Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done Questions in English

(A) The formula for the magnetic potential $V$ due to a magnetic dipole at a point on its axis at a distance $r$ from the center is given by:
$V = \frac{\mu_0}{4 \pi} \frac{M}{r^2}$
Given values:
$V = 1.5 \times 10^{-5} \ T \cdot m$
$r = 20 \ cm = 0.2 \ m = 2 \times 10^{-1} \ m$
$\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m \cdot A^{-1}$
Substituting these values into the formula:
$1.5 \times 10^{-5} = 10^{-7} \times \frac{M}{(2 \times 10^{-1})^2}$
$1.5 \times 10^{-5} = 10^{-7} \times \frac{M}{4 \times 10^{-2}}$
$M = \frac{1.5 \times 10^{-5} \times 4 \times 10^{-2}}{10^{-7}}$
$M = \frac{6 \times 10^{-7}}{10^{-7}} = 6 \ A \cdot m^2$
Thus,the magnetic moment of the dipole is $6 \ A \cdot m^2$.

(B) The potential energy of a magnetic dipole in a magnetic field is given by $U = -mB \cos \theta$.
At the most stable equilibrium position,the angle $\theta = 0^{\circ}$,so $U_i = -mB \cos 0^{\circ} = -mB$.
At the most unstable equilibrium position,the angle $\theta = 180^{\circ}$,so $U_f = -mB \cos 180^{\circ} = +mB$.
The work done $W$ is equal to the change in potential energy: $W = \Delta U = U_f - U_i$.
$W = mB - (-mB) = 2mB$.
Given $m = 0.5 \text{ A m}^2$ and $B = 8 \times 10^{-2} \text{ T}$.
$W = 2 \times 0.5 \times 8 \times 10^{-2} = 1 \times 8 \times 10^{-2} = 8 \times 10^{-2} \text{ J}$.

(D) The torque $\tau$ on a magnetic dipole is given by $\tau = MB \sin \theta$.
Given $\tau = 80 \sqrt{3} \ N \ m$ and $\theta = 60^{\circ}$.
$80 \sqrt{3} = MB \sin 60^{\circ} = MB \left( \frac{\sqrt{3}}{2} \right)$.
Multiplying both sides by $\frac{2}{\sqrt{3}}$,we get $MB = 160 \ J$.
The potential energy $U$ of the dipole is given by $U = -M \cdot B = -MB \cos \theta$.
Substituting the values,$U = -160 \cos 60^{\circ} = -160 \times \frac{1}{2} = -80 \ J$.

(A) The work done $W$ in rotating a magnetic dipole in a uniform magnetic field from an angle $\theta_1$ to $\theta_2$ is given by the formula:
$W = MB(\cos \theta_1 - \cos \theta_2)$
Given:
Magnetic moment $M = 4 \, A-m^2$
Magnetic field $B = 5 \times 10^{-4} \, T$
Initial angle $\theta_1 = 30^{\circ}$
Final angle $\theta_2 = 45^{\circ}$
Substituting the values:
$W = (4) \times (5 \times 10^{-4}) \times (\cos 30^{\circ} - \cos 45^{\circ})$
$W = 20 \times 10^{-4} \times (\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}})$
$W = 20 \times 10^{-4} \times (0.866 - 0.707)$
$W = 20 \times 10^{-4} \times (0.159)$
$W = 3.18 \times 10^{-4} \, J \approx 3.2 \times 10^{-4} \, J$

(C) Given: Magnetic moment $M = 16 \ Am^2$,distance $r = 20 \ cm = 0.2 \ m$. The constant $\frac{\mu_0}{4\pi} = 10^{-7} \ T \cdot m/A$.
For an axial point: $B_{\text{axial}} = \frac{\mu_0}{4\pi} \frac{2M}{r^3} = 10^{-7} \times \frac{2 \times 16}{(0.2)^3} = 10^{-7} \times \frac{32}{0.008} = 10^{-7} \times 4000 = 4 \times 10^{-4} \ T$.
For an equatorial point: $B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \frac{M}{r^3} = 10^{-7} \times \frac{16}{(0.2)^3} = 10^{-7} \times \frac{16}{0.008} = 10^{-7} \times 2000 = 2 \times 10^{-4} \ T$.
Thus,the values are $4 \times 10^{-4} \ T$ and $2 \times 10^{-4} \ T$.

(B) The work done $W$ in rotating a magnetic dipole in a uniform magnetic field $B$ from an angle $\theta_1$ to $\theta_2$ is given by $W = MB(\cos \theta_1 - \cos \theta_2)$.
Assuming the initial position is the stable equilibrium position,$\theta_1 = 0^{\circ}$.
For the first case,$\theta_2 = 90^{\circ}$:
$W_1 = MB(\cos 0^{\circ} - \cos 90^{\circ}) = MB(1 - 0) = MB$.
For the second case,$\theta_2 = 60^{\circ}$:
$W_2 = MB(\cos 0^{\circ} - \cos 60^{\circ}) = MB(1 - 0.5) = 0.5MB$.
According to the problem,$W_1 = n \times W_2$.
Substituting the values: $MB = n \times (0.5MB)$.
$1 = n \times 0.5$.
$n = 1 / 0.5 = 2$.

