Magnetic Moment of Current Carrying Coil Questions in English

(D) The magnetic moment $(M)$ of a current-carrying 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.
For a circular coil of radius $r$,the area $A = \pi r^2$.
Substituting the area into the formula,we get $M = ni(\pi r^2)$.
Since $n$,$i$,and $\pi$ are constants for a given coil,the magnetic moment $M$ is directly proportional to the square of the radius,i.e.,$M \propto r^2$.

(C) Let the length of each wire be $l$.
For the square,the side length is $a = l/4$. The area is $A_{square} = a^2 = (l/4)^2 = l^2/16$.
For the circle,the circumference is $2\pi r = l$,so the radius is $r = l/(2\pi)$. The area is $A_{circle} = \pi r^2 = \pi (l/(2\pi))^2 = l^2/(4\pi)$.
The magnetic moment is given by $M = iA$. Since the current $i$ is the same for both,
$\frac{M_{square}}{M_{circle}} = \frac{A_{square}}{A_{circle}} = \frac{l^2/16}{l^2/(4\pi)} = \frac{4\pi}{16} = \frac{\pi}{4}$.
Thus,the ratio is $\pi : 4$.

(C) The electric current $i$ produced by an electron moving in a circular orbit of radius $r$ with speed $v$ is given by $i = \frac{ev}{2\pi r}$.
The magnetic field $B$ at the centre of the circular current loop is given by $B = \frac{\mu_0 i}{2r}$.
Substituting the value of $i$ into the expression for $B$,we get $B = \frac{\mu_0}{2r} \cdot \frac{ev}{2\pi r} = \frac{\mu_0 ev}{4\pi r^2}$.
Rearranging the formula to solve for $r^2$,we get $r^2 = \frac{\mu_0 ev}{4\pi B}$.
Taking the square root of both sides,we find $r = \sqrt{\frac{\mu_0 ev}{4\pi B}}$.
Since $\mu_0$,$e$,and $4\pi$ are constants,the radius $r$ is proportional to $\sqrt{v/B}$.

(B) The magnetic dipole moment $M$ of a coil 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 area of the coil.
Given:
$N = 20$
$i = 3\, A$
$r = 4\, cm = 4 \times 10^{-2}\, m$
Area $A = \pi r^2 = \pi \times (4 \times 10^{-2})^2 = 16\pi \times 10^{-4}\, m^2$.
Using $\pi \approx 3.14$:
$A \approx 3.14 \times 16 \times 10^{-4} = 50.24 \times 10^{-4}\, m^2$.
Now,$M = 20 \times 3 \times 50.24 \times 10^{-4} = 60 \times 50.24 \times 10^{-4} = 3014.4 \times 10^{-4} \approx 0.3\, A \cdot m^2$.
Thus,the correct option is $B$.

(C) The magnetic moment $M$ of a current-carrying loop is defined as the product of the current $i$ and the area $A$ enclosed by the loop.
For a circular loop of radius $r$,the area $A$ is given by $A = \pi r^2$.
Therefore,the magnetic moment is $M = i \times A = i \pi r^2$.
Thus,the correct option is $C$.

(D) The magnetic dipole moment $(M)$ of a current-carrying coil is defined as the product of the number of turns $(N)$,the current $(I)$ flowing through the coil,and the area $(A)$ of the coil.
Mathematically,it is expressed as:
$M = NIA$
Therefore,the correct option is $D$.

(B) The magnetic moment $M$ of a current loop is given by $M = iA$,where $i$ is the current and $A$ is the area of the loop.
For an electron moving in a circle of radius $r$ with speed $v$,the time period $T$ is $T = \frac{2\pi r}{v}$.
The equivalent current $i$ is $i = \frac{e}{T} = \frac{ev}{2\pi r}$.
The area of the circle is $A = \pi r^2$.
Substituting these into the formula for magnetic moment:
$M = \left( \frac{ev}{2\pi r} \right) \times (\pi r^2) = \frac{1}{2}evr$.

(C) current-carrying circular loop behaves like a magnetic dipole or a bar magnet.
When a magnetic dipole is freely suspended in a magnetic field (like the Earth's magnetic field),it experiences a torque that aligns its magnetic moment vector with the external magnetic field.
The magnetic moment vector of a current loop is perpendicular to the plane of the loop.
For the magnetic moment to align with the Earth's North-South magnetic field,the plane of the loop must be perpendicular to the North-South direction,which means the plane of the loop will point in the East-West direction.

(A) The magnetic moment $M$ of a current-carrying loop is given by the product of the current $i$ and the area $A$ of the loop.
Mathematically,$M = iA$.
To find the current $i$,we rearrange the formula:
$i = M/A$.
Therefore,the correct option is $A$.

