}\, m^{-1} = 3000\, m^{-1}$
The magnetic field $B$ inside a long solenoid near its centre is given by the formula:
$B = \mu_{0} n I$
Substituting the values $\mu_{0} = 4\pi \times 10^{-7}\, T\cdot m/A$, $n = 3000\, m^{-1}$, and $I = 6\, A$:
$B = (4\pi \times 10^{-7}) \times 3000 \times 6$
$B = 72\pi \times 10^{-4}\, T$
Since $1\, T = 10^4\, G$ (Gauss), we have:
$B = 72\pi \times 10^{-4} \times 10^4\, G = 72\pi\, G$

(B) Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero,expressed as $\oint_S B \cdot dS = 0$.
In the case of an ideal solenoid,the magnetic field is assumed to be uniform inside and zero outside.
If we consider a Gaussian surface (a cylinder) partially inside and partially outside the solenoid,the flux through the end caps would be non-zero because the field lines are confined within the solenoid and do not exit through the sides,implying a violation of the divergence-free nature of magnetic fields $(
abla \cdot B = 0)$.
Therefore,the assumption of a perfectly confined magnetic field in an ideal solenoid contradicts Gauss's law for magnetism.

(D) According to Ampere's circuital law,the magnetic field $B$ at a distance $r$ from the axis of a current-carrying conductor is given by $\oint B \cdot dl = \mu_0 I_{enclosed}$.
For point $P$,which is in the hollow cavity,the enclosed current $I_{enclosed} = 0$,so the magnetic field $B_P = 0$.
For point $Q$,which is inside the material of the conducting pipe,the enclosed current is non-zero,so the magnetic field $B_Q \neq 0$.
For point $R$,which is outside the pipe,the total current enclosed by an Amperian loop passing through $R$ is the total current $I$ flowing through the pipe. However,for a long straight cylindrical conductor,the magnetic field outside is $B = \frac{\mu_0 I}{2 \pi r}$. This is only zero if $I=0$ or $r \to \infty$. Thus,$B_R \neq 0$.
Therefore,the magnetic field is zero only at point $P$.

(A) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n i$.
Here, $n$ is the number of turns per unit length.
Given: $n = 20 \text{ turns/cm} = 2000 \text{ turns/m}$.
Given: $B = 20 \text{ mT} = 20 \times 10^{-3} \text{ T}$.
Given: $\mu_0 = 4\pi \times 10^{-7} \text{ T} \cdot \text{m/A}$.
Substituting the values into the formula:
$20 \times 10^{-3} = (4\pi \times 10^{-7}) \times 2000 \times i$
$20 \times 10^{-3} = 8\pi \times 10^{-4} \times i$
$i = \frac{20 \times 10^{-3}}{8\pi \times 10^{-4}} = \frac{200}{8\pi} = \frac{25}{\pi} \approx \frac{25}{3.14} \approx 7.96 \text{ A}$.
Rounding to the nearest integer, the current is approximately $8 \text{ A}$.