(C) The time period $T$ of a magnetic needle performing simple harmonic motion in a magnetic field is given by the formula:
$T = 2 \pi \sqrt{\frac{I}{MB}}$
Given:
Magnetic moment $M = 6 \times 10^{-2} \text{ A m}^2$
Moment of inertia $I = 9.6 \times 10^{-5} \text{ kg m}^2$
Magnetic field $B = 0.01 \text{ T}$
Substituting the values:
$T = 2 \times 3.14 \times \sqrt{\frac{9.6 \times 10^{-5}}{6 \times 10^{-2} \times 0.01}}$
$T = 6.28 \times \sqrt{\frac{9.6 \times 10^{-5}}{6 \times 10^{-4}}}$
$T = 6.28 \times \sqrt{0.16}$
$T = 6.28 \times 0.4 = 2.512 \text{ s}$
The time taken for $10$ oscillations is $10 \times T = 10 \times 2.512 = 25.12 \text{ s}$.

(A) The potential energy $U$ of a magnetic dipole in an external magnetic field is given by the formula $U = -\vec{M} \cdot \vec{B} = -MB \cos \theta$,where $\theta$ is the angle between the magnetic moment $\vec{M}$ and the magnetic field $\vec{B}$.
For the minimum potential energy,$\cos \theta$ must be maximum,which occurs at $\theta = 0^{\circ}$. Thus,$U_{\text{min}} = -MB \cos 0^{\circ} = -MB(1) = -MB$.
For the maximum potential energy,$\cos \theta$ must be minimum,which occurs at $\theta = 180^{\circ}$. Thus,$U_{\text{max}} = -MB \cos 180^{\circ} = -MB(-1) = +MB$.
Therefore,the minimum and maximum values are $-MB$ and $+MB$ respectively.

(C) The potential energy of a magnetic dipole in a magnetic field is given by $U = -M B \cos \theta$.
Initially,the dipole is oriented along the field,so $\theta_1 = 0^\circ$. The initial potential energy is $U_i = -M B \cos(0^\circ) = -M B$.
To perform maximum work,the dipole must be rotated to the position of maximum potential energy,which is $\theta_2 = 180^\circ$. The final potential energy is $U_f = -M B \cos(180^\circ) = M B$.
The work done by an external agent is $W = U_f - U_i = M B - (-M B) = 2 M B$.

(A) Let the point be at a distance $d$ from the center of the magnets along the axis of magnet $A$.
The magnetic field due to magnet $A$ (axial position) is $B_{1} = \frac{\mu_{0}}{4\pi} \frac{2M_{1}}{d^{3}}$.
The magnetic field due to magnet $B$ (equatorial position) at the same point is $B_{2} = \frac{\mu_{0}}{4\pi} \frac{M_{2}}{d^{3}}$.
Since the axes are perpendicular,the resultant field makes an angle $\theta$ with the axis of magnet $A$ such that $\tan \theta = \frac{B_{2}}{B_{1}}$.
Given $\theta = 45^{\circ}$,so $\tan 45^{\circ} = 1$.
Therefore,$1 = \frac{(\mu_{0} M_{2}) / (4\pi d^{3})}{(2\mu_{0} M_{1}) / (4\pi d^{3})} = \frac{M_{2}}{2M_{1}}$.
This gives $\frac{M_{2}}{M_{1}} = 2$,or $M_{2} : M_{1} = 2 : 1$.

(D) For a bar magnet of magnetic moment $M$,the magnetic field at an axial point at distance $d$ is $B_{axial} = \frac{\mu_{0}}{4 \pi} \frac{2M}{d^{3}}$.
For the same magnet,the magnetic field at an equatorial point at distance $d$ is $B_{equatorial} = \frac{\mu_{0}}{4 \pi} \frac{M}{d^{3}}$.
Since the magnets are perpendicular,the point at distance $d$ from both centers acts as an axial point for one magnet and an equatorial point for the other.
The resultant magnetic field is $B_{net} = \sqrt{B_{axial}^{2} + B_{equatorial}^{2}}$.
Substituting the values: $B_{net} = \sqrt{\left(\frac{\mu_{0}}{4 \pi} \frac{2M}{d^{3}}\right)^{2} + \left(\frac{\mu_{0}}{4 \pi} \frac{M}{d^{3}}\right)^{2}}$.
$B_{net} = \frac{\mu_{0} M}{4 \pi d^{3}} \sqrt{2^{2} + 1^{2}} = \frac{\mu_{0}}{4 \pi} \frac{M}{d^{3}} \sqrt{5}$.

(A) The torque $\tau$ acting on a magnetic dipole in a magnetic field is given by $\tau = MB \sin \theta$.
Given,$\tau_1 = 1.732 \times 10^{-5} \text{ Nm}$ at $\theta_1 = 90^{\circ}$.
So,$\tau_1 = MB \sin 90^{\circ} = MB(1) = MB$.
Therefore,$MB = 1.732 \times 10^{-5} \text{ Nm}$.
Now,for $\theta_2 = 60^{\circ}$,the torque $\tau_2$ is:
$\tau_2 = MB \sin 60^{\circ} = (1.732 \times 10^{-5}) \times \frac{\sqrt{3}}{2}$.
Since $\sqrt{3} = 1.732$,we have:
$\tau_2 = 1.732 \times 10^{-5} \times \frac{1.732}{2} = 1.732 \times 10^{-5} \times 0.866 = 1.4999 \times 10^{-5} \approx 1.5 \times 10^{-5} \text{ Nm}$.

