Magnetic Moment of Current Carrying Coil Questions in English

(C) The magnetic moment $M$ of a circular loop is given by the formula $M = I A$,where $I$ is the current and $A$ is the area of the loop.
For a circular loop of radius $R$,the area is $A = \pi R^2$.
Thus,$M = I (\pi R^2)$.
The length of the wire $L$ forms the circumference of the loop,so $L = 2 \pi R$,which implies $R = \frac{L}{2 \pi}$.
Substituting the value of $R$ into the magnetic moment equation:
$M = I \pi \left( \frac{L}{2 \pi} \right)^2$
$M = I \pi \left( \frac{L^2}{4 \pi^2} \right)$
$M = \frac{I L^2}{4 \pi}$
Rearranging to solve for $L$:
$L^2 = \frac{4 M \pi}{I}$
$L = \sqrt{\frac{4 M \pi}{I}}$ or $\left[ \frac{4 M \pi}{I} \right]^{\frac{1}{2}}$.

(A) Let $r$ be the radius of the circular loop.
Since the area $A = \pi r^2$,we have $r = \sqrt{\frac{A}{\pi}}$.
The magnetic field at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2r}$.
Substituting the value of $r$,we get $B = \frac{\mu_0 I}{2 \sqrt{A/\pi}}$.
Solving for current $I$,we get $I = \frac{2B}{\mu_0} \sqrt{\frac{A}{\pi}}$.
The magnetic moment $M$ of the loop is defined as $M = I A$.
Substituting the expression for $I$,we get $M = \left( \frac{2B}{\mu_0} \sqrt{\frac{A}{\pi}} \right) A = \frac{2 B A^{3/2}}{\mu_0 \sqrt{\pi}}$.
Thus,the correct option is $A$.

(A) Let the length of the wire be $l$. For a square loop,the side of the square is $a = l/4$.
Hence,its magnetic moment is $M_{sq} = i A = i (l/4)^2 = i l^2 / 16$.
For a circular loop,the circumference is $2 \pi r = l$,so the radius is $r = l / (2 \pi)$.
Hence,its magnetic moment is $M_{cir} = i A = i \pi r^2 = i \pi (l / (2 \pi))^2 = i \pi (l^2 / 4 \pi^2) = i l^2 / (4 \pi)$.
The ratio of the magnetic moment of the circle to that of the square is:
$\frac{M_{cir}}{M_{sq}} = \frac{i l^2 / (4 \pi)}{i l^2 / 16} = \frac{16}{4 \pi} = \frac{4}{\pi}$.

(A) The area of the circular coil is given by $A = \pi R^2$,where $R$ is the radius of the coil.
From this,the radius is $R = \sqrt{\frac{A}{\pi}}$.
The magnetic field at the centre of a circular coil carrying current $I$ is $B = \frac{\mu_0 I}{2 R}$.
Rearranging for current $I$,we get $I = \frac{2 B R}{\mu_0}$.
The magnetic moment $M$ of the coil is defined as $M = I A$.
Substituting the expression for $I$ and $R$ into the formula for $M$:
$M = \left( \frac{2 B R}{\mu_0} \right) A = \frac{2 B A}{\mu_0} \sqrt{\frac{A}{\pi}}$.
Simplifying this,we get $M = \frac{2 B A^{3/2}}{\mu_0 \sqrt{\pi}}$.

(D) The length of the conductor $L$ forms the circumference of the circular loop.
$L = 2 \pi r$,where $r$ is the radius of the loop.
Therefore,$r = \frac{L}{2 \pi}$.
The area $A$ of the circular loop is given by $A = \pi r^2$.
Substituting the value of $r$,we get $A = \pi \left( \frac{L}{2 \pi} \right)^2 = \pi \left( \frac{L^2}{4 \pi^2} \right) = \frac{L^2}{4 \pi}$.
The magnetic moment $M$ of a current-carrying loop is given by $M = I \times A$.
Substituting the value of $A$,we get $M = I \left( \frac{L^2}{4 \pi} \right) = \frac{IL^2}{4 \pi}$.

