Magnetic Equipment's (Tangent galvanometer, Vibration magnetometer and Neutral point) Questions in English

(B) The principle of a tangent galvanometer is given by the relation $i = K \tan \theta$,where $K$ is the reduction factor of the galvanometer.
This implies $i \propto \tan \theta$.
Given: $i_1 = \sqrt{3} \text{ A}$,$\theta_1 = 30^{\circ}$ and $i_2 = 3 \text{ A}$.
Using the ratio formula: $\frac{i_1}{i_2} = \frac{\tan \theta_1}{\tan \theta_2}$.
Substituting the values: $\frac{\sqrt{3}}{3} = \frac{\tan 30^{\circ}}{\tan \theta_2}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{1/\sqrt{3}}{\tan \theta_2}$.
This simplifies to $\tan \theta_2 = 1$.
Therefore,$\theta_2 = 45^{\circ}$.

(B) The time period of a magnetic needle oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field strength.
For a needle of mass $m$ and length $l$,the moment of inertia is $I = \frac{ml^2}{12}$ and the magnetic moment is $M = q_m \cdot l$,where $q_m$ is the pole strength.
Substituting these,$T = 2\pi \sqrt{\frac{ml^2/12}{q_m \cdot l \cdot B}} = 2\pi \sqrt{\frac{ml}{12 q_m B}}$.
Since $T \propto \sqrt{ml}$,when the needle is cut in half,the new mass $m' = m/2$ and the new length $l' = l/2$. The pole strength $q_m$ remains the same.
Thus,$T' = T \sqrt{\frac{m'l'}{ml}} = T \sqrt{\frac{(m/2)(l/2)}{ml}} = T \sqrt{\frac{1}{4}} = \frac{T}{2}$.
Given $T = 1.0 \, s$,the new time period $T' = \frac{1.0}{2} = 0.5 \, s$.

(D) The time period of a freely suspended magnetic dipole is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field.
For a magnet of length $L$,mass $m$,and pole strength $q_m$,the magnetic moment is $M = q_m L$ and the moment of inertia is $I = \frac{mL^2}{12}$.
When the magnet is broken into two equal parts along its length,the new length is $L' = L/2$,the new mass is $m' = m/2$,and the new pole strength remains $q_m$.
The new magnetic moment is $M' = q_m (L/2) = M/2$.
The new moment of inertia is $I' = \frac{(m/2)(L/2)^2}{12} = \frac{mL^2}{8 \times 12} = I/8$.
The new time period $T'$ is $T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{I/8}{(M/2)B}} = 2\pi \sqrt{\frac{I}{4MB}} = \frac{1}{2} T$.
Given $T = 2 \ s$,the new time period is $T' = \frac{2}{2} = 1 \ s$.

(D) The period of oscillation $T$ for a vibration magnetometer is given by the formula:
$T = 2\pi \sqrt{\frac{I}{MH}}$
where:
$I$ = Moment of inertia of the magnet about the axis of suspension.
$M$ = Magnetic moment of the magnet.
$H$ = External magnetic field.
From the formula,it is clear that the period of oscillation $T$ depends on all three factors: $I$,$M$,and $H$.

(D) For a deflection magnetometer,the condition for no deflection (null point) is given by the balance of magnetic fields: $\frac{M_1}{d_1^3} = \frac{M_2}{d_2^3}$.
Given the ratio of magnetic moments is $\frac{M_1}{M_2} = \frac{27}{8}$.
Let $M_1$ be the stronger magnet and $M_2$ be the weaker magnet. Thus,$d_2 = 0.12 \ m$.
Substituting the values into the formula: $\frac{27}{8} = \left( \frac{d_1}{d_2} \right)^3$.
Taking the cube root on both sides: $\frac{3}{2} = \frac{d_1}{0.12}$.
Solving for $d_1$: $d_1 = \frac{3}{2} \times 0.12 = 1.5 \times 0.12 = 0.18 \ m$.