(C) The current $i$ produced by the revolving electron is given by $i = qf$,where $q = 1.6 \times 10^{-19} \, \text{C}$ and $f = 6.6 \times 10^{15} \, \text{Hz}$.
$i = 1.6 \times 10^{-19} \times 6.6 \times 10^{15} = 10.56 \times 10^{-4} \, \text{A}$.
The area $A$ of the orbit is $\pi R^2$,where $R = 0.528 \, \mathring{A} = 0.528 \times 10^{-10} \, \text{m}$.
$A = 3.142 \times (0.528 \times 10^{-10})^2 = 3.142 \times 0.278784 \times 10^{-20} \approx 0.876 \times 10^{-20} \, \text{m}^2$.
The magnetic moment $M$ is given by $M = iA$.
$M = (10.56 \times 10^{-4}) \times (0.876 \times 10^{-20}) \approx 9.25 \times 10^{-24} \, \text{A} \cdot \text{m}^2$.
Rounding to the nearest order of magnitude provided in the options,$M \approx 1 \times 10^{-23} \, \text{A} \cdot \text{m}^2$.

(A) The magnetic moment $M$ of a current-carrying 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.
Given:
Number of turns $N = 24$
Current $i = 0.75\,A$
Diameter $d = 7\,cm = 7 \times 10^{-2}\,m$
Radius $r = d/2 = 3.5 \times 10^{-2}\,m$
Area $A = \pi r^2 = 3.14 \times (3.5 \times 10^{-2})^2\,m^2$
Substituting the values:
$M = 24 \times 0.75 \times 3.14 \times (3.5 \times 10^{-2})^2$
$M = 18 \times 3.14 \times 12.25 \times 10^{-4}$
$M = 692.37 \times 10^{-4} \approx 6.9 \times 10^{-2}\,A\cdot m^2$

(D) The magnetic dipole moment $M$ of a current-carrying loop is given by the formula $M = NiA$,where $N$ is the number of turns,$i$ is the current,and $A$ is the area vector of the loop.
The magnitude of the magnetic dipole moment is directly proportional to the current $i$ and the area $A$ of the loop.
The direction of the magnetic dipole moment vector is always perpendicular to the plane of the loop,determined by the right-hand rule.
Therefore,the correct statement is that it is perpendicular to the plane of the loop and proportional to the area of the loop.

(D) The magnetic moment $M$ of a current-carrying loop is given by the formula $M = iA$,where $i$ is the current and $A$ is the area of the loop.
Given: Current $i = 0.1\, A$,Radius $r = 5\, cm = 0.05\, m$.
The area $A = \pi r^2 = \pi \times (0.05)^2 = 3.14159 \times 0.0025 = 7.854 \times 10^{-3}\, m^2$.
Substituting the values into the formula:
$M = 0.1 \times (7.854 \times 10^{-3}) = 7.854 \times 10^{-4}\, A\cdot m^2$.
Rounding to two decimal places,we get $M = 7.85 \times 10^{-4}\, A\cdot m^2$.

(B) The magnetic field at the center of a circular loop of radius $R$ carrying current $i$ is given by $B = \frac{{\mu _0 i}}{{2R}}$.
From this,we can express the current $i$ as $i = \frac{{2RB}}{{\mu _0}}$.
The magnetic moment $M$ of a current loop is given by $M = i \times A$,where $A$ is the area of the loop.
For a circular loop,$A = \pi R^2$.
Substituting the expression for $i$ and $A$ into the formula for $M$:
$M = \left( \frac{{2RB}}{{\mu _0}} \right) \times (\pi R^2) = \frac{{2\pi B R^3}}{{\mu _0}}$.

(C) The magnetic moment $M$ of a current-carrying coil is given by $M = NiA$,where $N$ is the number of turns,$i$ is the current,and $A$ is the area of the coil.
For a circular coil of radius $r$,the area $A = \pi r^2$.
The length of the wire $l$ used to make $N$ turns is $l = N(2\pi r)$,which implies $r = \frac{l}{2\pi N}$.
Substituting $A$ in the formula for $M$: $M = Ni(\pi r^2) = Ni\pi (\frac{l}{2\pi N})^2 = Ni\pi \frac{l^2}{4\pi^2 N^2} = \frac{il^2}{4\pi N}$.
Assuming the number of turns $N$ is constant,we find that $M \propto l^2$.
Therefore,the magnetic moment is directly proportional to the square of the length of the wire.

