Torque , Potential Energy and Work Done in Mangetic Field Questions in English

(A) The magnetic field $B$ produced by the larger ring at its center is given by $B = \frac{\mu_0 I}{2R}$.
The magnetic moment of the smaller coil with $N$ turns,current $i$,and radius $r$ is $M = N i A = N i (\pi r^2)$.
The torque $\tau$ acting on a magnetic dipole in an external magnetic field is given by $\tau = M B \sin \theta$,where $\theta$ is the angle between the magnetic moment vector and the magnetic field vector.
Since the planes of the two rings are perpendicular,the magnetic moment vector of the smaller ring (which is perpendicular to its plane) is perpendicular to the magnetic field vector produced by the larger ring (which is perpendicular to the larger ring's plane). Thus,$\theta = 90^{\circ}$.
Substituting the values,we get $\tau = (N i \pi r^2) \times \left( \frac{\mu_0 I}{2R} \right) \times \sin 90^{\circ}$.
Since $\sin 90^{\circ} = 1$,the torque is $\tau = Ni\pi r^2 \left( \frac{\mu_0 I}{2R} \right)$.

(C) The torque $\tau$ acting on a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin \theta$,where $\theta$ is the angle between the normal to the coil and the magnetic field.
Given:
Number of turns $N = 30$
Radius $r = 8.0\, cm = 0.08\, m$
Current $I = 6.0\, A$
Magnetic field $B = 1.0\, T$
Angle $\theta = 60^o$
Area $A = \pi r^2 = 3.14 \times (0.08)^2 = 3.14 \times 0.0064 = 0.020096\, m^2$
Substituting the values:
$\tau = 30 \times 6.0 \times 0.020096 \times 1.0 \times \sin 60^o$
$\tau = 180 \times 0.020096 \times \frac{\sqrt{3}}{2}$
$\tau = 90 \times 0.020096 \times 1.732$
$\tau \approx 3.133\, Nm$
Rounding to two significant figures,the magnitude of the counter torque is $3.1\, Nm$.

(A) The potential energy $U$ of a magnetic dipole (current-carrying coil) in a magnetic field $\vec{B}$ is given by $U = -\vec{M} \cdot \vec{B} = -MB \cos \theta$,where $\theta$ is the angle between the magnetic moment $\vec{M}$ (which is in the direction of the area vector $\hat{n}$) and the magnetic field $\vec{B}$.
Given that the magnetic field $\vec{B}$ is directed towards the right:
In position $I$,$\hat{n}$ is to the left,so $\theta = 180^\circ$,$U = -MB \cos(180^\circ) = +MB$ (Maximum).
In position $II$,$\hat{n}$ is downwards,so $\theta = 90^\circ$,$U = -MB \cos(90^\circ) = 0$.
In position $III$,$\hat{n}$ is at an obtuse angle,so $\cos \theta$ is negative,making $U$ positive.
In position $IV$,$\hat{n}$ is at an acute angle,so $\cos \theta$ is positive,making $U$ negative (Minimum).
Comparing the angles: $\theta_I = 180^\circ$,$\theta_{III} > 90^\circ$,$\theta_{II} = 90^\circ$,$\theta_{IV} < 90^\circ$.
Thus,the potential energy order is $U_I > U_{III} > U_{II} > U_{IV}$.

(A) The magnetic field is given as $B = 3000\,G = 3000 \times 10^{-4}\,T = 0.3\,T$ in the positive $z-$ direction.
The area vector $\vec{A}$ of the loop placed in the $xy-$ plane is perpendicular to the plane,i.e.,along the $z-$ axis.
The magnetic moment $\vec{M}$ is given by $\vec{M} = I\vec{A}$. Since the current loop is in the $xy-$ plane,its area vector is along the $z-$ axis.
The torque $\vec{\tau}$ on a current loop in a magnetic field is given by $\vec{\tau} = \vec{M} \times \vec{B}$.
Since both the magnetic moment vector $\vec{M}$ and the magnetic field vector $\vec{B}$ are directed along the $z-$ axis,the angle $\theta$ between them is $0^\circ$.
Therefore,the magnitude of the torque is $\tau = MB \sin(0^\circ) = 0$.

(D) The torque on a current-carrying coil in a uniform magnetic field is given by $\vec{\tau} = \vec{m} \times \vec{B}$,where $\vec{m}$ is the magnetic moment and $\vec{B}$ is the magnetic field.
The coil experiences zero torque when the magnetic moment $\vec{m}$ is parallel to the magnetic field $\vec{B}$.
The magnetic moment vector $\vec{m}$ is always perpendicular to the plane of the coil.
For $\vec{m}$ to be parallel to $\vec{B}$,the plane of the coil must be perpendicular to the magnetic field $\vec{B}$.

(A) The torque (couple) on a current-carrying loop in a magnetic field is given by $C = NiAB \sin \theta$.
Since the current $i$,magnetic field $B$,and angle $\theta$ are the same for both,the torque is proportional to the product of the number of turns $N$ and the area $A$ of the loop: $C \propto N \cdot A$.
Let $L$ be the length of the wire. For loop $A$ with $N_A = 1$ turn,$L = 2 \pi r_A$,so $r_A = L / (2 \pi)$. The area $A_A = \pi r_A^2 = \pi (L / 2 \pi)^2 = L^2 / (4 \pi)$.
For loop $B$ with $N_B = 2$ turns,$L = 2 \times (2 \pi r_B)$,so $r_B = L / (4 \pi)$. The area $A_B = \pi r_B^2 = \pi (L / 4 \pi)^2 = L^2 / (16 \pi)$.
Now,calculate the ratio of the torques:
$\frac{C_A}{C_B} = \frac{N_A A_A}{N_B A_B} = \frac{1 \times (L^2 / 4 \pi)}{2 \times (L^2 / 16 \pi)} = \frac{1/4}{2/16} = \frac{1/4}{1/8} = 2$.
Thus,$C_A = 2 C_B$,which means the couple on loop $A$ is twice that on loop $B$,or the couple on loop $A$ is greater than on loop $B$.