(B) The time period of oscillation of a magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{M B_H}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.
This implies $T \propto \frac{1}{\sqrt{B_H}}$,or $B_H \propto \frac{1}{T^2}$.
At the first place,the frequency is $40 \text{ oscillations/min}$,so the time period $T_1 = \frac{60}{40} = 1.5 \,s$. The magnetic field is $(B_H)_1 = 0.1 \times 10^{-5} \,T = 10^{-6} \,T$.
At the second place,the time period $T_2 = 2.5 \,s$.
Using the ratio $\frac{(B_H)_2}{(B_H)_1} = \left( \frac{T_1}{T_2} \right)^2$,we get:
$(B_H)_2 = (B_H)_1 \times \left( \frac{1.5}{2.5} \right)^2$
$(B_H)_2 = 10^{-6} \times \left( \frac{3}{5} \right)^2 = 10^{-6} \times \frac{9}{25} = 10^{-6} \times 0.36 = 0.36 \times 10^{-6} \,T$.

(C) In a tangent galvanometer,the current $i$ is given by $i = k \tan \phi$,where $k$ is the reduction factor and $\phi$ is the deflection.
Differentiating both sides with respect to $\phi$,we get $\frac{di}{d\phi} = k \sec^2 \phi$.
Thus,the fractional error is given by $\frac{di}{i} = \frac{k \sec^2 \phi \cdot d\phi}{k \tan \phi} = \frac{d\phi}{\sin \phi \cos \phi}$.
Multiplying the numerator and denominator by $2$,we get $\frac{di}{i} = \frac{2 d\phi}{2 \sin \phi \cos \phi} = \frac{2 d\phi}{\sin 2\phi}$.
For the error $\frac{di}{i}$ to be minimum,$\sin 2\phi$ must be maximum.
The maximum value of $\sin 2\phi$ is $1$,which occurs when $2\phi = 90^o$,or $\phi = 45^o$.

(A) The tangent galvanometer works on the tangent law,which states that $F = H \tan \theta$,where $F$ and $H$ are two uniform magnetic fields at right angles to each other.
One of these fields is the horizontal component of the Earth's magnetic field $(H)$,which acts along the magnetic meridian (North-South direction).
To perform the experiment,the coil of the tangent galvanometer must be placed in the magnetic meridian (vertically North-South) so that the magnetic field produced by the current in the coil is perpendicular to the horizontal component of the Earth's magnetic field.
This alignment ensures that the magnetic needle,which is initially in the magnetic meridian,deflects according to the tangent law when current flows through the coil.

(A) The time period of a bar magnet oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field.
When the magnet is cut into two equal parts perpendicular to its length,the new magnetic moment $M' = M/2$ and the new moment of inertia $I' = I/8$ (since $I = \frac{1}{12}mL^2$ and mass $m$ becomes $m/2$,length $L$ becomes $L/2$).
The new time period $T'$ is $T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{I/8}{(M/2)B}} = 2\pi \sqrt{\frac{I}{4MB}} = \frac{1}{2} \left( 2\pi \sqrt{\frac{I}{MB}} \right) = \frac{T}{2}$.

(C) The frequency of oscillation $f$ of a magnet in a vibration magnetometer is given by $f = \frac{1}{2\pi} \sqrt{\frac{\mu B_H}{I}}$,where $\mu$ is the magnetic moment,$B_H$ is the horizontal component of the earth's magnetic field,and $I$ is the moment of inertia.
Since $f \propto \sqrt{B_H}$,we have $\frac{f_B}{f_A} = \sqrt{\frac{(B_H)_B}{(B_H)_A}}$.
Given $f_A = 10 \text{ oscillations/min}$ and $f_B = 20 \text{ oscillations/min}$.
Substituting the values: $\frac{20}{10} = \sqrt{\frac{(B_H)_B}{36 \times 10^{-6} \ T}}$.
Squaring both sides: $2^2 = \frac{(B_H)_B}{36 \times 10^{-6} \ T}$.
$4 = \frac{(B_H)_B}{36 \times 10^{-6} \ T}$.
$(B_H)_B = 4 \times 36 \times 10^{-6} \ T = 144 \times 10^{-6} \ T$.