(B) Let the radius of the circle be $r$. The circumference of the circle is equal to the length of the wire,so $2\pi r = L$.
Thus,the radius is $r = \frac{L}{2\pi}$.
The area $A$ of the circle is given by $A = \pi r^2 = \pi \left(\frac{L}{2\pi}\right)^2 = \frac{\pi L^2}{4\pi^2} = \frac{L^2}{4\pi}$.
The magnetic moment $M$ is defined as $M = IA$.
Substituting the value of $A$,we get $M = I \left(\frac{L^2}{4\pi}\right) = \frac{IL^2}{4\pi}$.

(A) The current $i$ produced by the rotating charge $q$ is given by $i = \frac{q}{T} = \frac{q}{2\pi / \omega} = \frac{q\omega}{2\pi}$.
The magnetic moment $M$ of the system is $M = iA$,where $A = \pi R^2$ is the area of the circle traced by the particles.
$M = \left( \frac{q\omega}{2\pi} \right) (\pi R^2) = \frac{q\omega R^2}{2}$.
The angular momentum $L$ of the system about the centre is $L = I\omega$,where $I$ is the moment of inertia.
For two particles of mass $m$ at distance $R$ from the axis,$I = mR^2 + mR^2 = 2mR^2$.
Thus,$L = (2mR^2)\omega = 2mR^2\omega$.
The ratio of the magnitudes is $\frac{M}{L} = \frac{q\omega R^2 / 2}{2mR^2\omega} = \frac{q}{4m}$.
Wait,re-evaluating the current: Since there are two charges,the total current $i = \frac{2q}{T} = \frac{2q\omega}{2\pi} = \frac{q\omega}{\pi}$.
Then $M = iA = (\frac{q\omega}{\pi})(\pi R^2) = q\omega R^2$.
Then $\frac{M}{L} = \frac{q\omega R^2}{2mR^2\omega} = \frac{q}{2m}$.

(B) The magnetic moment $M$ of a current loop is given by $M = iA$,where $i$ is the current and $A$ is the area of the loop.
For a ring of radius $R$,the area $A = \pi R^2$.
The current $i$ produced by the rotating charge $Q$ is $i = \frac{Q}{T}$,where $T$ is the time period of one revolution.
Since the angular speed is $\omega$,the time period $T = \frac{2\pi}{\omega}$.
Thus,$i = \frac{Q}{(2\pi / \omega)} = \frac{Q\omega}{2\pi}$.
Substituting these into the formula for magnetic moment:
$M = \left( \frac{Q\omega}{2\pi} \right) \times (\pi R^2) = \frac{1}{2}Q\omega R^2$.

(C) The initial magnetic moment of the square loop is $\mu_1 = iA = iL^2$,where the direction is perpendicular to the plane of the loop.
When the loop is folded at the middle,it forms two smaller rectangular loops,each of area $A' = L \times (L/2) = L^2/2$.
The magnetic moment of each half is $M = iA' = iL^2/2 = \mu_1/2$.
These two halves lie in perpendicular planes. Let the magnetic moment of the horizontal half be $\overrightarrow{M_h}$ and the vertical half be $\overrightarrow{M_v}$.
Both have magnitude $M = \mu_1/2$. The resultant magnetic moment $\overrightarrow{\mu_2}$ is the vector sum of these two: $\mu_2 = \sqrt{M^2 + M^2} = M\sqrt{2}$.
Substituting $M = \mu_1/2$,we get $\mu_2 = (\mu_1/2) \times \sqrt{2} = \mu_1/\sqrt{2}$.
Therefore,the ratio $\frac{|\overrightarrow{\mu_1}|}{|\overrightarrow{\mu_2}|} = \frac{\mu_1}{\mu_1/\sqrt{2}} = \sqrt{2}$.

(D) The magnetic dipole moment $M$ is defined as the product of current $I$ and the area $A$ of the loop.
Mathematically,$M = I \times A$.
The $SI$ unit of current is Ampere $(A)$ and the $SI$ unit of area is square meter $(m^2)$.
Therefore,the $SI$ unit of magnetic dipole moment is $A \cdot m^2$.

(A) current-carrying loop placed in a magnetic field experiences a torque and behaves like a magnetic dipole.
The magnetic moment $M$ of a current loop 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 the loop.
Therefore,the correct option is $A$.