(D) The magnetic field of a short bar magnet at an axial point at distance '$r$' is given by $B_{axial} = \frac{\mu_0}{4\pi} \frac{2M}{r^3}$.
At the equatorial point at the same distance '$r$',the magnetic field is given by $B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{r^3}$.
As the point moves from the axial position to the equatorial position along a circular path of radius '$r$',the angle $\theta$ with the magnetic moment vector changes from $0^\circ$ to $90^\circ$.
The general formula for the magnetic field at any point $(r, \theta)$ is $B = \frac{\mu_0}{4\pi} \frac{M}{r^3} \sqrt{1 + 3\cos^2\theta}$.
As $\theta$ increases from $0^\circ$ to $90^\circ$,$\cos^2\theta$ decreases from $1$ to $0$,and therefore the magnitude of the magnetic field '$B$' will go on decreasing.

(D) The torque (couple) acting on a bar magnet in a uniform magnetic field is given by $T = MB \sin \theta$,where $M$ is the magnetic moment,$B$ is the magnetic field,and $\theta$ is the angle between the magnet and the field.
Initially,the magnet is perpendicular to the field,so $\theta = \frac{\pi}{2} = 90^{\circ}$.
The initial torque is $T = MB \sin(90^{\circ}) = MB$.
We want the new torque $T'$ to be half of the initial torque,so $T' = \frac{T}{2} = \frac{MB}{2}$.
Let the new angle be $\theta'$. Then $T' = MB \sin \theta' = \frac{MB}{2}$.
Dividing both sides by $MB$,we get $\sin \theta' = \frac{1}{2} = 0.5$.
Therefore,the angle $\theta'$ is $\sin^{-1}(0.5) = 30^{\circ}$.

(B) The magnetic field on the axis of a bar magnet is given by $B = \frac{\mu_0}{4\pi} \frac{2Md}{(d^2-l^2)^2}$,where $d$ is the distance from the center and $2l$ is the magnetic length.
Given $d_1 = 10 \ cm$ and $d_2 = 10 + 10 = 20 \ cm$.
The ratio is $\frac{B_1}{B_2} = \frac{d_1}{d_2} \left( \frac{d_2^2 - l^2}{d_1^2 - l^2} \right)^2 = \frac{25}{2}$.
Substituting the values: $\frac{10}{20} \left( \frac{20^2 - l^2}{10^2 - l^2} \right)^2 = \frac{25}{2}$.
$\frac{1}{2} \left( \frac{400 - l^2}{100 - l^2} \right)^2 = \frac{25}{2} \Rightarrow \left( \frac{400 - l^2}{100 - l^2} \right)^2 = 25$.
Taking the square root: $\frac{400 - l^2}{100 - l^2} = 5$.
$400 - l^2 = 500 - 5l^2 \Rightarrow 4l^2 = 100 \Rightarrow l^2 = 25$.
So,$l = 5 \ cm$.
The magnetic length is $2l = 2 \times 5 \ cm = 10 \ cm$.

(B) The magnetic potential $V$ due to a magnetic dipole of magnetic moment $M$ at a point $(r, \theta)$ is given by $V = \frac{\mu_{0}}{4 \pi} \frac{M \cos \theta}{r^{2}}$.
On the equatorial line (right bisector),the angle $\theta$ between the position vector and the dipole axis is $90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the magnetic potential $V$ is zero at all points on the equatorial line.
Therefore,the correct option is $B$.

(D) The torque $\tau$ acting on a magnetic dipole in a uniform magnetic field $B$ is given by the formula: $\tau = m B \sin \theta$.
Given:
Torque $\tau = 4.5 \times 10^{-2} \ J$
Magnetic field $B = 0.5 \ T$
Angle $\theta = 30^{\circ}$
Substituting the values into the formula:
$4.5 \times 10^{-2} = m \times 0.5 \times \sin 30^{\circ}$
Since $\sin 30^{\circ} = 0.5$:
$4.5 \times 10^{-2} = m \times 0.5 \times 0.5$
$4.5 \times 10^{-2} = m \times 0.25$
$m = \frac{4.5 \times 10^{-2}}{0.25}$
$m = 18 \times 10^{-2} \ J \ T^{-1}$.

(A) The magnetic field $B$ on the equatorial line of a short bar magnet is given by the formula:
$B = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$
Given:
Magnetic moment $m = 0.75 \ A \ m^2$
Distance $r = 75 \ cm = 0.75 \ m$
Constant $\frac{\mu_0}{4\pi} = 10^{-7} \ T \ m/A$
Substituting the values:
$B = 10^{-7} \times \frac{0.75}{(0.75)^3}$
$B = 10^{-7} \times \frac{1}{(0.75)^2}$
$B = 10^{-7} \times \frac{1}{0.5625}$
$B \approx 1.78 \times 10^{-7} \ T$

(C) The torque $\tau$ experienced by a magnetic dipole in a uniform magnetic field $B$ is given by $\tau = mB \sin \theta$,where $m$ is the magnetic moment and $\theta$ is the angle between the magnetic moment vector and the magnetic field vector.
Assuming the maximum torque condition (where $\sin \theta = 1$),we have $\tau = mB$.
Given $\tau = 0.64 \ J$ and $B = 0.32 \ T$.
Substituting the values: $0.64 = m \times 0.32$.
Therefore,$m = \frac{0.64}{0.32} = 2 \ Am^2$.
Thus,the magnetic moment of the magnet is $2 \ Am^2$.

