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

(B) The magnetic intensity $H$ inside a solenoid is given by the formula $H = nI$,where $n$ is the number of turns per unit length and $I$ is the current flowing through the solenoid.
Given:
Number of turns per meter,$n = 1000 \ m^{-1}$.
Current,$I = 2 \ A$.
Substituting these values into the formula:
$H = 1000 \times 2 = 2000 \ \frac{A}{m}$.
This can be written in scientific notation as $2 \times 10^3 \ \frac{A}{m}$.
Therefore,the correct option is $B$.

(B) The magnetic intensity $H$ inside a solenoid is given by the formula $H = nI$,where $n$ is the number of turns per unit length and $I$ is the current flowing through the solenoid.
Given:
Number of turns per unit length $n = 1000 \text{ m}^{-1}$
Current $I = 2 \text{ A}$
Relative permeability $\mu_r = 400$ (Note: This value is not required to calculate magnetic intensity $H$,as $H$ depends only on the geometry and current).
Substituting the values into the formula:
$H = 1000 \times 2$
$H = 2000 \text{ A m}^{-1}$
$H = 2 \times 10^3 \text{ A m}^{-1}$
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 $n = N/l$ is the number of turns per unit length.
Given:
Length $l = 0.5 \ m$
Number of turns $N = 250$
Current $I = 5 \ A$
Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$
Substituting the values:
$B = \frac{\mu_0 N I}{l}$
$B = \frac{4 \pi \times 10^{-7} \times 250 \times 5}{0.5}$
$B = \frac{4 \times 3.14159 \times 10^{-7} \times 1250}{0.5}$
$B = 8 \pi \times 10^{-7} \times 1250$
$B = 10000 \pi \times 10^{-7} \ T$
$B = \pi \times 10^{-3} \ T \approx 3.14 \times 10^{-3} \ T$.

(C) The magnetic field inside a solenoid with a magnetic core is given by the formula:
$B = \mu_0 \mu_r n I$
Where:
$\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$
$\mu_r = 400$
$n = 1000 \text{ turns/m}$
$I = 1 \text{ A}$
Substituting the values:
$B = (4\pi \times 10^{-7}) \times 400 \times 1000 \times 1$
$B = 4\pi \times 4 \times 10^5 \times 10^{-7}$
$B = 16\pi \times 10^{-2} \text{ T}$

(A) The magnetic field $B$ inside a long solenoid is given by the formula $B = \mu_0 n I$, where $n = N/l$ is the number of turns per unit length.
Given:
Length $l = 0.25 \ m$
Number of turns $N = 500$
Current $I = 2.5 \ A$
Permeability $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
Calculation:
$n = \frac{500}{0.25} = 2000 \ turns/m$
$B = (4 \pi \times 10^{-7}) \times 2000 \times 2.5$
$B = 4 \times 3.14 \times 10^{-7} \times 5000$
$B = 12.56 \times 10^{-4} \ T = 6.28 \times 2 \times 10^{-4} \ T = 6.28 \times 10^{-3} \ T$.

(A) The magnetic field $B$ inside a long solenoid is given by the formula: $B = \mu_0 n I$,where $n = \frac{N}{l}$ is the number of turns per unit length.
Given:
Length $l = 0.5 \ m$
Number of turns $N = 1000$
Current $I = 10 \ A$
Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$
Substituting the values:
$n = \frac{1000}{0.5} = 2000 \ turns/m$
$B = (4\pi \times 10^{-7}) \times 2000 \times 10$
$B = 8\pi \times 10^{-3} \ T$
$B \approx 25.12 \times 10^{-3} \ T = 2.51 \times 10^{-2} \ T$.

(B) 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 and $I$ is the current.
Given:
Number of turns per unit length $n = 50 \text{ turns/cm} = 50 \times 10^2 \text{ turns/m} = 5000 \text{ turns/m}$.
Current $I = 2.5 \ A$.
Permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$.
Substituting the values:
$B = (4 \pi \times 10^{-7}) \times (5000) \times (2.5)$
$B = 4 \pi \times 10^{-7} \times 12500$
$B = 4 \pi \times 1.25 \times 10^{-3}$
$B = 5 \pi \times 10^{-3} \ T$.

(D) For a long straight wire of radius $r$ carrying a uniform current $I$,the magnetic field $B$ at a distance $a$ from the axis inside the wire $(a < r)$ is given by Ampere's Law.
Using Ampere's circuital law: $\oint B \cdot dl = \mu_0 I_{enclosed}$.
For a point inside the wire,the enclosed current $I_{enclosed} = I \times (\frac{\pi a^2}{\pi r^2}) = I \frac{a^2}{r^2}$.
Thus,$B(2 \pi a) = \mu_0 I \frac{a^2}{r^2}$.
Solving for $B$,we get $B = \frac{\mu_0 I a}{2 \pi r^2}$.
Since $\mu_0$,$I$,and $r$ are constants,$B \propto a$.