(D) The magnetic moment $M$ of a current loop is given by $M = I \times A$.
Here, the current $I$ due to the revolving electron is $I = qf = ef$, where $f$ is the frequency of revolution.
The area $A$ of the circular orbit is $A = \pi r^2$.
Substituting these into the formula, we get $M = (ef)(\pi r^2)$.
Given: $r = 0.05 \,nm = 0.05 \times 10^{-9} \,m = 5 \times 10^{-11} \,m$, $f = 10^{16} \,Hz$, and $e = 1.6 \times 10^{-19} \,C$.
$M = (1.6 \times 10^{-19} \,C) \times (10^{16} \,s^{-1}) \times (3.14) \times (5 \times 10^{-11} \,m)^2$.
$M = 1.6 \times 10^{-3} \times 3.14 \times 25 \times 10^{-22}$.
$M = 1.6 \times 3.14 \times 25 \times 10^{-25}$.
$M = 125.6 \times 10^{-25} = 1.256 \times 10^{-23} \,A-m^2 \approx 1.26 \times 10^{-23} \,A-m^2$.

(B) Let $r$ be the radius of the circular loop. The area of the loop is $A = \pi r^{2}$.
Therefore,$r = \sqrt{\frac{A}{\pi}}$.
The magnetic field $B$ at the centre of the circular loop is given by $B = \frac{\mu_{0} I}{2 r}$.
Substituting the value of $r$,we get $B = \frac{\mu_{0} I}{2 \sqrt{\frac{A}{\pi}}}$.
Solving for current $I$,we get $I = \frac{2 B}{\mu_{0}} \sqrt{\frac{A}{\pi}}$.
The magnetic moment $M$ is given by $M = I A$.
Substituting the value of $I$,we get $M = \left( \frac{2 B}{\mu_{0}} \sqrt{\frac{A}{\pi}} \right) A$.
$M = \frac{2 B}{\mu_{0}} \cdot \frac{A^{1/2}}{\pi^{1/2}} \cdot A = \frac{2 B A^{3/2}}{\mu_{0} \pi^{1/2}}$.

(B) The orbital magnetic moment $M$ of an electron revolving in a circular orbit is given by $M = iA$,where $i$ is the current and $A$ is the area of the orbit.
The current $i$ is given by $i = ef$,where $e$ is the charge of the electron and $f$ is the frequency of revolution.
The area of the circular orbit is $A = \pi r^2$.
Thus,$M = (ef)(\pi r^2)$.
Since $e$,$\pi$,and $r$ are constants,we have $M \propto f$.
If the frequency $f$ is doubled $(f' = 2f)$,the new magnetic moment $M'$ will be $M' = e(2f)(\pi r^2) = 2(ef\pi r^2) = 2M$.

(C) magnetic moment is generated by the motion of electric charges,which creates a current loop or a magnetic field.
$1$. $A$ stationary charge produces only an electric field and does not create a magnetic field or a magnetic moment.
$2$. $A$ charge moving with constant velocity creates both an electric field and a magnetic field (and thus a magnetic moment).
$3$. Accelerated or retarded charges create time-varying electric and magnetic fields,which also involve magnetic effects.
Therefore,a stationary charge is the only option that does not produce a magnetic moment.

(A) The electric current $I$ is defined as the rate of flow of charge,given by $I = \frac{q}{t}$.
In the case of an electron revolving in an orbit,the charge $q$ is the elementary charge $e = 1.6 \times 10^{-19} \ C$,and the time $t$ is the time period $T = 1.5 \times 10^{-16} \ s$.
Substituting the values:
$I = \frac{1.6 \times 10^{-19} \ C}{1.5 \times 10^{-16} \ s}$
$I = 1.066 \times 10^{-3} \ A$.