(D) The torque acting on a current-carrying loop in a uniform magnetic field is given by $\tau = NIAB \sin \theta$,where $N$ is the number of turns,$I$ is the current,$A$ is the area of the loop,$B$ is the magnetic field,and $\theta$ is the angle between the normal to the loop and the magnetic field.
For a given perimeter (length of wire $L = 2.0\,m$),the area $A$ enclosed by a loop is maximum for a circle.
Since the torque $\tau$ is directly proportional to the area $A$ $(\tau \propto A)$,the torque will be maximum for the loop with the largest area.
Among the given shapes,the circular loop $S$ encloses the maximum area for a fixed perimeter.
Therefore,the torque is maximum on the loop $S$.

(B) The magnetic torque on the loop is given by $\vec{\tau} = \vec{M} \times \vec{B}$.
For a small angular displacement $\theta$,the magnitude of the torque is $\tau = MB \sin \theta \approx MB \theta$.
The restoring torque is $\tau = -MB \theta$.
Using Newton's second law for rotation,$\tau = I_{moment} \alpha$,where $I_{moment}$ is the moment of inertia of the circular loop about its diameter,$I_{moment} = \frac{ma^2}{2}$.
Thus,$\frac{ma^2}{2} \alpha = - (I \pi a^2) B \theta$.
$\alpha = - \frac{2 I \pi B}{m} \theta$.
Comparing this with the $SHM$ equation $\alpha = - \omega^2 \theta$,we get $\omega^2 = \frac{2 I \pi B}{m}$.
Therefore,$\omega = \sqrt{\frac{2 I \pi B}{m}}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{m}{2 I \pi B}} = \sqrt{\frac{4 \pi^2 m}{2 I \pi B}} = \sqrt{\frac{2 \pi m}{IB}}$.

(N/A) No,because that would require the torque $\tau$ to be in the vertical direction. Since the area vector $\vec{A}$ of the horizontal loop is in the vertical direction,the torque $\vec{\tau} = I(\vec{A} \times \vec{B})$ would always lie in the horizontal plane of the loop for any uniform magnetic field $\vec{B}$. Thus,it cannot rotate about the vertical axis.
$(b)$ The orientation of stable equilibrium is one where the area vector $\vec{A}$ of the loop is parallel to the external magnetic field $\vec{B}$. In this orientation,the magnetic field produced by the loop is in the same direction as the external field,both being normal to the plane of the loop. This alignment maximizes the total magnetic flux through the loop.
$(c)$ $A$ flexible current-carrying loop in a magnetic field tends to maximize the magnetic flux through it to reach a state of minimum potential energy. For a fixed perimeter,a circular shape encloses the maximum area. Therefore,the loop adopts a circular shape with its plane normal to the magnetic field to maximize the total flux.

(B) The side length of the square coil is $l = 10 \; cm = 0.1 \; m$.
The area of the coil is $A = l^2 = (0.1 \; m)^2 = 0.01 \; m^2$.
The number of turns is $n = 20$,and the current is $I = 12 \; A$.
The magnetic field strength is $B = 0.80 \; T$.
The angle between the normal to the plane of the coil and the magnetic field is $\theta = 30^{\circ}$.
The magnitude of the magnetic torque $\tau$ experienced by the coil is given by the formula $\tau = n I A B \sin \theta$.
Substituting the values: $\tau = 20 \times 12 \times 0.01 \times 0.80 \times \sin 30^{\circ}$.
Since $\sin 30^{\circ} = 0.5$,we have $\tau = 20 \times 12 \times 0.01 \times 0.80 \times 0.5$.
$\tau = 0.96 \; N \; m$.

(A) Number of turns on the circular coil,$n=30$.
Radius of the coil,$r=8.0 \; cm = 0.08 \; m$.
Area of the coil,$A = \pi r^2 = \pi(0.08)^2 \approx 0.0201 \; m^2$.
Current flowing in the coil,$I=6.0 \; A$.
Magnetic field strength,$B = 1.0 \; T$.
Angle between the field lines and the normal to the coil surface,$\theta = 60^{\circ}$.
The coil experiences a magnetic torque $\tau = n I A B \sin \theta$. To prevent the coil from turning,an equal and opposite counter-torque must be applied.
$\tau = 30 \times 6.0 \times 0.0201 \times 1.0 \times \sin 60^{\circ}$.
$\tau = 30 \times 6.0 \times 0.0201 \times 1.0 \times \frac{\sqrt{3}}{2} \approx 3.133 \; Nm$.
$(b)$ The torque on a current-carrying coil in a magnetic field is given by $\vec{\tau} = \vec{m} \times \vec{B}$,where the magnetic moment $\vec{m} = n I \vec{A}$. Since the magnitude of the torque depends only on the area $A$ and not on the specific shape of the coil,the answer would not change if the circular coil were replaced by a planar coil of any irregular shape that encloses the same area.

(E) Magnetic field strength,$B = 3000 \; G = 3000 \times 10^{-4} \; T = 0.3 \; T$.
Length of the rectangular loop,$l = 10 \; cm = 0.1 \; m$.
Width of the rectangular loop,$b = 5 \; cm = 0.05 \; m$.
Area of the loop,$A = l \times b = 0.1 \times 0.05 = 50 \times 10^{-4} \; m^2$.
Current in the loop,$I = 12 \; A$.
The magnetic moment is $\vec{m} = I \vec{A}$. The torque is $\vec{\tau} = \vec{m} \times \vec{B}$.
In a uniform magnetic field,the net force on a current loop is always zero.
$(a)$ $\vec{A}$ is along the $x-$axis. $\vec{m} = I A \hat{i}$. $\vec{\tau} = (I A \hat{i}) \times (B \hat{k}) = -I A B \hat{j} = -1.8 \times 10^{-2} \hat{j} \; Nm$.
$(b)$ Similar to $(a)$,$\vec{\tau} = -1.8 \times 10^{-2} \hat{j} \; Nm$.
$(c)$ $\vec{A}$ is along the $y-$axis. $\vec{m} = I A \hat{j}$. $\vec{\tau} = (I A \hat{j}) \times (B \hat{k}) = I A B \hat{i} = 1.8 \times 10^{-2} \hat{i} \; Nm$.
$(d)$ The angle between $\vec{m}$ and $\vec{B}$ is $60^{\circ}$. $|\vec{\tau}| = m B \sin(60^{\circ}) = 1.8 \times 10^{-2} \times \frac{\sqrt{3}}{2} \approx 1.56 \times 10^{-2} \; Nm$.
$(e)$ $\vec{m}$ is along the $z-$axis. $\vec{m} = I A \hat{k}$. $\vec{\tau} = (I A \hat{k}) \times (B \hat{k}) = 0$. This is stable equilibrium as $\vec{m} \parallel \vec{B}$.
$(f)$ $\vec{m}$ is along the $-z-$axis. $\vec{\tau} = 0$. This is unstable equilibrium as $\vec{m}$ is anti-parallel to $\vec{B}$.