(D) The current $i$ in a tangent galvanometer is related to the deflection $\phi$ by the formula $i = K \tan \phi$,where $K$ is the reduction factor of the galvanometer.
Therefore,$i \propto \tan \phi$.
Given $i_1 = 2 \ A$ and $\phi_1 = 30^\circ$.
We need to find $i_2$ for $\phi_2 = 60^\circ$.
Using the ratio: $\frac{i_1}{i_2} = \frac{\tan \phi_1}{\tan \phi_2}$.
Substituting the values: $\frac{2}{i_2} = \frac{\tan 30^\circ}{\tan 60^\circ}$.
Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$ and $\tan 60^\circ = \sqrt{3}$,we have $\frac{2}{i_2} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}$.
Thus,$i_2 = 2 \times 3 = 6 \ A$.

(A) The correct statement is that the plane of the coil must be set parallel to the earth's magnetic meridian,not at right angles.
When the plane of the coil is parallel to the magnetic meridian,the magnetic field produced by the coil is perpendicular to the horizontal component of the earth's magnetic field $(B_H)$.
This satisfies the condition for the tangent law,$B = B_H \tan \theta$.
Therefore,the statement in option $A$ is false.

(C) The time period $T$ of a magnet oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field.
Since $M = m \times 2l$ (where $m$ is the pole strength and $2l$ is the length of the magnet),we have $T \propto \frac{1}{\sqrt{M}}$.
Since $M \propto m$,it follows that $T \propto \frac{1}{\sqrt{m}}$.
Given that the pole strength $m$ becomes $4$ times its original value $(m' = 4m)$,the new time period $T'$ is:
$T' = T \times \sqrt{\frac{m}{m'}} = 2 \times \sqrt{\frac{m}{4m}} = 2 \times \sqrt{\frac{1}{4}} = 2 \times \frac{1}{2} = 1 \, s$.

(D) In the deflection magnetometer (equal distance method),the magnetic moment $M$ is proportional to $\tan \theta$. Thus,$\frac{M_1}{M_2} = \frac{\tan \theta_1}{\tan \theta_2}$.
Since magnetic moment $M = m \times L$ (where $m$ is pole strength and $L$ is length),we have $\frac{m_1 L_1}{m_2 L_2} = \frac{\tan \theta_1}{\tan \theta_2}$.
Given $\frac{L_1}{L_2} = \frac{1}{2}$,$\theta_1 = 45^\circ$,and $\theta_2 = 30^\circ$.
Substituting the values: $\frac{m_1}{m_2} \times \frac{1}{2} = \frac{\tan 45^\circ}{\tan 30^\circ} = \frac{1}{1/\sqrt{3}} = \sqrt{3}$.
Therefore,$\frac{m_1}{m_2} = 2\sqrt{3} : 1$.

(B) In a tangent galvanometer,the magnetic field $B$ produced by the coil is related to the horizontal component of the Earth's magnetic field $B_H$ and the deflection angle $\theta$ by the formula:
$B = B_H \tan \theta$
Given:
$B_H = 0.34 \times 10^{-4} \, T$
$\theta = 30^\circ$
Substituting the values:
$B = (0.34 \times 10^{-4}) \times \tan 30^\circ$
Since $\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$:
$B = 0.34 \times 10^{-4} \times 0.577$
$B \approx 0.196 \times 10^{-4} \, T$
$B = 1.96 \times 10^{-5} \, T$

(B) The principle of a tangent galvanometer is given by the formula $I = K \tan \phi$,where $I$ is the current,$K$ is the reduction factor,and $\phi$ is the deflection angle.
Since $K$ is constant for a given galvanometer,we have $I \propto \tan \phi$.
Therefore,the ratio is $\frac{I_1}{I_2} = \frac{\tan \phi_1}{\tan \phi_2}$.
Given $I_1 = 0.1 \, A$,$\phi_1 = 30^\circ$,and $\phi_2 = 60^\circ$.
Substituting the values: $\frac{0.1}{I_2} = \frac{\tan 30^\circ}{\tan 60^\circ}$.
We know $\tan 30^\circ = \frac{1}{\sqrt{3}}$ and $\tan 60^\circ = \sqrt{3}$.
So,$\frac{0.1}{I_2} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}$.
Thus,$I_2 = 0.1 \times 3 = 0.3 \, A$.