(A) The correct answer is $A$.
For a bar magnet,the magnetic field $\overrightarrow{B}$ at a point on its axial line at a distance $Z$ from the center is given by the formula:
$\overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{2\overrightarrow{M}}{Z^3}$
where $\overrightarrow{M}$ is the magnetic dipole moment of the magnet.
Since the expression for the magnetic field is directly proportional to the magnetic dipole moment vector $\overrightarrow{M}$,the direction of the magnetic field $\overrightarrow{B}$ is the same as the direction of the magnetic dipole moment $\overrightarrow{M}$.

(B) When a magnetic dipole with magnetic moment $\vec{M}$ is placed in an external magnetic field $\vec{B}$,it experiences a torque $\vec{\tau}$.
The torque is defined as the cross product of the magnetic moment and the magnetic field.
Mathematically,the torque is given by $\vec{\tau} = \vec{M} \times \vec{B}$.
Therefore,the correct option is $B$.

(C) The torque $\tau$ experienced by a magnetic dipole in an external magnetic field $B$ is given by the formula: $\tau = m B \sin \theta$,where $m$ is the magnetic moment and $\theta$ is the angle between the axis of the magnet and the magnetic field.
Given values are: $\tau = 4.5 \times 10^{-2} \ J$,$B = 0.25 \ T$,and $\theta = 30^{\circ}$.
Rearranging the formula to solve for $m$: $m = \frac{\tau}{B \sin \theta}$.
Substituting the values: $m = \frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}$.
Since $\sin 30^{\circ} = 0.5$,we have: $m = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5} = \frac{4.5 \times 10^{-2}}{0.125}$.
Calculating the result: $m = 36 \times 10^{-2} = 0.36 \ J \ T^{-1}$.

(A) The torque $\tau$ acting on a magnetic dipole of magnetic moment $M$ in a uniform magnetic field $B$ is given by the formula $\tau = M B \sin \theta$,where $\theta$ is the angle between the dipole axis and the magnetic field.
For the torque to be zero,we must have $\tau = 0$.
Substituting the formula,we get $M B \sin \theta = 0$.
Since $M$ and $B$ are non-zero,we must have $\sin \theta = 0$.
This implies $\theta = 0^{\circ}$ or $\theta = 180^{\circ}$.
Among the given options,the correct angle is $0^{\circ}$.

(C) magnetic dipole of magnetic moment $M$ and moment of inertia $I$ placed in a uniform magnetic field $B$ experiences a restoring torque $\tau = -MB \sin \theta$.
For small oscillations, $\sin \theta \approx \theta$, so $\tau = -MB \theta$.
Comparing this with the equation for angular $SHM$, $\tau = -C \theta$, where $C = MB$.
The time period $T$ of an angular $SHM$ is given by $T = 2 \pi \sqrt{\frac{I}{C}}$.
Substituting $C = MB$, we get $T = 2 \pi \sqrt{\frac{I}{MB}}$.

(C) Given: Magnetic moment $ M = 6 \times 10^{-2} \text{ A m}^2 $, Moment of inertia $ I = 12 \times 10^{-6} \text{ kg m}^2 $, Magnetic field $ B = 2 \times 10^{-2} \text{ T} $.
The time period $ t $ of a magnetic dipole oscillating in a magnetic field is given by $ t = 2 \pi \sqrt{\frac{I}{MB}} $.
Substituting the values: $ t = 2 \pi \sqrt{\frac{12 \times 10^{-6}}{(6 \times 10^{-2}) \times (2 \times 10^{-2})}} $.
$ t = 2 \pi \sqrt{\frac{12 \times 10^{-6}}{12 \times 10^{-4}}} = 2 \pi \sqrt{10^{-2}} = 2 \pi \times 10^{-1} \text{ s} $.
Given $ \pi \simeq 3 $, so $ t = 2 \times 3 \times 0.1 = 0.6 \text{ s} $.
The time taken to complete $ 20 $ oscillations is $ T = 20 \times t = 20 \times 0.6 = 12 \text{ s} $.

(D) The torque $\tau$ acting on a magnetic dipole in a uniform magnetic field $B$ is given by the formula: $\tau = M B \sin \theta$,where $M$ is the magnetic moment and $\theta$ is the angle between the magnetic moment vector and the magnetic field vector.
Given values are: $\tau = 0.36 \sqrt{2} \ Nm$,$B = 2 \ T$,and $\theta = 45^{\circ}$.
Substituting these values into the formula: $0.36 \sqrt{2} = M \times 2 \times \sin(45^{\circ})$.
Since $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$,we have: $0.36 \sqrt{2} = M \times 2 \times \frac{1}{\sqrt{2}}$.
Simplifying the equation: $0.36 \sqrt{2} = M \times \sqrt{2}$.
Dividing both sides by $\sqrt{2}$,we get: $M = 0.36 \ J \ T^{-1}$.