(C) The initial magnetic moment of the square loop is $\mu_1 = iL^2$,directed perpendicular to the plane of the loop.
When the loop is folded by $90^\circ$,it consists of two perpendicular loops,each of area $A' = L \times (L/2) = L^2/2$.
The magnetic moment of each half-loop is $M = iA' = iL^2/2 = \mu_1/2$.
These two magnetic moments are perpendicular to each other,one in the horizontal plane and one in the vertical plane.
The resultant magnetic moment $\mu_2$ is the vector sum of these two moments: $\mu_2 = \sqrt{M^2 + M^2} = M\sqrt{2}$.
Substituting $M = \mu_1/2$,we get $\mu_2 = (\mu_1/2) \times \sqrt{2} = \mu_1/\sqrt{2}$.
Therefore,the ratio is $\frac{|\overrightarrow {{\mu _1}} |}{|\overrightarrow {{\mu _2}} |} = \frac{\mu_1}{\mu_1/\sqrt{2}} = \sqrt{2}$.

(C) Consider a thin elemental ring of radius $x$ and width $dx$ on the disk.
The area of this elemental ring is $dA = 2 \pi x dx$.
The charge on this elemental ring is $dq = \sigma dA = 2 \pi \sigma x dx$.
Since the disk rotates with angular speed $\omega$,the time period of rotation is $T = \frac{2 \pi}{\omega}$.
The equivalent current $dI$ due to this rotating charge is $dI = \frac{dq}{T} = \frac{dq \cdot \omega}{2 \pi}$.
Substituting $dq$,we get $dI = \frac{(2 \pi \sigma x dx) \omega}{2 \pi} = \sigma \omega x dx$.
The magnetic moment $dM$ of this elemental ring is $dM = dI \cdot A_{ring} = (\sigma \omega x dx) \cdot (\pi x^2) = \pi \sigma \omega x^3 dx$.
To find the total magnetic moment $M$,integrate $dM$ from $x = 0$ to $x = R$:
$M = \int_{0}^{R} \pi \sigma \omega x^3 dx = \pi \sigma \omega \left[ \frac{x^4}{4} \right]_{0}^{R} = \frac{1}{4} \pi \sigma \omega R^4$.

(D) The magnetic moment $\mu$ of a current loop is given by $\mu = I A$.
Here,the current $I$ produced by the revolving charge $q$ is $I = \frac{q}{T}$,where $T$ is the time period of revolution.
The time period $T$ for a particle moving in a circle of radius $R$ with speed $v$ is $T = \frac{2 \pi R}{v}$.
Substituting $T$ into the expression for $I$,we get $I = \frac{q}{2 \pi R / v} = \frac{qv}{2 \pi R}$.
The area $A$ of the circular path is $A = \pi R^2$.
Therefore,the magnetic moment is $\mu = I A = \left( \frac{qv}{2 \pi R} \right) (\pi R^2) = \frac{qvR}{2}$.

(B) The coil is placed in the $N-S$ vertical plane. According to the right-hand thumb rule,for a clockwise current,the magnetic field produced by the coil at its center is directed from East to West.
Earth's horizontal magnetic field is directed from South to North.
The magnetic needle aligns itself along the direction of the resultant magnetic field.
The resultant vector is the vector sum of the coil's magnetic field (pointing West) and the Earth's magnetic field (pointing North).
Therefore,the north pole of the magnetic needle will point in the West-North direction.

(B) The convection current $I$ is defined as the charge per unit time.
Since the charge $q$ completes one revolution in time $T = \frac{2\pi r}{v}$,the current is:
$I = \frac{q}{T} = \frac{q}{2\pi r / v} = \frac{qv}{2\pi r}$
The magnetic moment $M$ of a current loop is given by the product of current $I$ and the area $A = \pi r^2$:
$M = I \times A = \left( \frac{qv}{2\pi r} \right) \times (\pi r^2)$
Simplifying the expression:
$M = \frac{qv \pi r^2}{2\pi r} = \frac{qvr}{2}$

(A) The area of the square loop is $A = (10\,cm)^2 = (0.1\,m)^2 = 0.01\,m^2$.
The magnitude of the magnetic moment is $M = iA = 10\,A \times 0.01\,m^2 = 0.1\,A\cdot m^2$.
According to the right-hand rule,the direction of the area vector $\vec{A}$ is perpendicular to the plane of the loop.
The loop lies in a plane such that its normal makes an angle of $60^{\circ}$ with the $x$-axis and $30^{\circ}$ with the $-z$-axis (based on the geometry provided in the figure).
Thus,the magnetic moment vector is $\vec{M} = M \cos(60^{\circ}) \hat{i} - M \cos(30^{\circ}) \hat{k}$.
Substituting the values: $\vec{M} = 0.1 \times (1/2) \hat{i} - 0.1 \times (\sqrt{3}/2) \hat{k}$.
$\vec{M} = 0.05 \hat{i} - 0.05\sqrt{3} \hat{k} = 0.05 (\hat{i} - \sqrt{3} \hat{k})\,A\cdot m^2$.