(D) The magnetic moment $M$ of a current loop is given by $M = I \cdot A$,where $I$ is the current and $A$ is the area. The $SI$ unit is $A \cdot m^2$.
Alternatively,for a magnetic dipole in an external magnetic field $B$,the potential energy $U$ is given by $U = -M \cdot B$.
Thus,$M = U / B$.
The unit of energy $U$ is Joule $(J)$ and the unit of magnetic field $B$ is Tesla $(T)$.
Therefore,the unit of magnetic moment is $J \cdot T^{-1}$.

(A) The magnetic dipole moment of a current-carrying loop is given by $m = NIA$,where $N$ is the number of turns,$I$ is the current,and $A$ is the area of the loop.
For the first case,$N_1 = 1$,$R_1 = R$,and $A_1 = \pi R^2$. Thus,$m = I \pi R^2$.
When the wire of length $L$ is converted into a ring of $N_2 = 2$ turns,the circumference of the new ring is $2 \times (2 \pi R_2) = L$. Since $L = 2 \pi R$,we have $4 \pi R_2 = 2 \pi R$,which gives $R_2 = \frac{R}{2}$.
The new area is $A_2 = \pi R_2^2 = \pi (\frac{R}{2})^2 = \frac{\pi R^2}{4}$.
The new magnetic dipole moment $m'$ is $m' = N_2 I A_2 = 2 \times I \times \frac{\pi R^2}{4} = \frac{I \pi R^2}{2}$.
Since $m = I \pi R^2$,we get $m' = \frac{m}{2}$.

(A) The magnetic dipole moment $M$ of a current loop is given by $M = I A$,where $I$ is the current and $A$ is the area of the loop.
For a circular loop of radius $r$,the area is $A = \pi r^2$.
Thus,$M = I (\pi r^2) \quad \dots(1)$
The magnetic field $B$ at the centre of a circular loop carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.
Rearranging this formula to solve for current $I$,we get $I = \frac{2rB}{\mu_0}$.
Substituting the expression for $I$ into equation $(1)$:
$M = \left( \frac{2rB}{\mu_0} \right) (\pi r^2)$
$M = \frac{2 \pi B r^3}{\mu_0}$.

(C) Let the total length of the wire be $L$.
For a single-turn coil $(n_1 = 1)$,the circumference is $L = 2\pi R_1$,so $R_1 = \frac{L}{2\pi}$.
The magnetic field at the centre is $B_1 = \frac{\mu_0 I}{2R_1} = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 I \pi}{L} = B$.
For a two-turn coil $(n_2 = 2)$,the total length $L = n_2(2\pi R_2) = 2(2\pi R_2) = 4\pi R_2$.
Thus,the new radius is $R_2 = \frac{L}{4\pi} = \frac{R_1}{2}$.
The magnetic field at the centre for $n$ turns is $B_n = \frac{n \mu_0 I}{2R_n}$.
For $n=2$,$B_2 = \frac{2 \mu_0 I}{2R_2} = \frac{\mu_0 I}{R_2}$.
Substituting $R_2 = \frac{R_1}{2}$,we get $B_2 = \frac{\mu_0 I}{R_1/2} = \frac{2 \mu_0 I}{R_1}$.
Since $B = \frac{\mu_0 I}{2R_1}$,we have $\frac{\mu_0 I}{R_1} = 2B$.
Therefore,$B_2 = 2(2B) = 4B$.

(D) Given:
Number of turns,$N = 1000$
Area of cross-section,$A = 2 \times 10^{-4} \ m^2$
Current,$I = 5.0 \ A$
The magnetic moment $\mu$ of a solenoid is given by the formula:
$\mu = N \cdot I \cdot A$
Substituting the given values:
$\mu = 1000 \times 5.0 \times (2 \times 10^{-4})$
$\mu = 5000 \times 2 \times 10^{-4}$
$\mu = 10000 \times 10^{-4}$
$\mu = 1 \ A \ m^2$
Therefore,the correct option is $D$.