(A) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the formula: $W = MB(\cos \theta_1 - \cos \theta_2)$.
Here,the magnetic moment $M = 10^4 \,J \,T^{-1}$,the magnetic field $B = 4 \times 10^{-5} \,T$,the initial angle $\theta_1 = 0^{\circ}$,and the final angle $\theta_2 = 60^{\circ}$.
Substituting the values into the formula:
$W = (10^4) \times (4 \times 10^{-5}) \times (\cos 0^{\circ} - \cos 60^{\circ})$
$W = 0.4 \times (1 - 0.5)$
$W = 0.4 \times 0.5 = 0.2 \,J$.
Therefore,the work done is $0.2 \,J$.

(A) The torque $\vec{\tau}$ on a magnetic dipole $\vec{M}$ in a magnetic field $\vec{B}$ is given by $\vec{\tau} = \vec{M} \times \vec{B}$.
Let the magnitude of the magnetic moment be $M$. The unit vector in the direction of the needle is $\hat{u} = \frac{\sqrt{3}\hat{i} + \hat{j}}{2}$. Thus,$\vec{M} = \frac{M}{2}(\sqrt{3}\hat{i} + \hat{j})$.
Given $\vec{B}_1 = B\hat{i}$,the torque is $\vec{\tau}_1 = \frac{M}{2}(\sqrt{3}\hat{i} + \hat{j}) \times B\hat{i} = \frac{MB}{2}(\sqrt{3}(\hat{i} \times \hat{i}) + (\hat{j} \times \hat{i})) = \frac{MB}{2}(0 - \hat{k}) = -\frac{MB}{2}\hat{k}$.
Given $|\vec{\tau}_1| = 0.06 \ N-m$,we have $\frac{MB}{2} = 0.06$,so $MB = 0.12 \ N-m$.
In the second case,the needle is oriented in the direction $(\hat{i} + \sqrt{3}\hat{j})$,so $\vec{M}_2 = \frac{M}{2}(\hat{i} + \sqrt{3}\hat{j})$. The magnetic field is $\vec{B}_2 = 2B\hat{j}$.
The torque is $\vec{\tau}_2 = \vec{M}_2 \times \vec{B}_2 = \frac{M}{2}(\hat{i} + \sqrt{3}\hat{j}) \times 2B\hat{j} = MB(\hat{i} \times \hat{j} + \sqrt{3}(\hat{j} \times \hat{j})) = MB(\hat{k} + 0) = MB\hat{k}$.
Thus,$|\vec{\tau}_2| = MB = 0.12 \ N-m$.

(B) The torque $\tau$ acting on a magnetic needle of magnetic moment $M$ in a uniform magnetic field $B$ is given by the formula: $\tau = M B \sin \theta$.
Here,$\theta$ is the angle between the magnetic moment and the magnetic field.
It is given that the needle remains perpendicular to the magnetic field,so $\theta = 90^{\circ}$.
Since $\sin 90^{\circ} = 1$,the torque becomes $\tau = M B$.
For the two different places with magnetic fields $B_1$ and $B_2$,the torques are $\tau_1 = M B_1$ and $\tau_2 = M B_2$.
Taking the ratio of the two torques: $\frac{\tau_1}{\tau_2} = \frac{M B_1}{M B_2} = \frac{B_1}{B_2}$.
Therefore,the ratio $B_1 : B_2$ is equal to $\tau_1 : \tau_2$.

(B) The work done in rotating a magnetic dipole in a uniform magnetic field is given by $W = MB(\cos \theta_1 - \cos \theta_2)$.
Initially,the magnet is in the direction of the field,so $\theta_1 = 0^{\circ}$ and $\theta_2 = 45^{\circ}$.
$W = MB(\cos 0^{\circ} - \cos 45^{\circ}) = MB(1 - \frac{1}{\sqrt{2}}) = MB(\frac{\sqrt{2}-1}{\sqrt{2}})$.
Thus,$MB = \frac{W\sqrt{2}}{\sqrt{2}-1}$.
Now,the additional work $W'$ to rotate it further by $15^{\circ}$ means rotating from $\theta_1 = 45^{\circ}$ to $\theta_2 = 45^{\circ} + 15^{\circ} = 60^{\circ}$.
$W' = MB(\cos 45^{\circ} - \cos 60^{\circ}) = MB(\frac{1}{\sqrt{2}} - \frac{1}{2}) = MB(\frac{\sqrt{2}-1}{2\sqrt{2}})$.
Substituting the value of $MB$:
$W' = (\frac{W\sqrt{2}}{\sqrt{2}-1}) \times (\frac{\sqrt{2}-1}{2\sqrt{2}}) = \frac{W}{2}$.