(A) The magnetic moment $\vec{P}_m$ of a current-carrying loop is given by $\vec{P}_m = I_2 \vec{A}$,where $\vec{A}$ is the area vector.
For a square loop of side $l$ in the $xy$-plane,the area vector is $\vec{A} = l^2 \hat{k}$ (assuming the current $I_2$ flows in the counter-clockwise direction as per the right-hand rule).
Thus,the magnetic moment is $\vec{P}_m = I_2 l^2 \hat{k}$.
Comparing this with the given options,option $A$ is correct.

(B) The magnetic field at the centre of a circular loop of radius $R$ carrying current $I$ is given by $B_1 = \frac{\mu_0 I}{2R}$.
The magnetic dipole moment of the loop is $m_1 = I A = I \pi R^2$.
When the dipole moment is doubled $(m_2 = 2m_1)$ while keeping the current $I$ constant,the area $A$ must double. Since $A = \pi R^2$,the new radius $R'$ must satisfy $\pi (R')^2 = 2 \pi R^2$,which gives $R' = \sqrt{2} R$.
The new magnetic field at the centre is $B_2 = \frac{\mu_0 I}{2R'} = \frac{\mu_0 I}{2(\sqrt{2} R)}$.
Taking the ratio $\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I}{2\sqrt{2}R}} = \sqrt{2}$.

(B) Consider a small element of length $dx$ at a distance $x$ from the axis of rotation. The charge on this element is $dq = \frac{Q}{l} dx$.
Since the rod rotates with frequency $f$,the current $di$ associated with this rotating charge element is $di = dq \cdot f = \frac{Q}{l} f dx$.
The magnetic moment $d\mu$ of this circular loop of radius $x$ is given by $d\mu = (di) \cdot A$,where $A = \pi x^2$ is the area of the loop.
Substituting the values,we get $d\mu = \left( \frac{Q f}{l} dx \right) (\pi x^2) = \frac{\pi f Q}{l} x^2 dx$.
To find the total magnetic moment $\mu$,we integrate $d\mu$ from $x = 0$ to $x = l$:
$\mu = \int_0^l \frac{\pi f Q}{l} x^2 dx = \frac{\pi f Q}{l} \left[ \frac{x^3}{3} \right]_0^l = \frac{\pi f Q}{l} \cdot \frac{l^3}{3} = \frac{1}{3} \pi f Q l^2$.

(C) The magnetic moment $\vec \mu$ of a current-carrying loop is given by $\vec \mu = i \vec A$,where $i$ is the current and $\vec A$ is the area vector.
The area vector $\vec A$ has a magnitude equal to the area of the loop $(A = a \times b)$ and its direction is perpendicular to the plane of the loop.
The loop lies in a plane that makes an angle of $30^\circ$ with the $xy$-plane,or more specifically,the normal to the loop makes an angle with the axes.
Looking at the geometry,the side of length $b$ is along the $y$-axis. The side of length $a$ makes an angle of $30^\circ$ with the $x$-axis in the $xz$-plane.
The normal vector $\vec n$ to the loop can be found by taking the cross product of the two vectors representing the sides of the rectangle.
Vector along side $b$ is $\vec b = b \hat j$.
Vector along side $a$ is $\vec a = a \cos 30^\circ \hat i + a \sin 30^\circ \hat k$.
The area vector $\vec A = \vec a \times \vec b = (a \cos 30^\circ \hat i + a \sin 30^\circ \hat k) \times (b \hat j)$.
Using the cross product rules ($\hat i \times \hat j = \hat k$ and $\hat k \times \hat j = -\hat i$):
$\vec A = ab \cos 30^\circ \hat k - ab \sin 30^\circ \hat i$.
Wait,checking the orientation: The magnetic moment direction is determined by the right-hand rule. For the given current direction,the normal vector points in the direction of $(\sin 30^\circ \hat i + \cos 30^\circ \hat k)$.
Thus,$\vec \mu = iab(\sin 30^\circ \hat i + \cos 30^\circ \hat k)$.

(B) Given: Length of wire $l = 2 \, m$, Current $I = 1 \, A$.
When the wire is bent into a circle, the circumference of the circle equals the length of the wire: $2 \pi r = l$.
Substituting the value of $l$: $2 \pi r = 2 \implies r = 1 / \pi \, m$.
The area of the circular coil is $A = \pi r^2 = \pi (1 / \pi)^2 = 1 / \pi \, m^2$.
The magnetic moment $M$ is given by $M = I \times A$.
Substituting the values: $M = 1 \times (1 / \pi) = 1 / \pi \, A \cdot m^2$.