(A) Given:
Number of turns,$n = 20$
Radius,$r = 10 \; cm = 0.1 \; m$
Magnetic field,$B = 0.10 \; T$
Current,$I = 5.0 \; A$
Cross-sectional area,$A = 10^{-5} \; m^2$
Electron density,$n_e = 10^{29} \; m^{-3}$
Charge of electron,$e = 1.6 \times 10^{-19} \; C$
$(a)$ The torque $\tau$ on a current loop is given by $\tau = N I A B \sin \theta$. Since the magnetic field is normal to the plane of the coil,the angle $\theta$ between the magnetic field and the area vector is $0^\circ$. Thus,$\tau = N I A B \sin(0^\circ) = 0$.
$(b)$ The total force on a closed current loop in a uniform magnetic field is always zero,as the forces on opposite segments cancel each other out.
$(c)$ The magnetic force on an electron is $F = B e v_d$,where $v_d$ is the drift velocity. The drift velocity is given by $v_d = \frac{I}{n_e e A}$. Substituting this into the force equation:
$F = B e \left( \frac{I}{n_e e A} \right) = \frac{B I}{n_e A}$
$F = \frac{0.10 \times 5.0}{10^{29} \times 10^{-5}} = \frac{0.5}{10^{24}} = 5 \times 10^{-25} \; N$.
Thus,the average force on each electron is $5 \times 10^{-25} \; N$.

(A) The magnetic moment $M$ of the 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.
$M = 800 \times 3.0 \times 2.5 \times 10^{-4} = 0.6 \;A \cdot m^{2}$.
The torque $\tau$ on a magnetic dipole in a magnetic field $B$ is given by $\tau = M B \sin \theta$,where $\theta$ is the angle between the magnetic moment vector (axis of the solenoid) and the magnetic field.
Given $\theta = 30^{\circ}$ and $B = 0.25 \;T$.
$\tau = 0.6 \times 0.25 \times \sin 30^{\circ}$.
$\tau = 0.6 \times 0.25 \times 0.5 = 0.075 \;N \cdot m$.
$\tau = 7.5 \times 10^{-2} \;N \cdot m$.

(A) Number of turns on the solenoid,$n = 2000$.
Area of cross-section of the solenoid,$A = 1.6 \times 10^{-4} \;m^{2}$.
Current in the solenoid,$I = 4.0 \;A$.
$(a)$ The magnetic moment $M$ associated with the solenoid is given by $M = nIA$.
Substituting the values: $M = 2000 \times 4.0 \times 1.6 \times 10^{-4} = 1.28 \;A \cdot m^{2}$.
$(b)$ Magnetic field,$B = 7.5 \times 10^{-2} \;T$.
Angle between the magnetic field and the axis of the solenoid,$\theta = 30^{\circ}$.
Since the magnetic field is uniform,the net force on the solenoid is zero.
The torque $\tau$ on the solenoid is given by $\tau = MB \sin \theta$.
Substituting the values: $\tau = 1.28 \times 7.5 \times 10^{-2} \times \sin 30^{\circ} = 1.28 \times 7.5 \times 10^{-2} \times 0.5 = 4.8 \times 10^{-2} \;N \cdot m$.

(D) Number of turns in the circular coil,$N = 16$.
Radius of the coil,$r = 10 \;cm = 0.1 \;m$.
Area of the coil,$A = \pi r^2 = \pi \times (0.1)^2 = 0.01\pi \;m^2$.
Current in the coil,$I = 0.75 \;A$.
Magnetic field strength,$B = 5.0 \times 10^{-2} \;T$.
Frequency of oscillations,$\nu = 2.0 \;s^{-1}$.
Magnetic moment,$M = N I A = 16 \times 0.75 \times 0.01\pi = 0.12\pi \;A \;m^2 \approx 0.377 \;J \;T^{-1}$.
The frequency of oscillation for a magnetic dipole in a magnetic field is given by $\nu = \frac{1}{2\pi} \sqrt{\frac{MB}{I_{rot}}}$,where $I_{rot}$ is the moment of inertia.
Rearranging for $I_{rot}$: $I_{rot} = \frac{MB}{4\pi^2 \nu^2}$.
Substituting the values: $I_{rot} = \frac{0.377 \times 5.0 \times 10^{-2}}{4 \times \pi^2 \times (2.0)^2}$.
$I_{rot} = \frac{0.01885}{157.91} \approx 1.19 \times 10^{-4} \;kg \;m^2$.

(N/A) As shown in the figure,the plane $ABCD$ is not along the magnetic field but makes an angle with it.
We take the angle between the field and the normal to the coil to be angle $\theta$.
The forces on the arms $BC$ and $DA$ are equal,opposite,and act along the axis of the coil,which connects the centers of mass of $BC$ and $DA$. Being collinear along the axis,they cancel each other,resulting in no net force or torque.
The forces on arms $AB$ and $CD$ are $\overrightarrow{F}_{1}$ and $\overrightarrow{F}_{2}$ respectively.
They are also equal and opposite with magnitude $F_{1} = F_{2} = I b B$.
The figure illustrates a view of the arrangement from the $AD$ side,showing these two forces constituting a couple. The magnitude of the torque on the loop is,
$\tau = \tau_{1} + \tau_{2}$
$\tau = F_{1} \left( \frac{a}{2} \sin \theta \right) + F_{2} \left( \frac{a}{2} \sin \theta \right)$
$[\because \tau = (\text{magnitude of force}) \times (\text{perpendicular distance from reference points})]$
$\tau = (I b B) \left( \frac{a}{2} \sin \theta \right) + (I b B) \left( \frac{a}{2} \sin \theta \right)$
$\tau = I (a b) B \sin \theta$
$\tau = I A B \sin \theta$

(N/A) When a current-carrying loop of area $A$ and current $I$ is placed in a uniform magnetic field $B$,it experiences a torque $\tau$. The expression for the torque is given by:
$\tau = \vec{m} \times \vec{B}$
Where:
- $\vec{m}$ is the magnetic dipole moment of the loop,defined as $\vec{m} = I\vec{A}$.
- $\vec{B}$ is the external uniform magnetic field.
- The magnitude of the torque is $\tau = mB \sin \theta$,where $\theta$ is the angle between the area vector $\vec{A}$ (normal to the loop) and the magnetic field vector $\vec{B}$.