(C) Given: Radius of the coil $r = 10 \text{ cm} = 0.1 \text{ m}$.
Number of turns $N = 100$.
Current $I = 1 \text{ A}$.
The magnetic moment $M$ of a current-carrying coil is given by the formula $M = N I A$,where $A$ is the area of the coil.
The area $A = \pi r^{2} = \pi \times (0.1 \text{ m})^{2} = 0.01 \pi \text{ m}^{2}$.
Substituting the values into the formula:
$M = 100 \times 1 \times (0.01 \times 3.142) \text{ A m}^{2}$.
$M = 100 \times 0.03142 \text{ A m}^{2} = 3.142 \text{ A m}^{2}$.
Thus,the magnetic moment of the coil is $3.142 \text{ A m}^{2}$.

(A) The magnetic dipole moment $M$ of a current loop is given by the formula:
$M = NIA$
where:
$N$ is the number of turns in the loop,
$I$ is the current flowing through the loop,
$A$ is the area of the loop.
From this relation,it is evident that the magnetic dipole moment depends only on the properties of the loop itself $(N, I, A)$ and is independent of the external magnetic field in which the loop is placed.

(D) According to the Right-Hand Thumb Rule for a circular current-carrying loop,if the fingers of the right hand are curled in the direction of the current,the thumb points in the direction of the magnetic field lines inside the loop.
If the current flows in an anticlockwise direction when viewed from the top,the magnetic field lines point upwards (towards the observer).
Since magnetic field lines emerge from the North Pole,the face of the loop where the current appears anticlockwise acts as a North Pole.
Conversely,if the current flows in a clockwise direction,the face acts as a South Pole.
Therefore,the upper side of the loop acts as a North Pole if the current is anticlockwise.

(C) The magnetic field at the centre of a circular current-carrying ring is given by $B = \frac{\mu_0 I}{2r}$.
Since the three rings are mutually perpendicular and centered at the origin,their magnetic field vectors will be directed along the $x$,$y$,and $z$ axes respectively.
Thus,the magnetic field vectors are:
$\vec{B_1} = \frac{\mu_0 I}{2r} \hat{i}$
$\vec{B_2} = \frac{\mu_0 I}{2r} \hat{j}$
$\vec{B_3} = \frac{\mu_0 I}{2r} \hat{k}$
The resultant magnetic field $\vec{B_0}$ at the origin is the vector sum of these fields:
$\vec{B_0} = \vec{B_1} + \vec{B_2} + \vec{B_3} = \frac{\mu_0 I}{2r} (\hat{i} + \hat{j} + \hat{k})$
The magnitude of the resultant magnetic field is:
$B_0 = |\vec{B_0}| = \frac{\mu_0 I}{2r} \sqrt{1^2 + 1^2 + 1^2}$
$B_0 = \frac{\mu_0 I}{2r} \sqrt{3} = \sqrt{3} \frac{\mu_0 I}{2r}$

(C) The magnetic moment $M$ of a current loop is given by $M = I A$,where $I$ is the current and $A$ is the area of the loop.
For an electron moving in a circular orbit of radius $R$ with time period $T$,the equivalent current $I$ is the charge $e$ flowing per unit time $T$,so $I = \frac{e}{T}$.
The area of the circular orbit is $A = \pi R^2$.
Substituting these values into the formula for magnetic moment,we get $M = I A = \left( \frac{e}{T} \right) (\pi R^2) = \frac{\pi e R^2}{T}$.