(A) The magnetic moment $M$ of a current loop is given by $M = iA$,where $i$ is the current and $A$ is the area of the loop.
For a circular loop of radius $r$,the circumference is $2 \pi r = l$,which implies $r = \frac{l}{2 \pi}$.
The area of the loop is $A = \pi r^2 = \pi \left( \frac{l}{2 \pi} \right)^2 = \frac{\pi l^2}{4 \pi^2} = \frac{l^2}{4 \pi}$.
Substituting this into the formula for magnetic moment,we get $M = i \left( \frac{l^2}{4 \pi} \right) = \frac{i l^2}{4 \pi}$.

(D) $1$. $A$ uniformly charged ring possesses a static charge distribution,which creates an electrostatic field in the surrounding space.
$2$. When the ring rotates with a constant angular velocity $\omega$,the moving charges constitute a circular current $I = qf = q(\omega / 2\pi)$,where $q$ is the total charge on the ring.
$3$. This circulating current produces a magnetic field in the region around the ring.
$4$. $A$ current-carrying loop also possesses a magnetic dipole moment $\mu = IA$,where $A$ is the area of the ring.
$5$. Therefore,the rotating charged ring produces an electrostatic field,a magnetic field,and a magnetic moment simultaneously.

(A) The current $I$ generated by an electron moving in a circular orbit is defined as the charge passing through a point per unit time.
Since the electron completes $v$ revolutions per second,the current is $I = q \times v = ev$.
The area $A$ of the circular orbit of radius $r$ is $A = \pi r^2$.
The magnetic moment $m$ is given by the product of current and area: $m = I \times A$.
Substituting the values,we get $m = (ev) \times (\pi r^2) = \pi ve r^2$.
Therefore,the correct option is $A$.

(A) The magnetic moment $M$ of a current-carrying coil is given by the formula:
$M = N \cdot I \cdot A$
Where:
$N = 1000$ (number of turns)
$I = 2 \, A$ (current)
$A = 1.5 \times 10^{-4} \, m^2$ (area of cross-section)
Substituting the values into the formula:
$M = 1000 \times 2 \times 1.5 \times 10^{-4}$
$M = 2000 \times 1.5 \times 10^{-4}$
$M = 3000 \times 10^{-4}$
$M = 0.3 \, Am^2$
Therefore,the correct option is $A$.

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

(C) The magnetic moment $\vec{M}$ of a current loop is given by $\vec{M} = I\vec{A}$,where $I$ is the current and $\vec{A}$ is the area vector.
From the figure,the current in the outer loop $(r_2 = 0.3\,m)$ is in the counter-clockwise direction,so its magnetic moment is $\vec{M}_2 = I(\pi r_2^2)\hat{k}$.
The current in the inner loop $(r_1 = 0.2\,m)$ is in the clockwise direction,so its magnetic moment is $\vec{M}_1 = -I(\pi r_1^2)\hat{k}$.
The net magnetic moment is $\vec{M} = \vec{M}_2 + \vec{M}_1 = I\pi(r_2^2 - r_1^2)\hat{k}$.
Substituting the values: $I = 7\,A$,$r_2 = 0.3\,m$,$r_1 = 0.2\,m$.
$\vec{M} = 7 \times \pi \times (0.3^2 - 0.2^2)\hat{k}$
$\vec{M} = 7 \times \frac{22}{7} \times (0.09 - 0.04)\hat{k}$
$\vec{M} = 22 \times 0.05\hat{k} = 1.1\,\hat{k}\,A\cdot m^2$.

(B) The loop consists of four semi-circles of radius $r = a/2$. The total area $A$ enclosed by the loop can be calculated by considering the four semi-circles and the central square of side $a$.
The area of four semi-circles of radius $r = a/2$ is $4 \times \frac{1}{2} \pi r^2 = 2 \pi (a/2)^2 = 2 \pi (a^2/4) = \frac{\pi a^2}{2}$.
The central square has an area of $a^2$.
Total area $A = \frac{\pi a^2}{2} + a^2 = \left( \frac{\pi}{2} + 1 \right) a^2$.
Using the right-hand rule for the current direction shown in the figure (counter-clockwise),the magnetic moment vector $\vec{M} = I \vec{A}$ points in the positive $z$-direction.
Therefore,$\vec{M} = \left( \frac{\pi}{2} + 1 \right) I a^2 \hat{k}$.