(N/A) The torque $\vec{\tau}$ acting on a current-carrying loop placed in a uniform magnetic field $\vec{B}$ is given by the vector product of the magnetic dipole moment $\vec{m}$ and the magnetic field $\vec{B}$.
Mathematically,$\vec{\tau} = \vec{m} \times \vec{B}$.
If $\theta$ is the angle between the magnetic dipole moment vector $\vec{m}$ (which is perpendicular to the plane of the loop) and the magnetic field $\vec{B}$,the magnitude of the torque is given by $\tau = mB \sin \theta$.
If $\theta$ is defined as the angle between the plane of the loop and the magnetic field,the torque is given by $\tau = mB \cos \theta$.

(D) The torque on an electric dipole $\vec{p}$ in an electric field $\vec{E}$ is given by $\vec{\tau}_e = \vec{p} \times \vec{E}$,with magnitude $\tau_e = pE \sin \theta$.
The torque on a magnetic dipole $\vec{M}$ in a magnetic field $\vec{B}$ is given by $\vec{\tau}_m = \vec{M} \times \vec{B}$,with magnitude $\tau_m = MB \sin \theta$.
For the motions to be identical,the torques must be equal for the same angular displacement $\theta$,implying $\tau_e = \tau_m$.
Thus,$pE \sin \theta = MB \sin \theta$,which simplifies to $pE = MB$.
Using the relation between electric and magnetic fields in electromagnetic waves,$E = cB$,where $c$ is the speed of light.
Substituting $E = cB$ into the equation $pE = MB$,we get $p(cB) = MB$.
Therefore,the condition for identical motion is $p = \frac{M}{c}$ and the fields must be related by $E = cB$.

(N/A) Let the resistance of the thick wire be $R$ and the thin wire be $2R$. Both wires are connected in parallel to the voltage source $V_{0}$.
The current in the thick wire is $I_{1} = \frac{V_{0}}{R}$ and the current in the thin wire is $I_{2} = \frac{V_{0}}{2R}$.
The magnetic force on each wire is given by $\vec{F} = I(\vec{l} \times \vec{B})$. Since the wires are perpendicular to the magnetic field component in the plane of the loop,the magnitude of the force is $F = IlB \sin(90^{\circ}) = IlB$.
The forces on the two wires are $F_{1} = I_{1}lB = \frac{V_{0}lB}{R}$ and $F_{2} = I_{2}lB = \frac{V_{0}lB}{2R}$.
The axis of rotation passes through the centers of the rods of length $d$. The perpendicular distance of each wire from this axis is $r_{\perp} = \frac{d}{2} \cos(45^{\circ}) = \frac{d}{2\sqrt{2}}$.
The torque exerted by each force is $\tau = F \cdot r_{\perp}$. Both forces create a torque in the same direction about the axis.
$\tau_{net} = F_{1} \left( \frac{d}{2\sqrt{2}} \right) + F_{2} \left( \frac{d}{2\sqrt{2}} \right) = (F_{1} + F_{2}) \frac{d}{2\sqrt{2}}$.
Substituting the values: $\tau_{net} = \left( \frac{V_{0}lB}{R} + \frac{V_{0}lB}{2R} \right) \frac{d}{2\sqrt{2}} = \left( \frac{3V_{0}lB}{2R} \right) \frac{d}{2\sqrt{2}} = \frac{3V_{0}ldB}{4\sqrt{2}R}$.

(B) The torque $\tau$ acting on a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin(\theta)$.
Since the coil is kept parallel to the magnetic field,the angle between the normal to the coil and the magnetic field is $\theta = 90^{\circ}$,so $\sin(90^{\circ}) = 1$.
Given: $N = 500$,$A = 3 \times 10^{-4} \ m^{2}$,$I = 0.5 \ A$,and $\tau = 1.5 \ Nm$.
Substituting the values into the formula: $1.5 = 500 \times 0.5 \times (3 \times 10^{-4}) \times B$.
$1.5 = 250 \times 3 \times 10^{-4} \times B$.
$1.5 = 750 \times 10^{-4} \times B$.
$1.5 = 0.075 \times B$.
$B = \frac{1.5}{0.075} = \frac{1500}{75} = 20 \ T$.

(D) Given: Moment of inertia $I = 0.8 \, kg \cdot m^2$,Magnetic moment $M = 20 \, A \cdot m^2$,Magnetic field $B = 4 \, T$.
Initially,the coil is in a vertical position,so the angle between the magnetic moment vector (perpendicular to the plane of the coil) and the magnetic field (vertical) is $\theta_i = 90^{\circ}$.
After rotating by $60^{\circ}$,the new angle is $\theta_f = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Using the principle of conservation of energy: $U_i + K_i = U_f + K_f$.
Potential energy $U = -\vec{M} \cdot \vec{B} = -MB \cos \theta$.
$K_i = 0$ (starts from rest).
$U_i = -MB \cos 90^{\circ} = 0$.
$U_f = -MB \cos 30^{\circ} = -20 \times 4 \times \frac{\sqrt{3}}{2} = -40\sqrt{3} \, J$.
$K_f = \frac{1}{2} I \omega^2 = \frac{1}{2} (0.8) \omega^2 = 0.4 \omega^2$.
Equating energies: $0 + 0 = -40\sqrt{3} + 0.4 \omega^2$.
$0.4 \omega^2 = 40\sqrt{3} \implies \omega^2 = 100\sqrt{3}$.
$\omega = 10(3)^{1/4} \, rad \cdot s^{-1}$.
Note: The provided options do not match the calculated result. Based on standard physics problem patterns,if the rotation was $90^{\circ}$ instead of $60^{\circ}$,$\omega$ would be $20 \, rad \cdot s^{-1}$.