(C) Given: Magnetic moment,$m = 0.4 \,Am^2$
Distance,$r = 50 \,cm = 0.5 \,m$
The formula for the magnitude of the axial magnetic field of a short bar magnet is:
$B_{\text{axial}} = \frac{\mu_0}{4\pi} \left( \frac{2m}{r^3} \right)$
Substituting the values:
$B_{\text{axial}} = (10^{-7}) \times \frac{2 \times 0.4}{(0.5)^3}$
$B_{\text{axial}} = 10^{-7} \times \frac{0.8}{0.125}$
$B_{\text{axial}} = 10^{-7} \times 6.4 = 6.4 \times 10^{-7} \,T$

(A) The work done $W$ to rotate a magnetic needle from an angle $\theta_1$ to $\theta_2$ in a magnetic field $B$ is given by $W = MB(\cos \theta_1 - \cos \theta_2)$.
Given $\theta_1 = 0^{\circ}$ and $\theta_2 = 60^{\circ}$,we have:
$W = MB(\cos 0^{\circ} - \cos 60^{\circ}) = MB(1 - 0.5) = 0.5 MB$.
Thus,$MB = 2W$.
The torque $\tau$ required to maintain the needle at an angle $\theta = 60^{\circ}$ is given by $\tau = MB \sin \theta$.
Substituting the values,$\tau = MB \sin 60^{\circ} = (2W) \times \frac{\sqrt{3}}{2} = \sqrt{3} W$.

(A) Given,length of bar magnet,$l = 10 \text{ cm} = 10^{-1} \text{ m}$.
Pole strength,$m = 10^{-3} \text{ A-m}$.
Magnetic induction,$B = 4 \pi \times 10^{-3} \text{ T}$.
Angle,$\theta = 30^{\circ}$.
First,calculate the magnetic dipole moment $M$:
$M = m \times l = 10^{-3} \text{ A-m} \times 10^{-1} \text{ m} = 10^{-4} \text{ A-m}^2$.
The torque $\tau$ acting on the magnet is given by the formula:
$\tau = M B \sin \theta$.
Substituting the values:
$\tau = (10^{-4}) \times (4 \pi \times 10^{-3}) \times \sin 30^{\circ}$.
Since $\sin 30^{\circ} = 0.5 = \frac{1}{2}$:
$\tau = 4 \pi \times 10^{-7} \times \frac{1}{2} = 2 \pi \times 10^{-7} \text{ N-m}$.

(D) The torque $\tau$ experienced by a bar magnet in a uniform magnetic field $B$ is given by $\tau = MB \sin \theta$,where $M$ is the magnetic dipole moment and $\theta$ is the angle between the magnet and the field.
Let the initial angle be $\theta_1$. Then the initial torque is $\tau_1 = MB \sin \theta_1$ $(i)$.
If the angle is doubled,the new angle is $\theta_2 = 2\theta_1$. The new torque is $\tau_2 = MB \sin \theta_2 = MB \sin 2\theta_1$ (ii).
Given that the torque increases by $41.4 \%$,we have $\tau_2 = \tau_1 + 0.414 \tau_1 = 1.414 \tau_1$.
Since $\sqrt{2} \approx 1.414$,we can write $\tau_2 = \sqrt{2} \tau_1$.
Substituting the expressions for $\tau_1$ and $\tau_2$ from $(i)$ and (ii):
$MB \sin 2\theta_1 = \sqrt{2} MB \sin \theta_1$
Using the trigonometric identity $\sin 2\theta = 2 \sin \theta \cos \theta$:
$2 \sin \theta_1 \cos \theta_1 = \sqrt{2} \sin \theta_1$
Assuming $\sin \theta_1 \neq 0$,we divide both sides by $\sin \theta_1$:
$2 \cos \theta_1 = \sqrt{2}$
$\cos \theta_1 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$
Therefore,$\theta_1 = 45^{\circ}$.
Thus,the correct option is $D$.

(D) The potential energy of a magnetic dipole in a magnetic field is given by $U = -MB \cos \theta$.
Initially, the dipole is in the East-West direction $(\theta_1 = 90^{\circ})$, so $U_i = -MB \cos 90^{\circ} = 0$.
Finally, the dipole is in the North-South position $(\theta_2 = 0^{\circ})$, so $U_f = -MB \cos 0^{\circ} = -MB$.
By the law of conservation of energy, the loss in potential energy is equal to the gain in kinetic energy:
$KE = U_i - U_f = 0 - (-MB) = MB$.
Given $M = 2.5 \text{ A m}^2$ and $B_H = 3 \times 10^{-5} \text{ T}$.
$KE = 2.5 \times 3 \times 10^{-5} = 7.5 \times 10^{-5} \text{ J}$.
Converting to microjoules: $7.5 \times 10^{-5} \text{ J} = 75 \times 10^{-6} \text{ J} = 75 \mu\text{J}$.

(A) The time period $T$ of a magnet suspended in a uniform magnetic field $B$ is given by the formula $T = 2 \pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.
Since $T \propto \frac{1}{\sqrt{M}}$,we have $\frac{T_2}{T_1} = \sqrt{\frac{M_1}{M_2}}$.
Given that the magnetic moment is reduced by $19 \%$,the new magnetic moment $M_2 = M_1 - 0.19 M_1 = 0.81 M_1$.
Substituting this into the ratio,we get $\frac{T_2}{T_1} = \sqrt{\frac{M_1}{0.81 M_1}} = \sqrt{\frac{1}{0.81}} = \frac{1}{0.9} \approx 1.111$.
Thus,$T_2 \approx 1.11 T_1$.
The percentage increase in the time period is $\frac{T_2 - T_1}{T_1} \times 100 = (1.11 - 1) \times 100 = 11 \%$.
Therefore,the time period increases by approximately $11 \%$.