(A) The magnetic moment $\mu$ is defined as the product of the current $I$ and the area $A$ of the loop,i.e.,$\mu = IA$.
The current $I$ generated by a rotating charge $q$ with frequency $f$ is $I = qf$.
Since the angular velocity is $\omega$,the frequency $f = \frac{\omega}{2\pi}$.
Thus,the current is $I = q \left( \frac{\omega}{2\pi} \right)$.
The area of the loop is $A = \pi r^2$.
Substituting these into the formula for magnetic moment:
$\mu = \left( \frac{q \omega}{2\pi} \right) (\pi r^2)$.
Simplifying the expression,we get:
$\mu = \frac{1}{2} q \omega r^2$.

(C) Consider a small element of length $dx$ at a distance $x$ from the origin. The charge on this element is $dq = \rho(x) dx = \rho_0 \frac{x}{l} dx$.
As the rod rotates with frequency $n$ (rotations per second),the angular velocity is $\omega = 2\pi n$. The current $di$ produced by this rotating charge element is $di = \frac{dq}{T} = dq \cdot n = \rho_0 \frac{x}{l} dx \cdot n$.
The magnetic moment $dM$ of this circular current loop of radius $x$ is $dM = di \cdot A = di \cdot (\pi x^2)$.
Substituting $di$: $dM = (\rho_0 \frac{x}{l} dx \cdot n) \cdot \pi x^2 = \frac{\pi n \rho_0}{l} x^3 dx$.
Integrating from $x = 0$ to $x = l$: $M = \int_0^l \frac{\pi n \rho_0}{l} x^3 dx = \frac{\pi n \rho_0}{l} [\frac{x^4}{4}]_0^l = \frac{\pi n \rho_0}{l} \cdot \frac{l^4}{4} = \frac{\pi}{4} n \rho_0 l^3$.

(C) For a square loop of side $l$,the perimeter is $P = 4l$ and the area is $A_s = l^2$. The magnetic dipole moment is $m = I A_s = I l^2$.
When reshaped into a circular loop of radius $r$,the perimeter remains the same: $2\pi r = 4l$,which gives $r = \frac{2l}{\pi}$.
The area of the circular loop is $A_c = \pi r^2 = \pi \left(\frac{2l}{\pi}\right)^2 = \frac{4l^2}{\pi}$.
The new magnetic dipole moment $m'$ is $m' = I A_c = I \left(\frac{4l^2}{\pi}\right)$.
Since $m = I l^2$,we substitute this into the expression for $m'$:
$m' = \frac{4}{\pi} (I l^2) = \frac{4m}{\pi}$.

(A) The magnetic field at the center of a circular coil with $N$ turns,current $I$,and radius $R$ is given by $B = \frac{\mu_0 NI}{2R}$.
The magnetic moment of the coil is $M = NIA = NI(\pi R^2)$.
The ratio $\alpha$ is given by $\frac{B}{M} = \frac{\mu_0 NI}{2R} \times \frac{1}{NI\pi R^2} = \frac{\mu_0}{2\pi R^3}$.
From this expression,we see that $\alpha \propto \frac{1}{R^3}$,and it is independent of the current $I$.
If the radius $R$ is doubled $(R' = 2R)$,the new ratio $\alpha'$ becomes:
$\alpha' = \frac{\mu_0}{2\pi (2R)^3} = \frac{\mu_0}{2\pi (8R^3)} = \frac{1}{8} \left( \frac{\mu_0}{2\pi R^3} \right) = \frac{\alpha}{8}$.
Thus,the new ratio is $\frac{\alpha}{8}$.

(D) Let the linear charge density be $\lambda = \frac{q}{l}$.
Consider a small element of length $dx$ at a distance $x$ from the pivot.
The charge on this element is $dq = \lambda dx = \frac{q}{l} dx$.
As the rod rotates with frequency $f$,this charge element moves in a circle of radius $x$ with time period $T = \frac{1}{f}$.
The equivalent current due to this rotating charge is $dI = \frac{dq}{T} = dq \cdot f = \frac{q}{l} f dx$.
The magnetic moment $dm$ of this current loop is $dm = dI \cdot A$,where $A = \pi x^2$ is the area of the circle.
$dm = (\frac{q}{l} f dx) (\pi x^2) = \frac{\pi qf}{l} x^2 dx$.
Integrating from $x = 0$ to $x = l$ to find the total magnetic moment $M$:
$M = \int_0^l \frac{\pi qf}{l} x^2 dx = \frac{\pi qf}{l} [\frac{x^3}{3}]_0^l = \frac{\pi qf}{l} \cdot \frac{l^3}{3} = \frac{1}{3} \pi qfl^2$.