(A) The time period of a bar magnet oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field.
For a bar magnet of mass $m$ and length $l$,the moment of inertia is $I = \frac{ml^2}{12}$.
Substituting this into the formula,we get $T = 2\pi \sqrt{\frac{ml^2}{12MB}}$.
This shows that $T \propto \sqrt{m}$.
If the mass $m$ is quadrupled $(m' = 4m)$,the new time period $T'$ becomes $T' = 2\pi \sqrt{\frac{4ml^2}{12MB}} = 2 \times (2\pi \sqrt{\frac{ml^2}{12MB}}) = 2T$.
The motion remains $S.H.M.$ because the restoring torque $\tau = -MB \sin \theta \approx -MB \theta$ for small oscillations.

(C) The time period of oscillation of a magnetic dipole in a uniform magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field.
For the original magnet of mass $m$ and length $l$,$I = \frac{ml^2}{12}$ and $M = m_s l$ (where $m_s$ is pole strength).
When the magnet is cut into two equal halves along its length,for each piece: mass $m' = \frac{m}{2}$,length $l' = \frac{l}{2}$,and pole strength $m_s' = m_s$.
New moment of inertia $I' = \frac{m' (l')^2}{12} = \frac{(m/2) (l/2)^2}{12} = \frac{1}{8} \left( \frac{ml^2}{12} \right) = \frac{I}{8}$.
New magnetic moment $M' = m_s l' = m_s (l/2) = \frac{M}{2}$.
The new time period $T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{I/8}{(M/2)B}} = 2\pi \sqrt{\frac{I}{4MB}} = \frac{1}{2} \left( 2\pi \sqrt{\frac{I}{MB}} \right) = \frac{T}{2}$.
Therefore,the ratio $\frac{T'}{T} = \frac{1}{2}$.

(B) The time period $T$ of an oscillating bar magnet in a magnetic field is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB_H}}$.
Here,$I$ is the moment of inertia of the bar magnet,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the earth's magnetic field.
For a rectangular bar magnet of mass $w$,length $L$,and breadth $b$,the moment of inertia is $I = \frac{w(L^2 + b^2)}{12}$.
Since $I \propto w$,we can write $T \propto \sqrt{I} \propto \sqrt{w}$.
If the mass $w$ is increased by a factor of $4$,the new time period $T'$ becomes $T' = T \sqrt{4} = 2T$.

(B) The time period of a magnet in a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.
Initially,$T = 2 \, s = 2\pi \sqrt{\frac{I}{MB}}$.
When the magnet is cut into three equal parts along its length,for each part,the mass $m' = m/3$ and length $l' = l/3$.
The moment of inertia of each part about the center is $I' = \frac{1}{12} m' (l')^2 = \frac{1}{12} (m/3) (l/3)^2 = \frac{1}{27} I$.
The magnetic moment of each part is $M' = M/3$.
When three such parts are placed on each other,the total moment of inertia $I_s = 3 \times I' = 3 \times (I/27) = I/9$.
The total magnetic moment $M_s = 3 \times M' = 3 \times (M/3) = M$.
The new time period $T_s = 2\pi \sqrt{\frac{I_s}{M_s B}} = 2\pi \sqrt{\frac{I/9}{MB}} = \frac{1}{3} \times 2\pi \sqrt{\frac{I}{MB}} = \frac{T}{3}$.
Substituting $T = 2 \, s$,we get $T_s = 2/3 \, s$.

(C) The time period of an oscillating magnet in a horizontal plane is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $B_H = B \cos \phi$ is the horizontal component of the Earth's magnetic field.
Thus,$T \propto \frac{1}{\sqrt{B \cos \phi}}$,which implies $B \propto \frac{1}{T^2 \cos \phi}$.
Given $T_1 = 2 \ s$,$\phi_1 = 30^o$ and $T_2 = 3 \ s$,$\phi_2 = 60^o$.
The ratio of the resultant magnetic fields is $\frac{B_1}{B_2} = \frac{T_2^2 \cos \phi_2}{T_1^2 \cos \phi_1}$.
Substituting the values: $\frac{B_1}{B_2} = \left( \frac{3}{2} \right)^2 \times \frac{\cos 60^o}{\cos 30^o} = \frac{9}{4} \times \frac{1/2}{\sqrt{3}/2} = \frac{9}{4\sqrt{3}}$.