(B) The magnetic field $B$ inside a long solenoid near its centre is given by the formula $B = \mu_0 n I$,where $n$ is the number of turns per unit length.
Given:
Length of solenoid $L = 120 \ cm = 1.2 \ m$.
Number of layers $= 4$.
Turns per layer $= 400$.
Total number of turns $N = 4 \times 400 = 1600$.
Current $I = 8.0 \ A$.
Number of turns per unit length $n = \frac{N}{L} = \frac{1600}{1.2} = \frac{4000}{3} \ m^{-1}$.
Substituting the values into the formula:
$B = (4 \pi \times 10^{-7} \ T \cdot m/A) \times (\frac{4000}{3} \ m^{-1}) \times (8.0 \ A)$.
$B = 4 \pi \times 10^{-7} \times \frac{32000}{3} \ T$.
$B = \frac{128000}{3} \pi \times 10^{-7} \ T$.
$B \approx 42666.67 \pi \times 10^{-7} \ T$.
$B \approx 4.266 \pi \times 10^{-3} \ T$.
Rounding to two decimal places,we get $B \approx 4.27 \pi \times 10^{-3} \ T$.

(C) 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 $I_{\text{enclosed}}$ passing through the surface bounded by the path: $\oint \vec{B} \cdot d\vec{l} = \mu_{0} I_{\text{enclosed}}$.
For an internal point inside the pipe where the radial distance $r < R$ (where $R$ is the radius of the pipe),any closed circular path drawn inside the pipe encloses zero net current because the current flows only through the walls of the pipe.
Thus,$I_{\text{enclosed}} = 0$.
Applying Ampere's law: $B(2 \pi r) = \mu_{0}(0) \implies B = 0$.
Therefore,the magnetic field is zero at any point inside the pipe.
For an external point where $r > R$,the path encloses the total current $I$,so $B(2 \pi r) = \mu_{0} I \implies B = \frac{\mu_{0} I}{2 \pi r}$.
Hence,option $(c)$ is correct.

(B) Given: Length of solenoid $\ell = 1 \ m$,Diameter $d = 4 \ cm$,Radius $r = 2 \ cm = 0.02 \ m$.
Total number of turns $N = 5 \times 1000 = 5000$.
Current $I = 7 \ A$.
The magnetic field at the centre of a finite solenoid is given by $B = \frac{\mu_0 N I}{2\ell} (\cos \theta_1 + \cos \theta_2)$.
Since the solenoid is long compared to its radius,we can approximate the field using the formula for an ideal solenoid $B = \mu_0 n I$,where $n = N/\ell$.
$B = \frac{\mu_0 N I}{\ell} = \frac{4\pi \times 10^{-7} \times 5000 \times 7}{1}$.
$B = 4\pi \times 10^{-7} \times 35000 = 14\pi \times 10^{-3} \ T$.
$B \approx 14 \times 3.14159 \times 10^{-3} \ T = 43.98 \times 10^{-3} \ T = 4.398 \times 10^{-2} \ T$.
Rounding to the given options,the correct value is $4.396 \times 10^{-2} \ T$.

(D) Given,length of solenoid,$l = 50 \,cm = 0.5 \,m$.
Number of turns,$N = 100$.
Current,$I = 2.5 \,A$.
Number of turns per unit length in the solenoid is given by $n = \frac{N}{l} = \frac{100}{0.5} = 200 \,turns/m$.
The magnetic field at one end of a long solenoid is given by the formula $B = \frac{\mu_0 n I}{2}$.
Substituting the values,we get $B = \frac{4\pi \times 10^{-7} \times 200 \times 2.5}{2}$.
$B = 2\pi \times 10^{-7} \times 500 = 1000\pi \times 10^{-7} = \pi \times 10^{-4} \,T$.
Using $\pi \approx 3.14$,we get $B = 3.14 \times 10^{-4} \,T$.

(C) The magnetic field $B$ due to a toroid is determined using Ampere's circuital law:
$(i)$ For $r < R_{1}$ (inside the hollow space of the toroid),the net current enclosed is zero,so $B = 0$.
(ii) For $R_{1} < r < R_{2}$ (inside the core of the toroid),the magnetic field is given by $B = \frac{\mu_{0} N I}{2 \pi r}$.
(iii) For $r > R_{2}$ (outside the toroid),the net current enclosed by the Amperian loop is zero,so $B = 0$.
Since the question asks for the variation of the magnetic field with radial distance $r$ within the core,and the options provided represent the behavior across the cross-section,the correct representation for the field inside the toroid is a curve proportional to $1/r$,which is shown in graph $C$.

(C) For a long cylindrical wire of radius $R$ carrying a uniform current $I$:
$1$. Inside the wire $(r < R)$,the magnetic field $B_{\text{in}}$ is given by $B_{\text{in}} = \frac{\mu_0 I r}{2 \pi R^2}$. Since $\mu_0, I, R$ are constants,we have $B_{\text{in}} \propto r$. This represents a straight line passing through the origin.
$2$. Outside the wire $(r > R)$,the magnetic field $B_{\text{out}}$ is given by $B_{\text{out}} = \frac{\mu_0 I}{2 \pi r}$. Since $\mu_0, I$ are constants,we have $B_{\text{out}} \propto \frac{1}{r}$. This represents a rectangular hyperbola.
$3$. At the surface $(r = R)$,the magnetic field is maximum,$B_{\text{max}} = \frac{\mu_0 I}{2 \pi R}$.
Combining these,the graph shows a linear increase for $r < R$ and a hyperbolic decrease for $r > R$,which corresponds to Graph $C$.