(A) The magnetic moment $M$ of a coil is given by $M = N I A$,where $N$ is the number of turns,$I$ is the current,and $A$ is the area of the coil.
For a circular coil of $N$ turns and radius $R$,the total length of the wire is $L = N(2 \pi R)$,which implies $R = \frac{L}{2 \pi N}$.
The area of the circular coil is $A = \pi R^2 = \pi \left( \frac{L}{2 \pi N} \right)^2 = \frac{L^2}{4 \pi N^2}$.
Substituting this into the magnetic moment formula: $M = N I \left( \frac{L^2}{4 \pi N^2} \right) = \frac{I L^2}{4 \pi N}$.
Since $I$,$L$,and $\pi$ are constants,$M$ is inversely proportional to $N$ $(M \propto \frac{1}{N})$.
To maximize $M$,we must minimize $N$. Therefore,$N = 1$ gives the maximum magnetic moment.
For a fixed perimeter,a circle encloses the maximum area compared to other shapes,so the coil must be circular.

(D) Given:
Length of wire $L = 2\,m$.
Current $I = 1\,A$.
Since the wire is bent into a circular loop,the circumference of the loop is equal to the length of the wire:
$2\pi r = L = 2\,m$
$r = 1/\pi \,m$.
The magnetic moment $M$ of a current loop is given by the product of current and the area of the loop:
$M = I \times A = I \times (\pi r^2)$.
Substituting the values:
$M = 1 \times \pi \times (1/\pi)^2 = \pi \times (1/\pi^2) = 1/\pi \,Am^2$.

(D) The magnetic field at the centre of the square is zero if the currents in the opposite segments produce equal and opposite magnetic fields at the centre.
In case $P$,the two paths are symmetric and consist of one thick and one thin wire each. Thus,the resistance of both paths is equal,leading to equal currents. The magnetic fields produced by these currents cancel each other out at the centre.
In case $Q$,the paths are asymmetric (one path has two thick wires,the other has two thin wires). The resistances are unequal,so the currents are unequal,and the magnetic fields do not cancel out.
In case $R$,the two paths are again symmetric,each consisting of one thick and one thin wire. Thus,the currents are equal,and the magnetic fields at the centre cancel out.
Therefore,the magnetic field at the centre is zero in cases $P$ and $R$.

(D) Let the length of the wire be $L$.
For the circular loop,the circumference is $L = 2\pi r$,so the radius is $r = \frac{L}{2\pi}$.
The magnetic moment of the circular loop is $M = I A = I(\pi r^2) = I \pi \left(\frac{L}{2\pi}\right)^2 = \frac{IL^2}{4\pi}$.
When the wire is reshaped into a square of side $a$,the perimeter is $4a = L$,so $a = \frac{L}{4}$.
The area of the square is $A' = a^2 = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16}$.
The new magnetic moment is $M' = I A' = I \frac{L^2}{16}$.
From the expression for $M$,we have $IL^2 = 4\pi M$.
Substituting this into the expression for $M'$,we get $M' = \frac{4\pi M}{16} = \frac{\pi}{4} M$.

(A) Consider a small element of length $dx$ at a distance $x$ from the axis of rotation.
The charge on this element is $dq = \frac{q}{l} dx$.
The time period of rotation is $T = \frac{2\pi}{\omega}$.
The current $di$ due to this rotating element is $di = \frac{dq}{T} = \frac{(q/l) dx}{2\pi/\omega} = \frac{q\omega}{2\pi l} dx$.
The magnetic moment $dM$ of this small current loop is $dM = (di) \cdot A = (di) \cdot (\pi x^2) = \left( \frac{q\omega}{2\pi l} dx \right) (\pi x^2) = \frac{q\omega}{2l} x^2 dx$.
Integrating from $x = 0$ to $x = l$ to find the total magnetic moment $M$:
$M = \int_{0}^{l} \frac{q\omega}{2l} x^2 dx = \frac{q\omega}{2l} \left[ \frac{x^3}{3} \right]_{0}^{l} = \frac{q\omega}{2l} \cdot \frac{l^3}{3} = \frac{q\omega l^2}{6}$.

(B) The magnetic dipole moment $M$ of a current-carrying coil is given by the formula:
$M = N \cdot I \cdot A$
Where:
$N = 20$ (number of turns)
$I = 3 \, A$ (current)
$A = \pi \cdot r^2$ (area of the coil)
Given radius $r = 4 \, cm = 0.04 \, m$.
Substituting the values:
$A = \pi \cdot (0.04)^2 = \pi \cdot 0.0016 \, m^2$
$M = 20 \times 3 \times \pi \times 0.0016$
$M = 60 \times 0.0050265$
$M \approx 0.3016 \, A \cdot m^2$
Rounding to the nearest provided option,the value is $0.3 \, A \cdot m^2$.