(D) Given: Earth's magnetic field $B_e = 4 \times 10^{-5} \ T$,distance $r = 40 \ cm = 0.4 \ m$.
The magnetic field $B$ due to a short bar magnet at a point on its equatorial line (normal bisector) is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3}$.
The resultant magnetic field makes an angle of $45^{\circ}$ with the earth's magnetic field. Since the magnet's axis is perpendicular to the earth's field,the field $B$ is perpendicular to $B_e$.
The angle $\theta$ of the resultant field with $B_e$ is given by $\tan \theta = \frac{B}{B_e}$.
Given $\theta = 45^{\circ}$,we have $\tan 45^{\circ} = 1$,so $B = B_e$.
Substituting the values: $4 \times 10^{-5} = 10^{-7} \times \frac{M}{(0.4)^3}$.
$M = \frac{4 \times 10^{-5} \times 0.064}{10^{-7}} = 4 \times 10^2 \times 0.064 = 400 \times 0.064 = 25.6 \ Am^2$.

(B) Given that, magnetic moment, $M = 2 \,J \,T^{-1}$.
Magnetic field, $B = 0.1 \,T$.
Since the magnetic moment is aligned in the direction of the magnetic field, the initial angle is $\theta_1 = 0^{\circ}$.
When the magnet is normal to the field, the final angle is $\theta_2 = 90^{\circ}$.
The work done $W$ in rotating a magnet in a magnetic field is given by the formula:
$W = MB(\cos \theta_1 - \cos \theta_2)$
Substituting the values:
$W = 2 \times 0.1 \times (\cos 0^{\circ} - \cos 90^{\circ})$
$W = 0.2 \times (1 - 0)$
$W = 0.2 \,J$.
Hence, the net work done is $0.2 \,J$.

(D) The work done to rotate a magnetic dipole in a magnetic field from an initial angle $\theta_1$ to a final angle $\theta_2$ is given by the formula:
$W = MB(\cos \theta_1 - \cos \theta_2)$
In this problem,the magnet is rotated through $360^{\circ}$,which means the initial angle $\theta_1 = 0^{\circ}$ and the final angle $\theta_2 = 360^{\circ}$.
Substituting these values into the formula:
$W = MB(\cos 0^{\circ} - \cos 360^{\circ})$
Since $\cos 0^{\circ} = 1$ and $\cos 360^{\circ} = 1$:
$W = MB(1 - 1) = 0$
Therefore,the total work done is $0$.

(C) The origin is on the axial line of the first magnet $(M_1)$ and on the equatorial line of the second magnet $(M_2)$.
Given: $M = 9 \text{ Am}^2$, $r = 3 \text{ cm} = 3 \times 10^{-2} \text{ m}$.
Magnetic field due to $M_1$ (axial point) at the origin:
$B_1 = \frac{\mu_0}{4 \pi} \times \frac{2M}{r^3} = 10^{-7} \times \frac{2 \times 9}{(3 \times 10^{-2})^3} = 10^{-7} \times \frac{18}{27 \times 10^{-6}} = \frac{2}{3} \times 10^{-1} \text{ T}$.
This field is directed along the positive $X$-axis.
Magnetic field due to $M_2$ (equatorial point) at the origin:
$B_2 = \frac{\mu_0}{4 \pi} \times \frac{M}{r^3} = 10^{-7} \times \frac{9}{(3 \times 10^{-2})^3} = 10^{-7} \times \frac{9}{27 \times 10^{-6}} = \frac{1}{3} \times 10^{-1} \text{ T}$.
Since the magnetic moment of $M_2$ is in the negative $X$-direction, the equatorial field at the origin points in the positive $X$-direction.
As both $B_1$ and $B_2$ point in the same direction, the resultant magnetic field is:
$B = B_1 + B_2 = \left(\frac{2}{3} + \frac{1}{3}\right) \times 10^{-1} \text{ T} = 1 \times 10^{-1} \text{ T} = 0.1 \text{ T}$.

(D) The magnetic field on the axial line of a short bar magnet is given by $B_{axial} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}$.
Since the axis of the magnet is perpendicular to the horizontal component of the earth's magnetic field $(B_H)$,the resultant magnetic field makes an angle of $45^{\circ}$ with $B_H$ when the magnitude of the axial field equals the magnitude of the horizontal component of the earth's magnetic field.
Therefore,$B_{axial} = B_H$.
Substituting the values: $\frac{10^{-7} \times 2 \times 0.21}{r^3} = 4.2 \times 10^{-5}$.
$r^3 = \frac{2 \times 0.21 \times 10^{-7}}{4.2 \times 10^{-5}} = \frac{0.42 \times 10^{-7}}{4.2 \times 10^{-5}} = 0.1 \times 10^{-2} = 10^{-3} \ m^3$.
$r = 0.1 \ m = 10 \ cm$.