(A) The magnetic field produced by the long wire at the center of the loop is $B = \frac{\mu_{0} I}{2 \pi b}$.
The magnetic moment of the square loop is $M = I \times \text{Area} = I \times (2a)^2 = 4a^2 I$.
The torque on the loop is given by $\tau = M \times B \times \sin(\theta)$.
Since the magnetic moment vector is along the $y$-axis and the magnetic field is along the $x$-axis,the angle between them is $90^{\circ}$.
Thus,$\tau = (4a^2 I) \times (\frac{\mu_{0} I}{2 \pi b}) \times \sin(90^{\circ}) = \frac{2 \mu_{0} I^{2} a^{2}}{\pi b}$.

(A) The magnetic field $B$ produced by the long wire at a distance $r$ is given by $B = \frac{\mu_{0} I}{2 \pi r}$.
For the two sides of the loop parallel to the $z$-axis,the distance from the wire is $r = \sqrt{b^2 + a^2}$.
The force on each of these sides is $F = B I (2a) = \frac{\mu_{0} I}{2 \pi \sqrt{b^2 + a^2}} \cdot I \cdot 2a = \frac{\mu_{0} I^2 a}{\pi \sqrt{b^2 + a^2}}$.
The torque $\tau$ about the $z$-axis is produced by the vertical components of these forces. The component of force perpendicular to the lever arm is $F \cos \theta$,where $\cos \theta = \frac{b}{\sqrt{b^2 + a^2}}$.
The total torque is $\tau = 2 \cdot (F \cos \theta) \cdot a = 2 \cdot \left( \frac{\mu_{0} I^2 a}{\pi \sqrt{b^2 + a^2}} \right) \cdot \left( \frac{b}{\sqrt{b^2 + a^2}} \right) \cdot a$.
Simplifying this,we get $\tau = \frac{2 \mu_{0} I^2 a^2 b}{\pi (a^2 + b^2)}$.

(C) The torque $\tau$ experienced by a current-carrying coil in a magnetic field is given by $\tau = |\vec{m} \times \vec{B}| = N I A B \sin \theta$,where $\theta$ is the angle between the area vector $\vec{A}$ and the magnetic field $\vec{B}$.
The coil is in the $y-z$ plane,so its area vector $\vec{A}$ is along the $x$-axis. However,the problem states the coil makes an angle of $30^{\circ}$ with the $x$-axis. This implies the angle between the normal to the coil (area vector) and the magnetic field (which is along the $x$-axis) is $\theta = 30^{\circ}$.
Given values: $N=100$,$I=1 \text{ A}$,$R=2 \text{ m}$,$B=\frac{1}{\pi} \text{ T}$.
Area $A = \pi R^2 = \pi (2)^2 = 4\pi \text{ m}^2$.
Substituting these into the torque formula:
$\tau = N I A B \sin 30^{\circ}$
$\tau = 100 \times 1 \times (4\pi) \times \frac{1}{\pi} \times \sin 30^{\circ}$
$\tau = 100 \times 4 \times \frac{1}{2}$
$\tau = 200 \text{ N} \cdot \text{m}$.

(A) The area of an equilateral triangle with side length $a = 10 \, cm = 0.1 \, m$ is given by $A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (0.1)^2 = \frac{\sqrt{3}}{4} \times 0.01 \, m^2$.
The magnetic moment of the coil is $M = I A = 0.2 \times \frac{\sqrt{3}}{4} \times 0.01 = 0.05 \times \sqrt{3} \times 0.01 = 5 \sqrt{3} \times 10^{-4} \, Am^2$.
The torque acting on the coil is $\tau = M B \sin \theta$,where $\theta$ is the angle between the normal to the coil and the magnetic field. When the plane of the coil is parallel to the magnetic field,the normal to the coil is perpendicular to the magnetic field,so $\theta = 90^{\circ}$.
Thus,$\tau = M B \sin 90^{\circ} = M B = (5 \sqrt{3} \times 10^{-4}) \times (20 \times 10^{-3} \, T) = 100 \sqrt{3} \times 10^{-7} = \sqrt{3} \times 10^{-5} \, Nm$.
Comparing this with $\sqrt{x} \times 10^{-5} \, Nm$,we get $\sqrt{x} = \sqrt{3}$,which implies $x = 3$.

(C) The torque $\tau$ on a current-carrying loop in a magnetic field is given by the formula $\vec{\tau} = \vec{M} \times \vec{B}$,where $\vec{M}$ is the magnetic dipole moment.
The magnitude of the magnetic moment is $M = i A$,where $A = l^2$ is the area of the square loop. Thus,$M = i l^2$.
The angle between the area vector $\vec{A}$ (which is perpendicular to the plane of the loop) and the magnetic field $\vec{B}$ is $\theta = 90^\circ - \alpha$,where $\alpha$ is the angle between the plane of the loop and the magnetic field.
The magnitude of the torque is $\tau = M B \sin \theta$.
Substituting the values,we get $\tau = (i l^2) B \sin(90^\circ - \alpha)$.
Since $\sin(90^\circ - \alpha) = \cos \alpha$,the torque is $\tau = B i l^2 \cos \alpha$.

(B) The area vector $\vec{A}$ of the loop in the $Y-Z$ plane,with current flowing anticlockwise when viewed from the negative $X$-axis,is directed along the negative $X$-axis: $\vec{A} = (0.2 \ m \times 0.1 \ m)(-\hat{i}) = 0.02(-\hat{i}) \ m^2$.
The magnetic moment $\vec{M}$ is given by $\vec{M} = i\vec{A} = 5 \ A \times 0.02(-\hat{i}) \ m^2 = 0.1(-\hat{i}) \ A-m^2$.
The magnetic field is $\vec{B} = 2 \times 10^{-3} \hat{j} \ T$.
The torque $\vec{\tau}$ is given by $\vec{\tau} = \vec{M} \times \vec{B}$.
$\vec{\tau} = [0.1(-\hat{i})] \times [2 \times 10^{-3} \hat{j}] = 0.1 \times 2 \times 10^{-3} \times (-\hat{i} \times \hat{j}) = 2 \times 10^{-4} \times (-\hat{k}) \ N-m$.
Thus,the torque is $2 \times 10^{-4} \ N-m$ along the negative $Z$-direction.