(B) The magnetic field $B$ inside a toroid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current.
Given: $n = 500 \text{ m}^{-1}$, $I = 2 \text{ A}$, $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
The magnetic energy density $u_B$ is given by $u_B = \frac{B^2}{2\mu_0}$.
Substituting $B = \mu_0 n I$ into the formula, we get $u_B = \frac{(\mu_0 n I)^2}{2\mu_0} = \frac{\mu_0 n^2 I^2}{2}$.
Calculating the value: $u_B = \frac{(4\pi \times 10^{-7}) \times (500)^2 \times (2)^2}{2}$.
$u_B = \frac{(4\pi \times 10^{-7}) \times 250000 \times 4}{2} = 2\pi \times 10^{-7} \times 10^6 = 2\pi \times 10^{-1} = 0.2 \times 3.14 = 0.628 \text{ J/m}^3$.

(A) Given: Number of turns per unit length $ n = 40 \text{ turns/cm} = 4000 \text{ turns/m} = 4 \times 10^3 \text{ m}^{-1} $.
Current $ I = 1 \text{ A} $.
The magnetic energy stored per unit volume $ u_m $ is given by the formula $ u_m = \frac{1}{2} \frac{B^2}{\mu_0} = \frac{1}{2} \mu_0 n^2 I^2 $.
Substituting the values:
$ u_m = \frac{1}{2} \times (4\pi \times 10^{-7} \text{ T m/A}) \times (4 \times 10^3 \text{ m}^{-1})^2 \times (1 \text{ A})^2 $.
$ u_m = \frac{1}{2} \times 4\pi \times 10^{-7} \times 16 \times 10^6 \times 1 $.
$ u_m = 2\pi \times 16 \times 10^{-1} $.
$ u_m = 32\pi \times 0.1 = 3.2\pi \text{ J m}^{-3} $.

(A) Given: Length of solenoid $l = 0.4 \,m$, number of turns $N = 400$, current $I = 5 \,A$.
The number of turns per unit length $n = \frac{N}{l} = \frac{400}{0.4} = 1000 \,m^{-1}$.
The magnetic field inside a long solenoid is given by $B = \mu_0 n I$.
Substituting the values: $B = (4\pi \times 10^{-7}) \times 1000 \times 5$.
$B = 4 \times 3.14159 \times 10^{-7} \times 5000$.
$B = 20 \times 3.14159 \times 10^{-4} = 62.83 \times 10^{-4} = 6.28 \times 10^{-3} \,T$.

(D) Given: Mass $m = 2.2 \times 10^{-30} \,kg$,Charge $q = 1.6 \times 10^{-19} \,C$,Velocity $v = 10 \,km/s = 10^4 \,m/s$,Radius $r = 2.8 \,cm = 2.8 \times 10^{-2} \,m$,Turns per unit length $n = 25 \,turns/cm = 2500 \,turns/m$.
For a charged particle in a magnetic field,the radius of the circular path is given by $r = \frac{mv}{Bq}$.
The magnetic field inside a solenoid is $B = \mu_0 n I$.
Substituting $B$ into the radius formula: $r = \frac{mv}{(\mu_0 n I)q}$.
Rearranging for current $I$: $I = \frac{mv}{\mu_0 n q r}$.
Substituting the values: $I = \frac{2.2 \times 10^{-30} \times 10^4}{4\pi \times 10^{-7} \times 2500 \times 1.6 \times 10^{-19} \times 2.8 \times 10^{-2}}$.
$I = \frac{2.2 \times 10^{-26}}{4 \times 3.14159 \times 10^{-7} \times 2500 \times 1.6 \times 10^{-19} \times 2.8 \times 10^{-2}} \approx 1.56 \times 10^{-3} \,A = 1.56 \,mA$.

(B) The magnetic field $B$ at the surface of a wire carrying current $I$ is given by the formula derived from Ampere's Law: $B = \frac{\mu_0 I}{2 \pi r}$.
Here,the current $I = 12 \ A$.
The diameter of the wire is $d = 1.2 \ mm = 1.2 \times 10^{-3} \ m$.
The radius $r = \frac{d}{2} = 0.6 \times 10^{-3} \ m$.
The permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$.
Substituting the values:
$B = \frac{(4 \pi \times 10^{-7}) \times 12}{2 \pi \times 0.6 \times 10^{-3}}$
$B = \frac{2 \times 10^{-7} \times 12}{0.6 \times 10^{-3}}$
$B = \frac{24 \times 10^{-7}}{0.6 \times 10^{-3}} = 40 \times 10^{-4} \ T = 4 \times 10^{-3} \ T = 4 \ mT$.
Thus,the maximum magnetic field is $4 \ mT$.

(B) Ampere's circuital law states that the line integral of the magnetic field $B$ around any closed loop is equal to $\mu_0$ times the total current $I$ passing through the surface enclosed by the loop.
Mathematically,this is expressed as $\oint B \cdot dl = \mu_0 I$.