(C) The magnetic moment $M$ of a solenoid is given by the formula $M = N \cdot I \cdot A$,where $N$ is the number of turns,$I$ is the current,and $A$ is the cross-sectional area of the solenoid.
Given:
Number of turns $N = 500$
Current $I = 15 \,A$
Radius $r = 2 \,cm = 0.02 \,m$
Area $A = \pi r^2 = \pi (0.02)^2 = 0.0004 \pi \,m^2$
Substituting these values into the formula:
$M = 500 \times 15 \times 0.0004 \pi$
$M = 7500 \times 0.0004 \pi$
$M = 3 \pi \,J \,T^{-1}$
Therefore,the correct option is $C$.

(B) The magnetic moment $M$ of a current loop is given by the product of the current $I$ and the area $A$ of the loop,$M = I A$.
When a wire of length $L$ is bent into a circle,its circumference is $2 \pi r = L$,which implies the radius $r = \frac{L}{2 \pi}$.
The area of this circular loop is $A = \pi r^2 = \pi \left( \frac{L}{2 \pi} \right)^2 = \frac{L^2}{4 \pi}$.
Substituting the area into the magnetic moment formula,we get $M = I \times \frac{L^2}{4 \pi} = \frac{L^2 I}{4 \pi}$.

(B) Let $l$ be the length of the wire.
When the wire is bent into a circular loop of radius $r$,the circumference is equal to the length of the wire:
$l = 2 \pi r \Rightarrow r = \frac{l}{2 \pi}$
The magnetic dipole moment $M$ of a current loop is given by:
$M = i A$
Substituting the area $A = \pi r^2$:
$M = i \cdot \pi r^2 = i \pi \left( \frac{l}{2 \pi} \right)^2$
$M = i \pi \cdot \frac{l^2}{4 \pi^2} = \frac{i l^2}{4 \pi}$
Rearranging for $l$:
$l^2 = \frac{4 \pi M}{i}$
$l = \sqrt{\frac{4 \pi M}{i}}$

(D) The magnetic field $B$ at the centre of a circular loop of radius $R$ carrying current $i$ is given by:
$B = \frac{\mu_0 i}{2 R}$ ...$(i)$
The magnetic moment $M$ of the loop is given by:
$M = i A$ ...(ii)
From equation $(i)$,we can express the current $i$ as:
$i = \frac{2 B R}{\mu_0}$
Substitute this into equation (ii):
$M = \left( \frac{2 B R}{\mu_0} \right) A$
Since the area of the loop is $A = \pi R^2$,we have $R = \sqrt{\frac{A}{\pi}}$.
Substituting $R$ into the expression for $M$:
$M = \frac{2 B A}{\mu_0} \sqrt{\frac{A}{\pi}}$
$M = \frac{2 B A \sqrt{A}}{\mu_0 \sqrt{\pi}}$
Therefore,the correct option is $D$.

(A) The magnetic dipole moment $M$ is given by $M = i A$,where $i$ is the current and $A$ is the area of the ring.
Since the charge $q$ is rotating with angular velocity $\omega$,the time period of one revolution is $T = \frac{2 \pi}{\omega}$.
The equivalent current $i$ is given by $i = \frac{q}{T} = \frac{q \omega}{2 \pi}$.
The area of the ring is $A = \pi R^2$.
Substituting these values into the formula for $M$:
$M = \left( \frac{q \omega}{2 \pi} \right) (\pi R^2) = \frac{1}{2} q \omega R^2$.

(D) Let the length of the wire be $L$.
For the circular loop,the circumference is $2 \pi r = L$,so $r = \frac{L}{2 \pi}$. The area is $A_1 = \pi r^2 = \pi (\frac{L}{2 \pi})^2 = \frac{L^2}{4 \pi}$.
For the square loop,the perimeter is $4a = L$,so $a = \frac{L}{4}$. The area is $A_2 = a^2 = (\frac{L}{4})^2 = \frac{L^2}{16}$.
The magnetic dipole moment is $M = iA$.
Since the current $i$ is the same,the ratio of the dipole moments is $\frac{M_1}{M_2} = \frac{i A_1}{i A_2} = \frac{A_1}{A_2}$.
Substituting the areas: $\frac{M_1}{M_2} = \frac{L^2 / 4 \pi}{L^2 / 16} = \frac{16}{4 \pi} = \frac{4}{\pi}$.