(D) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the change in potential energy:
$W = U_f - U_i$
Potential energy $U = -\vec{M} \cdot \vec{B} = -MB \cos \theta$
Initial orientation: $\vec{M}$ is along $\vec{B}$,so $\theta_i = 0^{\circ}$.
$U_i = -MB \cos 0^{\circ} = -MB$
Final orientation: $\vec{M}$ is perpendicular to $\vec{B}$,so $\theta_f = 90^{\circ}$.
$U_f = -MB \cos 90^{\circ} = 0$
Work done $W = 0 - (-MB) = MB$
Given: $N = 200$,$I = 100 \mu\text{A} = 100 \times 10^{-6} \text{ A}$,$A = 2.5 \times 10^{-4} \text{ m}^2$,$B = 1 \text{ T}$.
Magnetic moment $M = NIA = 200 \times (100 \times 10^{-6}) \times (2.5 \times 10^{-4}) = 5 \times 10^{-6} \text{ A m}^2$.
Work $W = MB = (5 \times 10^{-6}) \times 1 = 5 \times 10^{-6} \text{ J} = 5 \mu\text{J}$.

(A) The work done in rotating a magnetic dipole in a magnetic field is given by $W = U_f - U_i = -\vec{\mu} \cdot \vec{B}_f - (-\vec{\mu} \cdot \vec{B}_i)$.
Initially, the plane of the coil is perpendicular to the magnetic field, so the area vector $\vec{A}$ is parallel to $\vec{B}$. Thus, the angle $\theta_i = 0^{\circ}$.
After rotating by $90^{\circ}$, the plane of the coil becomes parallel to the magnetic field, so the area vector $\vec{A}$ is perpendicular to $\vec{B}$. Thus, $\theta_f = 90^{\circ}$.
The magnetic moment $\mu = N I A = 100 \times 1 \times 10^{-3} \times 5 \times 10^{-3} = 5 \times 10^{-4} \, A \cdot m^2$.
The work done is $W = -\mu B \cos(90^{\circ}) - (-\mu B \cos(0^{\circ})) = 0 + \mu B = \mu B$.
$W = (5 \times 10^{-4}) \times 0.20 = 1 \times 10^{-4} \, J$.
Since $1 \, J = 10^6 \, \mu J$, we have $W = 10^{-4} \times 10^6 \, \mu J = 100 \, \mu J$.

(B) The torque $\tau$ experienced by a current-carrying coil in a magnetic field is given by $\vec{\tau} = \vec{M} \times \vec{B}_0$,where $\vec{M}$ is the magnetic moment of the coil.
The magnitude of the torque is $\tau = M B_0 \sin(\theta)$. Since the plane of the coil is vertical and the magnetic field $B_0$ is horizontal and parallel to the plane of the coil,the angle between the area vector (normal to the plane) and the magnetic field is $\theta = 90^\circ$. Thus,$\sin(90^\circ) = 1$.
The magnetic moment of the coil is $M = N i A = N i (\pi R^2)$.
Therefore,the torque is $\tau = N i \pi R^2 B_0$.
From the angular impulse-momentum theorem,the change in angular momentum $L$ is given by $\Delta L = \int \tau dt$.
Substituting the expression for torque: $\Delta L = \int (N i \pi R^2 B_0) dt = N \pi R^2 B_0 \int i dt$.
Since the total charge $Q$ discharged through the coil is $Q = \int i dt$,we get:
$L = N \pi R^2 B_0 Q$.

(A) Let the loop make an angle $\theta$ with the vertical.
In equilibrium,the net torque about the axis $PQ$ is zero.
The magnetic torque is $\tau_m = M B_0 \sin(90^\circ - \theta) = M B_0 \cos \theta$,where $M = I A = I (\pi r^2)$ is the magnetic moment.
So,$\tau_m = I \pi r^2 B_0 \cos \theta$.
The gravitational torque is $\tau_g = mg \cdot r \sin \theta$,where $r \sin \theta$ is the horizontal distance of the center of mass from the axis $PQ$.
For equilibrium,$\tau_m = \tau_g$.
$I \pi r^2 B_0 \cos \theta = mg r \sin \theta$.
Dividing both sides by $mg r \cos \theta$,we get $\tan \theta = \frac{\pi r I B_0}{mg}$.

(D) The torque on a current loop in a magnetic field is given by $\vec{\tau} = \vec{m} \times \vec{B}$,where $\vec{m}$ is the magnetic dipole moment. Equilibrium occurs when $\vec{\tau} = 0$,which means $\vec{m}$ is parallel or anti-parallel to $\vec{B}$.
Stable equilibrium occurs when the potential energy $U = -\vec{m} \cdot \vec{B}$ is minimum,which happens when $\vec{m}$ is parallel to $\vec{B}$ (i.e.,the angle $\theta = 0^{\circ}$).
In case $(e)$,the current flows such that the area vector $\vec{A}$ (and thus $\vec{m}$) points in the positive $z-$direction,parallel to $\vec{B}$. This is a state of stable equilibrium.
In case $(f)$,the current flows such that $\vec{m}$ points in the negative $z-$direction,anti-parallel to $\vec{B}$. This is a state of unstable equilibrium.
Therefore,only case $(e)$ corresponds to stable equilibrium.

(B) The torque $\vec{\tau}$ on a current-carrying coil in a magnetic field is given by $\vec{\tau} = \vec{M} \times \vec{B}$.
Here,the magnetic moment $\vec{M} = N I A \hat{n}$,where $A = \pi R^2$.
Since the coil is in the $z-x$ plane,its area vector $\hat{n}$ is along the $y$-axis,so $\vec{M} = N I (\pi R^2) \hat{j}$.
The magnetic field is $\vec{B} = B \hat{k}$.
Therefore,$\vec{\tau} = (N I \pi R^2 \hat{j}) \times (B \hat{k})$.
Using the cross product rules $\hat{j} \times \hat{k} = \hat{i}$,we get $\vec{\tau} = N I \pi R^2 B \hat{i}$.
The magnitude of the torque is $N I \pi R^2 B$.