(B) The magnetic energy density $(u_B)$ stored in a magnetic field $B$ is given by the formula: $u_B = \frac{B^2}{2\mu_0}$.
We know that the speed of light in vacuum is $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$,which implies $c^2 = \frac{1}{\mu_0 \varepsilon_0}$.
From this,we can express the permeability of free space as $\mu_0 = \frac{1}{\varepsilon_0 c^2}$.
Substituting this value of $\mu_0$ into the energy density formula:
$u_B = \frac{B^2}{2(1 / \varepsilon_0 c^2)}$
$u_B = \frac{\varepsilon_0 c^2 B^2}{2}$.
Therefore,the correct option is $B$.

(A) 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.
Given:
Length $L = 1 \ m$
Current $I = 5 \ A$
Number of layers $= 5$
Turns per layer $= 700$
Total number of turns $N = 5 \times 700 = 3500$
Since the length is $1 \ m$,the number of turns per unit length $n = N / L = 3500 / 1 = 3500 \ m^{-1}$.
The permeability of free space $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$.
Substituting the values:
$B = (4\pi \times 10^{-7}) \times 3500 \times 5$
$B = 20\pi \times 3500 \times 10^{-7}$
$B = 70000\pi \times 10^{-7}$
$B = 7\pi \times 10^{-3} \ T$
Using $\pi \approx 3.14$,$B \approx 7 \times 3.14 \times 10^{-3} \ T = 21.98 \times 10^{-3} \ T \approx 22 \ mT$.
Thus,the correct option is $A$.

$(A)$ 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.
For the first solenoid: $B_1 = \mu_0 n_1 i_1 = 6.24 \times 10^{-2} \,T$,where $n_1 = 400$ and $i_1 = i$.
So,$\mu_0 (400) i = 6.24 \times 10^{-2} \,T$.
For the second solenoid: $n_2 = 200$ and $i_2 = \frac{i}{2}$.
The magnetic field $B_2 = \mu_0 n_2 i_2 = \mu_0 (200) \left( \frac{i}{2} \right) = \mu_0 (100) i$.
Comparing $B_2$ with $B_1$: $B_2 = \frac{\mu_0 (100) i}{\mu_0 (400) i} \times B_1 = \frac{1}{4} \times 6.24 \times 10^{-2} \,T$.
$B_2 = 1.56 \times 10^{-2} \,T$.

(D) The magnetic field $B$ inside a toroid is given by the formula:
$B = \mu_0 \left( \frac{N}{2 \pi R} \right) I$
where $N$ is the number of turns,$R$ is the average radius,and $I$ is the current.
From the formula,we see that $B \propto \frac{N}{R}$.
Given:
$N_1 = 400, R_1 = 30 \ cm$
$N_2 = 200, R_2 = 60 \ cm$
Since the current $I$ is the same for both,the ratio of the magnetic fields is:
$\frac{B_1}{B_2} = \left( \frac{N_1}{N_2} \right) \times \left( \frac{R_2}{R_1} \right)$
$\frac{B_1}{B_2} = \left( \frac{400}{200} \right) \times \left( \frac{60}{30} \right)$
$\frac{B_1}{B_2} = 2 \times 2 = 4$
Therefore,the ratio is $4:1$.

(D) The magnetizing field (or magnetic intensity) $H$ of a solenoid is given by the formula $H = nI$,where $n$ is the number of turns per unit length and $I$ is the current flowing through the windings.
Given:
$n = 500 \,m^{-1}$
$I = 4 \,A$
Substituting these values into the formula:
$H = 500 \times 4 = 2000 \,Am^{-1}$
$H = 2 \times 10^3 \,Am^{-1}$
Comparing this result with the given options,none of the options match $2000 \,Am^{-1}$. Therefore,the correct option is $D$.

(D) The mean radius $r$ of the toroid is given by the average of the inner and outer radii:
$r = \frac{r_1 + r_2}{2} = \frac{24 \ cm + 25 \ cm}{2} = 24.5 \ cm = 24.5 \times 10^{-2} \ m$
Given number of turns $N = 4900$ and current $I = 12 \ A$.
The magnetic field $B$ inside the core of a toroid is given by the formula:
$B = \frac{\mu_0 N I}{2 \pi r}$
Substituting the values:
$B = \frac{(4 \pi \times 10^{-7} \ T \cdot m/A) \times 4900 \times 12}{2 \pi \times 24.5 \times 10^{-2} \ m}$
$B = \frac{2 \times 10^{-7} \times 4900 \times 12}{24.5 \times 10^{-2}}$
$B = \frac{2 \times 4900 \times 12}{24.5} \times 10^{-5}$
$B = \frac{117600}{24.5} \times 10^{-5} = 4800 \times 10^{-5} \ T = 48 \times 10^{-3} \ T = 48 \ mT$.