(D) The magnetic moment $M$ of a circular coil is given by the formula $M = NIA$,where $N$ is the number of turns,$I$ is the current,and $A$ is the area of the coil.
Since the coil is circular,the area $A = \pi r^2$,where $r$ is the radius of the coil.
Thus,$M = NI(\pi r^2)$.
Given for coil $A$: $N_A = 300$,$M_A = M_1$,$r_A = r_1$.
Given for coil $B$: $N_B = 200$,$M_B = M_2$,$r_B = r_2$.
We are given $M_A / M_B = 1/2$ and $I_A = I_B = I$.
Taking the ratio of the magnetic moments:
$\frac{M_A}{M_B} = \frac{N_A I_A \pi r_A^2}{N_B I_B \pi r_B^2} = \frac{N_A}{N_B} \cdot \left(\frac{r_A}{r_B}\right)^2$.
Substituting the known values:
$\frac{1}{2} = \frac{300}{200} \cdot \left(\frac{r_A}{r_B}\right)^2$.
$\frac{1}{2} = \frac{3}{2} \cdot \left(\frac{r_A}{r_B}\right)^2$.
$\left(\frac{r_A}{r_B}\right)^2 = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$.
Therefore,$\frac{r_A}{r_B} = \frac{1}{\sqrt{3}}$.

(B) Given:
Area of coil, $A = 2 \,cm^2 = 2 \times 10^{-4} \,m^2$
Number of turns, $n = 1000$
Current, $I = 1 \,A$
The magnetic moment $M$ of a current-carrying coil is given by the formula:
$M = nIA$
Substituting the given values:
$M = (1000) \times (1) \times (2 \times 10^{-4})$
$M = 10^3 \times 2 \times 10^{-4}$
$M = 2 \times 10^{-1} \,Am^2$
$M = 0.2 \,Am^2$
Therefore, the magnetic moment is $0.2 \,Am^2$.

(A) The length of the wire $L$ forms the circumference of the circular loop,so $L = 2 \pi r$,where $r$ is the radius of the loop.
From this,the radius is $r = \frac{L}{2 \pi}$.
The magnetic moment $M$ of a current-carrying loop is given by $M = iA$,where $A$ is the area of the loop.
The area $A = \pi r^2 = \pi \left( \frac{L}{2 \pi} \right)^2 = \pi \left( \frac{L^2}{4 \pi^2} \right) = \frac{L^2}{4 \pi}$.
Substituting this into the formula for magnetic moment,we get $M = i \times \frac{L^2}{4 \pi} = \frac{i L^2}{4 \pi}$.

(D) The magnetic moment $\mu$ of a particle with charge $Q$ moving in a circular path of radius $R$ with speed $v$ is given by the formula:
$\mu = I A$
Where $I$ is the equivalent current,$I = \frac{Q}{T} = \frac{Q v}{2 \pi R}$,and $A$ is the area of the circular path,$A = \pi R^2$.
Substituting these values,we get:
$\mu = \left( \frac{Q v}{2 \pi R} \right) (\pi R^2) = \frac{Q v R}{2}$
From this expression,it is clear that the magnetic moment $\mu$ depends only on the charge $Q$,speed $v$,and radius $R$.
Since the mass $m$ of the particle does not appear in the formula for the magnetic moment,changing the mass while keeping $Q$,$v$,and $R$ constant will not affect the magnetic moment.
Therefore,the magnetic moment remains unchanged.