(A) Let the length of the wire be $L$. For the square loop, the perimeter is $4a = L$, so the side length $a = L/4$. The area $A_s = a^2 = (L/4)^2 = L^2/16$.
For the circular loop, the circumference is $2\pi r = L$, so the radius $r = L/(2\pi)$. The area $A_c = \pi r^2 = \pi (L/(2\pi))^2 = L^2/(4\pi)$.
Since $\pi \approx 3.14$, $4\pi \approx 12.56$, which is less than $16$. Therefore, $A_c > A_s$.
The torque experienced by a current-carrying loop in a magnetic field is given by $\tau = NIAB \sin \theta$. Since $N$, $I$, $B$, and $\theta$ are the same for both loops, the torque is directly proportional to the area $A$.
Since $A_c > A_s$, the torque experienced by the circular loop is greater.

(A) The length of the wire $L$ forms the circumference of the circular coil,so $L = 2 \pi r$,where $r$ is the radius of the coil.
Thus,$r = \frac{L}{2 \pi}$.
The area of the coil is $A = \pi r^2 = \pi \left( \frac{L}{2 \pi} \right)^2 = \frac{L^2}{4 \pi}$.
The magnetic moment of the coil is $M = I A = I \left( \frac{L^2}{4 \pi} \right)$.
The maximum torque $\tau$ on a current-carrying coil in a magnetic field $B$ is given by $\tau = M B \sin \theta$. For maximum torque,$\sin \theta = 1$.
Therefore,$\tau_{max} = M B = \left( \frac{I L^2}{4 \pi} \right) B = \frac{L^2 IB}{4 \pi}$.

(D) Let the length of each wire be $L$.
For the square loop, perimeter $4a = L$, so $a = L/4$. The area $A_s = a^2 = (L/4)^2 = L^2/16$.
For the circular loop, circumference $2\pi r = L$, so $r = L/(2\pi)$. The area $A_c = \pi r^2 = \pi (L/(2\pi))^2 = L^2/(4\pi)$.
Since $\pi \approx 3.14$, $4\pi \approx 12.56$, which is less than $16$. Therefore, $A_c > A_s$.
The torque on a current-carrying loop in a magnetic field is given by $\tau = NIAB \sin \theta$.
Since $N$, $I$, $B$, and $\theta$ are the same for both loops, the torque is directly proportional to the area $A$.
Because $A_c > A_s$, the torque experienced by the circular loop is greater than that experienced by the square loop.

(A) The length of the wire is $L$. When it is formed into a square coil of single turn,the perimeter of the square is $L$.
Let $a$ be the side of the square. Then $4a = L$,which implies $a = \frac{L}{4}$.
The area of the square coil is $A = a^2 = (\frac{L}{4})^2 = \frac{L^2}{16}$.
The torque $\tau$ on a current-carrying coil in a magnetic field is given by $\tau = NIAB \sin \theta$.
For maximum torque,$\sin \theta = 1$.
Given $N = 1$,the maximum torque is $\tau_{max} = IAB = I \times (\frac{L^2}{16}) \times B = \frac{IBL^2}{16}$.

(C) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I_s$, where $n = \frac{N_s}{L}$.
Given: $N_s = 500$, $L = 0.4 \ m$, $I_s = 3 \ A$.
$B = (4\pi \times 10^{-7}) \times (500 / 0.4) \times 3 = 4\pi \times 10^{-7} \times 1250 \times 3 = 15000\pi \times 10^{-7} = 1.5\pi \times 10^{-3} \ T$.
The torque $\tau$ on a coil in a magnetic field is $\tau = N_c I_c A B \sin \theta$.
Given: $N_c = 10$, $I_c = 0.4 \ A$, $r = 0.1 \ m$, $\theta = 90^{\circ}$.
Area $A = \pi r^2 = \pi (0.1)^2 = 0.01\pi \ m^2$.
$\tau = 10 \times 0.4 \times (0.01\pi) \times (1.5\pi \times 10^{-3}) \times \sin 90^{\circ}$.
$\tau = 4 \times 0.01 \times 1.5 \times \pi^2 \times 10^{-3} \times 1$.
Using $\pi^2 = 10$:
$\tau = 0.06 \times 10 \times 10^{-3} = 0.6 \times 10^{-3} = 6 \times 10^{-4} \ Nm$.

(A) The torque $\tau$ on a current-carrying loop in a magnetic field is given by $\tau = N i A B \sin \theta$,where $N$ is the number of turns,$i$ is the current,$A$ is the area,$B$ is the magnetic field,and $\theta$ is the angle between the normal to the loop and the magnetic field.
For a wire of length $L$ formed into a circular coil of $N$ turns with radius $r$,the circumference is $L = N(2 \pi r)$,so $r = \frac{L}{2 \pi N}$.
The area of the coil is $A = \pi r^2 = \pi \left( \frac{L}{2 \pi N} \right)^2 = \frac{L^2}{4 \pi N^2}$.
Substituting $A$ into the torque equation: $\tau = N i \left( \frac{L^2}{4 \pi N^2} \right) B \sin \theta = \frac{i L^2 B \sin \theta}{4 \pi N}$.
To maximize the torque,we set $\sin \theta = 1$ and choose the minimum number of turns,$N = 1$.
Therefore,the maximum torque is $\tau_{\max} = \frac{i L^2 B}{4 \pi}$.

(C) The torque $\tau$ acting on a current-carrying loop in a uniform magnetic field is given by the formula: $\tau = N i A B \sin \theta$.
Here,$N$ is the number of turns,$i$ is the current,$A$ is the area of the loop,$B$ is the magnetic field strength,and $\theta$ is the angle between the normal to the loop and the magnetic field.
From the formula,it is evident that the torque depends on $N, i, A, B,$ and $\theta$.
It does not depend on the shape of the loop,as long as the area $A$ remains constant.

(A) The length of the wire is $L$. When it is bent into a circular coil of $n$ turns,the circumference of one turn is $2\pi r = L/n$,where $r$ is the radius of the coil.
Thus,the radius $r = \frac{L}{2\pi n}$.
The area of one turn is $A = \pi r^2 = \pi \left(\frac{L}{2\pi n}\right)^2 = \frac{\pi L^2}{4\pi^2 n^2} = \frac{L^2}{4\pi n^2}$.
The total magnetic moment of the coil with $n$ turns is $M = nIA = nI \left(\frac{L^2}{4\pi n^2}\right) = \frac{IL^2}{4\pi n}$.
The maximum torque acting on a coil in a magnetic field is given by $\tau_{max} = MB$.
Substituting the value of $M$,we get $\tau_{max} = \left(\frac{IL^2}{4\pi n}\right)B = \frac{BIL^2}{4\pi n}$.