(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.
Given:
Length $L = 1 \,m$
Number of layers $= 5$
Turns per layer $= 500$
Total number of turns $N = 5 \times 500 = 2500$
Number of turns per unit length $n = N / L = 2500 / 1 = 2500 \,m^{-1}$
Magnetic field $B = 4.4 \,mT = 4.4 \times 10^{-3} \,T$
Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \,T \cdot m/A$
Substituting the values into the formula:
$4.4 \times 10^{-3} = (4\pi \times 10^{-7}) \times 2500 \times I$
$4.4 \times 10^{-3} = (4 \times 3.14159 \times 10^{-7}) \times 2500 \times I$
$4.4 \times 10^{-3} = 3.14159 \times 10^{-3} \times I$
$I = 4.4 / 3.14159 \approx 1.4 \,A$
Therefore, the current carried is $1.4 \,A$.

(A) According to Ampere's circuital law,$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$.
For a point inside the hollow cylinder $(r < R)$,the enclosed current $I_{\text{enclosed}} = 0$. Therefore,$B(2\pi r) = 0$,which implies $B = 0$ for $r < R$.
For a point outside the hollow cylinder $(r \geq R)$,the enclosed current $I_{\text{enclosed}} = i$. Therefore,$B(2\pi r) = \mu_0 i$,which implies $B = \frac{\mu_0 i}{2\pi r}$.
This shows that the magnetic field is zero inside the cylinder and decreases as $1/r$ outside the cylinder. Graph $(i)$ correctly represents this behavior.

(A) Given: Inner radius $r_1 = 20 \,cm = 0.2 \,m$, Outer radius $r_2 = 25 \,cm = 0.25 \,m$, Number of turns $N = 800$, Current $I = 12 \,A$.
The magnetic field $B$ inside a toroid at a distance $r$ from the center is given by $B = \frac{\mu_0 N I}{2 \pi r}$.
Since $B \propto \frac{1}{r}$, the magnetic field is maximum at the minimum radius $(r_1)$ and minimum at the maximum radius $(r_2)$.
Maximum magnetic field: $B_{\max} = \frac{4 \pi \times 10^{-7} \times 800 \times 12}{2 \pi \times 0.2} = \frac{2 \times 10^{-7} \times 9600}{0.2} = 9.6 \times 10^{-3} \,T = 9.6 \,mT$.
Minimum magnetic field: $B_{\min} = \frac{4 \pi \times 10^{-7} \times 800 \times 12}{2 \pi \times 0.25} = \frac{2 \times 10^{-7} \times 9600}{0.25} = 7.68 \times 10^{-3} \,T = 7.68 \,mT$.

(A) An ideal solenoid is defined as a coil where the length $L$ is much greater than its radius $R$ $(L \gg R)$ and the turns are closely wound. In such a configuration,the magnetic field inside the solenoid is almost uniform and parallel to the axis. Therefore,the statement that the turns are widely separated is incorrect.

(D) To find the magnetic field at a point $P$ at a distance $r = 3R/2$ from the axis,we use Ampere's circuital law: $\oint B \cdot dl = \mu_0 I_{\text{enclosed}}$.
Since the current $i$ is uniformly distributed over the cross-sectional area $A = \pi(2R)^2 - \pi R^2 = 3\pi R^2$,the current density $J = \frac{i}{3\pi R^2}$.
The current enclosed by the Amperian loop of radius $r = 3R/2$ is $I_{\text{enclosed}} = J \times \text{Area}_{\text{enclosed}} = J \times \pi(r^2 - R^2) = \frac{i}{3\pi R^2} \times \pi \left( \left(\frac{3R}{2}\right)^2 - R^2 \right) = \frac{i}{3R^2} \times \left( \frac{9R^2}{4} - R^2 \right) = \frac{i}{3R^2} \times \frac{5R^2}{4} = \frac{5i}{12}$.
Applying Ampere's law: $B(2\pi r) = \mu_0 I_{\text{enclosed}}$.
$B \times 2\pi \left(\frac{3R}{2}\right) = \mu_0 \left(\frac{5i}{12}\right)$.
$B(3\pi R) = \frac{5\mu_0 i}{12}$.
$B = \frac{5\mu_0 i}{36\pi R}$.

(B) The magnetic moment $M$ of a solenoid is given by the formula: $M = N I A$, where $N$ is the number of turns, $I$ is the current, and $A$ is the area of cross-section.
Given:
$N = 1200$
$A = 5 \,cm^2 = 5 \times 10^{-4} \,m^2$
$M = 1.2 \,J \,T^{-1}$
Substituting the values into the formula:
$1.2 = 1200 \times I \times (5 \times 10^{-4})$
$1.2 = 1200 \times 5 \times 10^{-4} \times I$
$1.2 = 6000 \times 10^{-4} \times I$
$1.2 = 0.6 \times I$
$I = \frac{1.2}{0.6} = 2 \,A$
Therefore, the current through the solenoid is $2 \,A$.

(C) toroid is essentially a solenoid bent into a circular shape. The magnetic field lines produced by a toroid are confined entirely within its core,forming concentric circles.
Because the magnetic field is confined within the toroid,the net magnetic field outside the toroid is zero.
If the magnetic moment $m$ were non-zero,the toroid would act like a magnetic dipole at large distances,producing a magnetic field that falls off as $\frac{1}{r^3}$.
Since the magnetic field outside an ideal toroid is zero,its magnetic moment $m$ must be zero.