(B) The magnetic induction $B$ at the centre of a circular loop with $n$ turns and radius $r$ carrying current $i$ is given by:
$B = \frac{\mu_0 n i}{2r}$
From this, we can express the current as:
$i = \frac{2 B r}{\mu_0 n}$
The magnetic moment $M$ of the loop is defined as:
$M = n i A$
Substituting the expression for $i$:
$M = n \left( \frac{2 B r}{\mu_0 n} \right) A = \frac{2 B r A}{\mu_0}$
Given area $A = \pi \ m^2$, we know $A = \pi r^2$, so $\pi r^2 = \pi$, which implies $r = 1 \ m$.
Substituting the values $B = 0.1 \ T$, $r = 1 \ m$, and $A = \pi \ m^2$:
$M = \frac{2 \times 0.1 \times 1 \times \pi}{\mu_0} = \frac{0.2 \pi}{\mu_0}$

(A) The magnetic moment $(\mu)$ of a current loop is given by $\mu = iA$,where $i$ is the current and $A$ is the area of the loop.
For a particle of charge $q$ moving with speed $V$ in a circle of radius $R$,the time period $T$ is $T = \frac{2\pi R}{V}$.
The equivalent current is $i = \frac{q}{T} = \frac{qV}{2\pi R}$.
The area of the circle is $A = \pi R^2$.
Thus,$\mu = iA = \left(\frac{qV}{2\pi R}\right) \times (\pi R^2) = \frac{qVR}{2}$.
The angular momentum $(L)$ of the particle about the center of the circle is $L = mVR$.
Taking the ratio of magnetic moment to angular momentum:
$\frac{\mu}{L} = \frac{qVR / 2}{mVR} = \frac{q}{2m}$.

(D) Given: Torque $\tau = 0.016 \ Nm$,angle $\theta = 30^{\circ}$,and magnetic field $B = 0.04 \ T$.
The torque on a magnetic dipole is given by $\tau = mB \sin \theta$.
Substituting the values: $0.016 = m \times 0.04 \times \sin 30^{\circ}$.
Since $\sin 30^{\circ} = 0.5$,we have $0.016 = m \times 0.04 \times 0.5 = m \times 0.02$.
Thus,the magnetic moment $m = \frac{0.016}{0.02} = 0.8 \ Am^2$.
For a solenoid,the magnetic moment is $m = NIA$.
Given: Area $A = 1 \ cm^2 = 10^{-4} \ m^2$ and number of turns $N = 1000$.
Equating the magnetic moments: $0.8 = 1000 \times I \times 10^{-4}$.
$0.8 = 0.1 \times I$.
Therefore,the current $I = \frac{0.8}{0.1} = 8 \ A$.

(A) The magnetic field $B$ at the centre of a circular loop of radius $a$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2a}$.
The magnetic moment $M$ of the loop is given by $M = I A = I(\pi a^2)$.
The ratio $x$ is given by:
$x = \frac{B}{M} = \frac{\mu_0 I}{2a} \times \frac{1}{I \pi a^2} = \frac{\mu_0}{2 \pi a^3}$.
When the current $I$ is doubled $(I' = 2I)$ and the radius $a$ is doubled $(a' = 2a)$,the new ratio $x'$ is:
$x' = \frac{\mu_0}{2 \pi (a')^3} = \frac{\mu_0}{2 \pi (2a)^3} = \frac{\mu_0}{2 \pi (8a^3)} = \frac{1}{8} \left( \frac{\mu_0}{2 \pi a^3} \right) = \frac{x}{8}$.