(A) The torque $\tau$ experienced by a magnetic dipole in a uniform magnetic field $B$ is given by the formula: $\tau = MB \sin \theta$.
Given:
Torque $\tau = 3.6 \times 10^{-5} \,J$
Magnetic field $B = 28.3 \times 10^{-3} \,T$
Angle $\theta = 45^{\circ}$
Rearranging the formula to solve for the magnetic moment $M$:
$M = \frac{\tau}{B \sin \theta}$
Substituting the values:
$M = \frac{3.6 \times 10^{-5}}{28.3 \times 10^{-3} \times \sin 45^{\circ}}$
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707$:
$M = \frac{3.6 \times 10^{-5}}{28.3 \times 10^{-3} \times 0.707} \approx \frac{3.6 \times 10^{-5}}{20.008 \times 10^{-3}} \approx 0.1799 \times 10^{-2} \approx 1.8 \times 10^{-3} \,J \,T^{-1}$.

(C) The magnetic field produced by a short bar magnet on its axis $(B_{\text{axis}})$ and on its equatorial line $(B_{\text{equator}})$ at a distance $r$ from the centre is given by:
$B_{\text{axis}} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}$
$B_{\text{equator}} = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3}$
Thus,$B_{\text{axis}} = 2 \times B_{\text{equator}}$ for the same distance $r$.
Given:
$B_{\text{equator}} = 6.4 \times 10^{-5} \,T$ at $r_2 = 20 \,cm = 0.2 \,m$
We need to find $B_{\text{axis}}$ at $r_1 = 40 \,cm = 0.4 \,m$.
Using the general formula $B \propto \frac{1}{r^3}$:
$\frac{B_{\text{axis}}(r_1)}{B_{\text{equator}}(r_2)} = \frac{\frac{\mu_0}{4\pi} \cdot \frac{2M}{r_1^3}}{\frac{\mu_0}{4\pi} \cdot \frac{M}{r_2^3}} = 2 \times \left(\frac{r_2}{r_1}\right)^3$
$B_{\text{axis}} = 2 \times B_{\text{equator}} \times \left(\frac{20}{40}\right)^3$
$B_{\text{axis}} = 2 \times (6.4 \times 10^{-5}) \times \left(\frac{1}{2}\right)^3$
$B_{\text{axis}} = 2 \times (6.4 \times 10^{-5}) \times \frac{1}{8}$
$B_{\text{axis}} = \frac{6.4 \times 10^{-5}}{4} = 1.6 \times 10^{-5} \,T$

(C) Given: Dipole moment $M = 1.25 \ A-m^2$ and distance $r = 0.5 \ m$.
The magnetic field $B$ on the axial position of a short bar magnet is given by the formula:
$B = \frac{\mu_0}{4 \pi} \times \frac{2 M}{r^3}$
Substituting the values:
$B = 10^{-7} \times \frac{2 \times 1.25}{(0.5)^3}$
$B = 10^{-7} \times \frac{2.5}{0.125}$
$B = 10^{-7} \times 20 = 2.0 \times 10^{-6} \ T$
Thus,the magnetic field is $2.0 \times 10^{-6} \ T$.

(D) Let the magnetic moment of the weaker magnet be $M_1 = M$ and the stronger magnet be $M_2 = \sqrt{3} M$.
Since they are joined in a cross (+) shape,the angle between them is $90^{\circ}$.
Let the weaker magnet make an angle $\theta$ with the earth's magnetic field $B_H$.
Then the stronger magnet makes an angle $(90^{\circ} - \theta)$ with $B_H$.
In equilibrium,the net torque acting on the system due to the earth's magnetic field must be zero.
$\tau_1 + \tau_2 = 0$
$M_1 B_H \sin(\theta) = M_2 B_H \sin(90^{\circ} - \theta)$
$M \sin(\theta) = \sqrt{3} M \cos(\theta)$
$\tan(\theta) = \sqrt{3}$
$\theta = 60^{\circ}$.

(C) The magnetic field due to a short bar magnet at a point on its equatorial line is given by $B = \frac{\mu_0}{4 \pi} \frac{M}{r^3}$.
Here,the distance between the centers is $20 \text{ cm}$,so the distance of the midpoint from each magnet is $r = 10 \text{ cm} = 0.1 \text{ m}$.
Since the north poles point towards the geographic south,the magnetic fields produced by both magnets at the midpoint will be in the same direction as the Earth's horizontal magnetic field $(B_H)$.
Thus,the resultant magnetic field $B_{net} = B_1 + B_2 + B_H$.
$B_1 = \frac{10^{-7} \times 1.2}{(0.1)^3} = \frac{1.2 \times 10^{-7}}{10^{-3}} = 1.2 \times 10^{-4} \text{ T}$.
$B_2 = \frac{10^{-7} \times 1.0}{(0.1)^3} = \frac{1.0 \times 10^{-7}}{10^{-3}} = 1.0 \times 10^{-4} \text{ T}$.
$B_H = 3.6 \times 10^{-5} = 0.36 \times 10^{-4} \text{ T}$.
$B_{net} = (1.2 + 1.0 + 0.36) \times 10^{-4} \text{ T} = 2.56 \times 10^{-4} \text{ T}$.

(A) When a magnet is displaced by a very small angle $\theta$,the restoring torque acting on the magnet is given by $\tau = -M B_H \sin \theta$.
The negative sign indicates the restoring nature of the torque.
Since $\tau = I \alpha$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration,we have $I \alpha = -M B_H \sin \theta$.
For small angular displacements,$\sin \theta \approx \theta$.
Therefore,$I \alpha = -M B_H \theta$.
The magnitude of the angular acceleration is $\alpha = \frac{M B_H \theta}{I}$.