(B) Given: mass $m = 1 \times 10^{-27} \ kg$,charge $q = 1 \times 10^{-16} \ C$,speed $v = 1000 \ m/s$,angle $\theta = 60^{\circ}$,number of turns $N = 5000$,current $I = 5 \ A$.
The velocity components are:
$v_{\parallel} = v \cos 60^{\circ} = 1000 \times 0.5 = 500 \ m/s$ (along the axis)
$v_{\perp} = v \sin 60^{\circ} = 1000 \times \frac{\sqrt{3}}{2} = 500\sqrt{3} \ m/s$ (perpendicular to the axis)
The magnetic field inside the solenoid is $B = \mu_0 n I = \mu_0 (N/L) I$.
The time taken to travel the length $L$ of the solenoid is $t = \frac{L}{v_{\parallel}} = \frac{L}{500}$.
The time period of one revolution in the magnetic field is $T = \frac{2\pi m}{qB}$.
The number of revolutions $n'$ is given by $n' = \frac{t}{T} = \frac{L/v_{\parallel}}{2\pi m / qB} = \frac{L \cdot q \cdot B}{v_{\parallel} \cdot 2\pi m}$.
Substituting $B = \frac{\mu_0 N I}{L}$:
$n' = \frac{L \cdot q \cdot (\mu_0 N I / L)}{v_{\parallel} \cdot 2\pi m} = \frac{q \mu_0 N I}{v_{\parallel} \cdot 2\pi m}$.
Plugging in the values:
$n' = \frac{10^{-16} \times (4\pi \times 10^{-7}) \times 5000 \times 5}{500 \times 2\pi \times 10^{-27}}$
$n' = \frac{10^{-16} \times 4\pi \times 10^{-7} \times 25000}{1000\pi \times 10^{-27}}$
$n' = \frac{10^{-16} \times 10^{-7} \times 10^5}{10^{-27} \times 10^2} = \frac{10^{-18}}{10^{-25}} = 10^7 / 10 = 10^6$.
Thus,the number of revolutions is $1 \times 10^6$.

(A) The magnetic field produced by the current in the toroid winding is given by $B = \mu_0 n I$.
Given: $\mu_0 = 4 \pi \times 10^{-7} \text{ H m}^{-1}$,$n = 1500 \text{ turns/m}$,and $I = 10 \text{ A}$.
$B = 4 \pi \times 10^{-7} \times 1500 \times 10 = 6 \pi \times 10^{-3} \text{ T} = 6 \pi \text{ mT}$.
The total internal magnetic field $B_0$ is given as $10 \pi \text{ mT}$.
The total magnetic field is the sum of the field due to the current and the field due to magnetization $(B_m)$:
$B_0 = B + B_m$.
Therefore,$B_m = B_0 - B = 10 \pi \text{ mT} - 6 \pi \text{ mT} = 4 \pi \text{ mT}$.

(C) The magnetic field inside an ideal solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.
Given: Length $L = 0.5 \ m$,Radius $R = 0.1 \ m$,Current $I = 2.5 \ A$.
Total turns $N = 2 \times 100 = 200$.
Number of turns per unit length $n = N / L = 200 / 0.5 = 400 \ turns/m$.
The magnetic field $B = \mu_0 n I = (4 \pi \times 10^{-7}) \times 400 \times 2.5$.
$B = (4 \pi \times 10^{-7}) \times 1000 = 4 \pi \times 10^{-4} \ T$.
Since the point is $5 \ cm$ from the axis,it lies inside the solenoid (as $5 \ cm < 10 \ cm$).
Thus,the magnetic field is $4 \pi \times 10^{-4} \ T$.

(A) The magnetic intensity $H$ inside a solenoid is given by the formula $H = nI$,where $n$ is the number of turns per unit length and $I$ is the current flowing through the solenoid.
Given:
Number of turns per unit length,$n = 1000 \ turns/m = 10^3 \ m^{-1}$.
Current,$I = 2 \ A$.
Relative permeability,$\mu_r = 400$ (Note: Magnetic intensity $H$ is independent of the core material).
Substituting the values:
$H = 10^3 \ m^{-1} \times 2 \ A = 2000 \ Am^{-1} = 2 \times 10^3 \ Am^{-1}$.
Thus,the magnetic intensity inside the solenoid is $2 \times 10^3 \ Am^{-1}$.

(B) Length of the solenoid, $L = 80 \,cm = 0.8 \,m$.
Total number of turns, $N = 5 \times 400 = 2000$.
Current, $i = 8 \,A$.
Number of turns per unit length, $n = \frac{N}{L} = \frac{2000}{0.8} = 2500 \,turns/m$.
The magnetic field $B$ inside a long solenoid near its centre is given by $B = \mu_0 ni$.
Substituting the values:
$B = (4 \pi \times 10^{-7} \,T \cdot m/A) \times (2500 \,m^{-1}) \times (8 \,A)$.
$B = 4 \times 3.14159 \times 10^{-7} \times 20000$.
$B = 12.566 \times 10^{-3} \times 2 = 2.513 \times 10^{-2} \,T$.
Thus, the magnitude of the magnetic field is approximately $2.5 \times 10^{-2} \,T$.