(C) The potential energy of a magnetic dipole in a magnetic field is given by $U = -\vec{m} \cdot \vec{B} = -mB \cos \theta$.
Initially, the axis of the coil is in the direction of the field, so $\theta_i = 0^{\circ}$.
$U_i = -mB \cos 0^{\circ} = -mB$.
After rotating through $90^{\circ}$, the final angle is $\theta_f = 90^{\circ}$.
$U_f = -mB \cos 90^{\circ} = 0$.
The change in potential energy is $\Delta U = U_f - U_i = 0 - (-mB) = mB$.
This change in potential energy is converted into rotational kinetic energy: $\Delta U = K_f - K_i$.
Since the coil starts from rest, $K_i = 0$, so $mB = \frac{1}{2} I \omega^2$.
Rearranging for $\omega$: $\omega = \sqrt{\frac{2mB}{I}}$.
Substituting the given values: $m = 10 \text{ A m}^2$, $B = 2 \text{ T}$, $I = 0.1 \text{ kg m}^2$.
$\omega = \sqrt{\frac{2 \times 10 \times 2}{0.1}} = \sqrt{\frac{40}{0.1}} = \sqrt{400} = 20 \text{ rad/s}$.

(D) The magnetic field is given by $\vec{B} = 0.3 \hat{k} \text{ T}$.
The area vector $\vec{A}$ of the loop placed in the $xy$-plane is perpendicular to the plane,so $\vec{A} = (10 \times 10^{-2} \text{ m} \times 5 \times 10^{-2} \text{ m}) \hat{k} = 50 \times 10^{-4} \hat{k} \text{ m}^2 = 5 \times 10^{-3} \hat{k} \text{ m}^2$.
The magnetic moment $\vec{m}$ is given by $\vec{m} = I \vec{A} = 12 \times 5 \times 10^{-3} \hat{k} = 60 \times 10^{-3} \hat{k} = 0.06 \hat{k} \text{ A m}^2$.
The torque $\vec{\tau}$ acting on the loop is given by $\vec{\tau} = \vec{m} \times \vec{B}$.
Substituting the values,$\vec{\tau} = (0.06 \hat{k}) \times (0.3 \hat{k})$.
Since the cross product of a vector with itself is zero $(\hat{k} \times \hat{k} = 0)$,the torque $\vec{\tau} = 0$.

(B) The magnetic moment of the coil is $M = N I A$.
Given: $N = 10$, $I = \frac{21}{44} \,A$, $A = 1 \,mm^{2} = 10^{-6} \,m^{2}$.
$M = 10 \times \frac{21}{44} \times 10^{-6} \,A-m^{2}$.
The magnetic field inside the solenoid is $B = \mu_{0} n I_{s}$.
Given: $n = 10^{3} \,turns/m$, $I_{s} = 2.5 \,A$, $\mu_{0} = 4\pi \times 10^{-7} \,T-m/A$.
$B = (4 \times \frac{22}{7} \times 10^{-7}) \times 10^{3} \times 2.5 \,T$.
The torque $\tau$ required to hold the coil with its axis perpendicular to the magnetic field is $\tau = M B \sin(90^{\circ}) = M B$.
$\tau = (10 \times \frac{21}{44} \times 10^{-6}) \times (4 \times \frac{22}{7} \times 10^{-7} \times 10^{3} \times 2.5)$.
$\tau = (10 \times \frac{21}{44} \times 10^{-6}) \times (4 \times \frac{22}{7} \times 2.5 \times 10^{-4})$.
$\tau = (10 \times \frac{21}{44} \times 10^{-6}) \times (22 \times 10^{-4}) = 1.5 \times 10^{-8} \,N-m$.

(C) The potential energy $U$ of a magnetic dipole in a magnetic field is given by $U = -\vec{M} \cdot \vec{B} = -MB \cos \theta$,where $\theta$ is the angle between the magnetic moment vector $\vec{M}$ and the magnetic field vector $\vec{B}$.
When the loop is rotated about an axis perpendicular to its plane,the direction of the magnetic moment vector $\vec{M}$ (which is always perpendicular to the plane of the loop) remains unchanged relative to the magnetic field $\vec{B}$.
Since the angle $\theta$ between $\vec{M}$ and $\vec{B}$ does not change,the potential energy of the loop remains constant.
Therefore,the work done $W = \Delta U = 0$.

(A) $1$. The circumference of the loop is $L = 2 \pi r$,so the radius $r = \frac{L}{2 \pi}$.
$2$. The loop rotates with angular velocity $\omega$,creating an equivalent current $I = \frac{q}{T} = \frac{q \omega}{2 \pi}$.
$3$. The magnetic moment of the loop is $M = I A = I (\pi r^2) = \left( \frac{q \omega}{2 \pi} \right) \pi \left( \frac{L}{2 \pi} \right)^2 = \frac{q \omega L^2}{8 \pi^2}$.
$4$. The magnetic torque is given by $\tau = |\vec{M} \times \vec{B}| = M B \sin \theta$. Since the magnetic field $B$ is parallel to the plane of the loop,the angle between the magnetic moment vector (perpendicular to the plane) and the magnetic field is $\theta = 90^\circ$.
$5$. Therefore,$\tau = M B \sin 90^\circ = M B = \frac{q \omega L^2 B}{8 \pi^2}$.

(B) The length of the wire is $L = 10 \ m$. When it is bent into a circular loop of radius $r$,the circumference is $2 \pi r = L = 10 \ m$.
Thus,$r = \frac{10}{2 \pi} \ m$.
The area of the loop is $A = \pi r^2 = \pi \left( \frac{10}{2 \pi} \right)^2 = \pi \left( \frac{100}{4 \pi^2} \right) = \frac{25}{\pi} \ m^2$.
The magnetic moment of the loop is $M = I \times A = 1 \times \frac{25}{\pi} = \frac{25}{\pi} \ A \ m^2$.
The maximum torque acting on the loop in a magnetic field $B$ is given by $\tau_{max} = M \times B$.
Substituting the values,$\tau_{max} = \left( \frac{25}{\pi} \right) \times (2 \pi \times 10^{-4}) = 50 \times 10^{-4} \ N \ m$.