(D) Given:
Radius $r = 0.05 \ nm = 0.05 \times 10^{-9} \ m$
Frequency $f = 10^{16} \ Hz$ (revolutions per second)
Charge of electron $e = 1.6 \times 10^{-19} \ C$
The magnetic moment $M$ of a current loop is given by $M = I \times A$,where $I$ is the current and $A$ is the area of the loop.
The current $I$ due to the revolving electron is $I = e \times f$.
The area of the circular orbit is $A = \pi r^{2}$.
Substituting these into the formula for magnetic moment:
$M = (e \times f) \times (\pi r^{2})$
$M = (1.6 \times 10^{-19} \ C) \times (10^{16} \ s^{-1}) \times (3.14) \times (0.05 \times 10^{-9} \ m)^{2}$
$M = 1.6 \times 10^{-19} \times 10^{16} \times 3.14 \times 0.0025 \times 10^{-18}$
$M = 1.6 \times 3.14 \times 0.0025 \times 10^{-21} \ m^{2} \cdot C/s$
$M = 0.01256 \times 10^{-21} \ A \ m^{2}$
$M = 1.26 \times 10^{-23} \ A \ m^{2}$

(B) The magnetic moment $m$ of a solenoid is given by the formula $m = N \cdot I \cdot A$.
Here,the number of turns $N = 800$,the current $I = 3.0 \text{ A}$,and the area of cross-section $A = 2.5 \times 10^{-4} \text{ m}^2$.
Substituting these values into the formula:
$m = 800 \times 3.0 \times 2.5 \times 10^{-4}$
$m = 2400 \times 2.5 \times 10^{-4}$
$m = 6000 \times 10^{-4}$
$m = 0.6 \text{ JT}^{-1}$.
Thus,the magnetic moment associated with the solenoid is $0.60 \text{ JT}^{-1}$.

(B) The magnetic moment $M$ of a coil is given by $M = N I A$. For a flat spiral coil where the turns are uniformly distributed between radii $r_1$ and $r_2$,the effective area $A$ is calculated by integrating the area of each turn: $A = \int_{r_1}^{r_2} \pi r^2 \frac{N}{r_2 - r_1} dr = \frac{N \pi}{r_2 - r_1} [\frac{r^3}{3}]_{r_1}^{r_2} = N \pi \frac{r_2^3 - r_1^3}{3(r_2 - r_1)} = N \pi \frac{r_1^2 + r_1 r_2 + r_2^2}{3}$.
Given $N = 200$,$r_1 = 0.03\text{ m}$,$r_2 = 0.06\text{ m}$,and $I = 20\text{ mA} = 0.02\text{ A}$.
Substituting these values: $M = 200 \times 0.02 \times \pi \times \frac{(0.03)^2 + (0.03)(0.06) + (0.06)^2}{3}$.
$M = 4 \times \pi \times \frac{0.0009 + 0.0018 + 0.0036}{3} = 4 \times \pi \times \frac{0.0063}{3} = 4 \times \pi \times 0.0021 = 0.0084 \times 3.14159 \approx 0.02639\text{ A.m}^2$.
This can be written as $2.639 \times 10^{-2}\text{ A.m}^2$. Rounding to two decimal places,$\alpha = 2.64$.

(C) The magnetic field $B$ at the centre of a circular coil is given by $B = \frac{\mu_0 N I}{2r}$.
Given: $B = 3.14 \times 10^{-3} \text{ T}$,$N = 100$,$r = 5 \text{ cm} = 0.05 \text{ m}$,$\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
Substituting the values: $3.14 \times 10^{-3} = \frac{4 \times 3.14 \times 10^{-7} \times 100 \times I}{2 \times 0.05}$.
$3.14 \times 10^{-3} = \frac{4 \times 3.14 \times 10^{-5} \times I}{0.1}$.
$10^{-3} = \frac{4 \times 10^{-5} \times I}{0.1} = 4 \times 10^{-4} \times I$.
$I = \frac{10^{-3}}{4 \times 10^{-4}} = \frac{10}{4} = 2.5 \text{ A}$.
The magnetic moment $M$ is given by $M = N I A = N I (\pi r^2)$.
$M = 100 \times 2.5 \times 3.14 \times (0.05)^2$.
$M = 250 \times 3.14 \times 0.0025 = 1.9625 \approx 2 \text{ A m}^2$.
Thus,the current is $2.5 \text{ A}$ and the magnetic moment is $2 \text{ A m}^2$.