(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 any point inside a thin-walled pipe,we can choose an Amperian loop (a circle) that lies entirely inside the pipe.
Since the current $i$ flows only through the walls of the pipe,the current enclosed by this loop is $i_{\text{enclosed}} = 0$.
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) Given:
Inner radius $r_1 = 24 \ cm = 0.24 \ m$
Outer radius $r_2 = 26 \ cm = 0.26 \ m$
Number of turns $N = 2000$
Current $I = 12 \ A$
The mean radius $r$ of the toroid is given by:
$r = \frac{r_1 + r_2}{2} = \frac{24 + 26}{2} = 25 \ cm = 0.25 \ m$
The magnetic field $B$ inside the core of a toroid is given by the formula:
$B = \mu_0 n I$
where $n$ is the number of turns per unit length,$n = \frac{N}{2 \pi r}$.
Substituting the values:
$B = \mu_0 \left( \frac{N}{2 \pi r} \right) I$
$B = (4 \pi \times 10^{-7}) \times \left( \frac{2000}{2 \pi \times 0.25} \right) \times 12$
$B = (2 \times 10^{-7}) \times \left( \frac{2000}{0.25} \right) \times 12$
$B = (2 \times 10^{-7}) \times 8000 \times 12$
$B = 192000 \times 10^{-7} \ T$
$B = 1.92 \times 10^{-2} \ T$

(B) Given:
$n = 70 \text{ turns } cm^{-1} = 7000 \text{ turns } m^{-1}$
$r = 2.5 \text{ cm} = 0.025 \text{ m}$
$v = 4.4 \times 10^6 \text{ m s}^{-1}$
$m = 9 \times 10^{-31} \text{ kg}$
$q = 1.6 \times 10^{-19} \text{ C}$

The centripetal force required for circular motion is provided by the magnetic Lorentz force:
$\frac{mv^2}{r} = qvB$
For a long solenoid, the magnetic field $B$ is given by:
$B = \mu_0 n I$
Substituting $B$ into the force equation:
$\frac{mv^2}{r} = qv(\mu_0 n I)$
Solving for current $I$:
$I = \frac{mv}{q \mu_0 n r}$
Substituting the values:
$I = \frac{(9 \times 10^{-31}) \times (4.4 \times 10^6)}{(1.6 \times 10^{-19}) \times (4 \pi \times 10^{-7}) \times (7000) \times (0.025)}$
$I = \frac{39.6 \times 10^{-25}}{1.6 \times 4 \times 3.14159 \times 10^{-7} \times 175}$
$I = \frac{39.6 \times 10^{-25}}{3518.57 \times 10^{-7}} \approx 0.1125 \text{ A} = 112.5 \text{ mA}$

(C) The magnetic field due to a long solenoid at any point inside it is directed along its axis: $B_s = \mu_0 n I_s$. Given $n = 10 \text{ turns/cm} = 1000 \text{ turns/m}$ and $I_s = 7 \times 10^{-3} \text{ A}$. Thus, $B_s = 4\pi \times 10^{-7} \times 1000 \times 7 \times 10^{-3} = 28\pi \times 10^{-7} \text{ T}$.
The magnetic field due to the straight conductor at a distance $r = 5 \text{ cm} = 0.05 \text{ m}$ is directed tangentially: $B_c = \frac{\mu_0 I_c}{2\pi r}$.
The resultant magnetic field makes an angle $\theta = 60^{\circ}$ with the axis. Therefore, $\tan(60^{\circ}) = \frac{B_c}{B_s}$.
$\sqrt{3} = \frac{\mu_0 I_c / (2\pi r)}{\mu_0 n I_s} = \frac{I_c}{2\pi r n I_s}$.
$I_c = \sqrt{3} \times 2\pi r n I_s = 1.7 \times 2 \times 3.14 \times 0.05 \times 1000 \times 7 \times 10^{-3}$.
$I_c = 1.7 \times 6.28 \times 0.05 \times 7 = 3.7366 \text{ A} \approx 3.74 \text{ A}$.

(C) The magnetic field intensity $(H)$ inside a long solenoid is given by the formula $H = n \cdot I$,where $n$ is the number of turns per unit length and $I$ is the current flowing through the solenoid.
Given:
$H = 1000 \ A/m$
$I = 2 \ A$
Using the formula $H = n \cdot I$:
$1000 = n \times 2$
$n = 500 \ \text{turns/meter}$.
To convert turns per meter to turns per centimeter:
$n = 500 \ \text{turns/m} = 500 \ \text{turns} / 100 \ \text{cm} = 5 \ \text{turns/cm}$.
Thus,the number of turns per centimeter is $5$.

(C) 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.
Since the magnetic field $B$ is directly proportional to the number of turns per unit length $n$ $(B \propto n)$,if the value of $n$ is doubled,the magnetic field $B$ will also double.

(B) According to Ampere's Circuital Law,for a long straight cylindrical conductor carrying a current $I$,the magnetic field $B$ at an external point $P$ at a distance $r$ from the axis of the conductor (where $r > R$) is given by:
$B = \frac{\mu_0 I}{2 \pi r}$
Here,$\mu_0$ is the permeability of free space,$I$ is the current,and $r$ is the radial distance from the center of the conductor to point $P$.

(B) The magnetic field at the center of a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length $(n = N/L)$.
Given:
Length $L = 50 \ cm = 0.5 \ m$
Number of turns $N = 400$
Magnetic field $B = 4 \pi \times 10^{-3} \ T$
Permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
First,calculate the number of turns per unit length:
$n = \frac{N}{L} = \frac{400}{0.5} = 800 \ turns/m$
Now,substitute the values into the formula $B = \mu_0 n I$:
$4 \pi \times 10^{-3} = (4 \pi \times 10^{-7}) \times 800 \times I$
$I = \frac{4 \pi \times 10^{-3}}{4 \pi \times 10^{-7} \times 800}$
$I = \frac{10^{-3}}{10^{-7} \times 800} = \frac{10^4}{800} = \frac{100}{8} = 12.5 \ A$.

(A) Given: Relative permeability, $\mu_r = \frac{800}{\pi}$.
Current, $I = 2 \text{ A}$.
Number of turns per unit length, $n = 1000 \text{ m}^{-1}$.
The magnetic field intensity $H$ inside the solenoid is given by $H = nI$.
$H = 1000 \times 2 = 2000 \text{ A/m} = 2 \times 10^3 \text{ A/m}$.
The magnetic field $B$ is given by $B = \mu H = \mu_0 \mu_r H$.
Substituting the values: $B = (4\pi \times 10^{-7}) \times (\frac{800}{\pi}) \times (2 \times 10^3)$.
$B = 4 \times 10^{-7} \times 800 \times 2 \times 10^3$.
$B = 6400 \times 10^{-4} \text{ T} = 0.64 \text{ T}$.
Since $1 \text{ T} = 1000 \text{ mT}$, $B = 0.64 \times 1000 \text{ mT} = 640 \text{ mT}$.

(A) The magnetic field at an axial point of a solenoid is given by $B = \frac{\mu_0 n I}{2}(\cos \theta_1 - \cos \theta_2)$.
For a point well inside the solenoid (at the center),$\theta_1 = 0^{\circ}$ and $\theta_2 = 180^{\circ}$.
Thus,$B_{\text{center}} = \frac{\mu_0 n I}{2}(\cos 0^{\circ} - \cos 180^{\circ}) = \frac{\mu_0 n I}{2}(1 - (-1)) = \mu_0 n I$.
For a point at the axial end of the solenoid,$\theta_1 = 90^{\circ}$ and $\theta_2 = 180^{\circ}$.
Thus,$B_{\text{end}} = \frac{\mu_0 n I}{2}(\cos 90^{\circ} - \cos 180^{\circ}) = \frac{\mu_0 n I}{2}(0 - (-1)) = \frac{\mu_0 n I}{2}$.
The ratio of the magnetic field at the center to the end is $\frac{B_{\text{center}}}{B_{\text{end}}} = \frac{\mu_0 n I}{\frac{\mu_0 n I}{2}} = 2$.

(B) The magnetic energy $U$ stored in a solenoid is given by $U = \frac{B^2 V}{2 \mu_0}$,where $B$ is the magnetic field,$V$ is the volume,and $\mu_0$ is the permeability of free space.
Given that the magnetic energy stored in both solenoids is the same,$U_X = U_Y$.
Therefore,$\frac{B_X^2 V_X}{2 \mu_0} = \frac{B_Y^2 V_Y}{2 \mu_0}$,which simplifies to $B_X^2 V_X = B_Y^2 V_Y$.
Since volume $V = A \times L$,we have $B_X^2 A_X L_X = B_Y^2 A_Y L_Y$.
Given $A_Y = 2 A_X$ and $L_Y = 2 L_X$,we substitute these into the equation:
$B_X^2 A_X L_X = B_Y^2 (2 A_X) (2 L_X)$.
$B_X^2 A_X L_X = 4 B_Y^2 A_X L_X$.
Dividing both sides by $A_X L_X$,we get $B_X^2 = 4 B_Y^2$.
Thus,$\frac{B_X^2}{B_Y^2} = 4$,which implies $\frac{|B_X|}{|B_Y|} = \sqrt{4} = 2$.
Therefore,the ratio is $2 : 1$.

(D) For a solenoid,the magnetic field is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.
Given: $I = 20 \ A$,$l = 2 \ m$,$B = 20 \ mT = 20 \times 10^{-3} \ T$,and diameter $d = 3 \ cm = 3 \times 10^{-2} \ m$.
Radius $r = \frac{d}{2} = 1.5 \times 10^{-2} \ m$.
Calculating the number of turns per unit length $n$:
$n = \frac{B}{\mu_0 I} = \frac{20 \times 10^{-3}}{4 \pi \times 10^{-7} \times 20} = \frac{10^4}{4 \pi} \ m^{-1}$.
Total number of turns $N = n \times l = \frac{10^4}{4 \pi} \times 2 = \frac{10^4}{2 \pi}$.
The length of the wire is the total number of turns multiplied by the circumference of one turn:
$L_{wire} = N \times (2 \pi r) = \left( \frac{10^4}{2 \pi} \right) \times (2 \pi \times 1.5 \times 10^{-2}) = 10^4 \times 1.5 \times 10^{-2} = 1.5 \times 10^2 \ m = 150 \ m$.