(C) The time period of a single bar magnet of magnetic moment $M$ and moment of inertia $I$ in a magnetic field $H$ is given by $T' = 2\pi \sqrt{\frac{I}{MH}}$.
For the combination of two identical magnets placed perpendicularly,the total moment of inertia is $I_{total} = I + I = 2I$.
The resultant magnetic moment is $M_{res} = \sqrt{M^2 + M^2} = \sqrt{2}M$.
The time period of the combination is $T = 2\pi \sqrt{\frac{I_{total}}{M_{res}H}} = 2\pi \sqrt{\frac{2I}{\sqrt{2}MH}} = 2\pi \sqrt{\frac{\sqrt{2}I}{MH}}$.
We can write this as $T = 2^{1/4} \times 2\pi \sqrt{\frac{I}{MH}} = 2^{1/4} T'$.
Therefore,$T' = \frac{T}{2^{1/4}} = 2^{-1/4} T$.

(B) The principle of a Tangent galvanometer is given by the formula: $B = B_H \tan \theta$,where $B$ is the magnetic field produced by the coil.
Substituting the formula for the magnetic field at the center of a circular coil: $\frac{\mu_0 n i}{2r} = B_H \tan \theta$.
Rearranging for current $i$: $i = \frac{2r B_H \tan \theta}{\mu_0 n}$.
Given values: $r = 0.1\,m$,$n = 10$,$B_H = 4 \times 10^{-5}\,T$,$\theta = 60^{\circ}$,and $\mu_0 = 4\pi \times 10^{-7}\,T\cdot m/A$.
Substituting these values: $i = \frac{2 \times 0.1 \times 4 \times 10^{-5} \times \tan(60^{\circ})}{10 \times 4\pi \times 10^{-7}}$.
Since $\tan(60^{\circ}) = \sqrt{3} \approx 1.732$,we have $i = \frac{0.8 \times 10^{-5} \times 1.732}{4\pi \times 10^{-6}} = \frac{0.8 \times 1.732}{4\pi \times 0.1} \approx 1.1\,A$.

(D) Initially,the neutral point is obtained on the equatorial line of the magnet. At the neutral point,the magnitude of the horizontal component of the Earth's magnetic field $(B_H)$ is equal to the magnetic field due to the bar magnet on its equatorial line $(B_e)$. Thus,$|B_H| = |B_e|$.
When the magnet is rotated by $90^o$,the point $P$ now lies on the axial line of the magnet. The magnetic field due to the bar magnet on its axial line is $B_a = 2 B_e$. Since $B_e = B_H$,we have $B_a = 2 B_H$.
The Earth's horizontal magnetic field $B_H$ remains in the same direction. The net magnetic field $B$ at point $P$ is the vector sum of $B_a$ and $B_H$,which are perpendicular to each other.
Therefore,the net magnetic induction is $B = \sqrt{B_a^2 + B_H^2} = \sqrt{(2 B_H)^2 + B_H^2} = \sqrt{4 B_H^2 + B_H^2} = \sqrt{5} B_H$.

(C) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$.
Initially,two identical magnets of magnetic moment $M$ and moment of inertia $I$ are placed perpendicular to each other.
The resultant magnetic moment is $M_1 = \sqrt{M^2 + M^2} = M\sqrt{2}$.
The total moment of inertia is $I_1 = I + I = 2I$.
The initial time period is $T_1 = 2\pi \sqrt{\frac{2I}{M\sqrt{2} B_H}} = 2\pi \sqrt{\frac{\sqrt{2}I}{MB_H}} = 2^{1/4} \cdot 2\pi \sqrt{\frac{I}{MB_H}}$.
Given $T_1 = 2^{5/4} \ s$,we have $2^{5/4} = 2^{1/4} \cdot T_0$,where $T_0 = 2\pi \sqrt{\frac{I}{MB_H}}$ is the time period of a single magnet.
Thus,$T_0 = \frac{2^{5/4}}{2^{1/4}} = 2^1 = 2 \ s$.
When one magnet is removed,the time period of the remaining magnet is $T_2 = T_0 = 2 \ s$.

(B) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MH}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $H$ is the magnetic field.
Thus,$T \propto \frac{1}{\sqrt{B_{net}}}$,where $B_{net}$ is the net magnetic field.
Initially,$T_1 = 2 \ s$ and $B_{net1} = H$.
When the magnet is brought near and parallel,the net field becomes $B_{net2} = H + F$,and the time period is $T_2 = 1 \ s$.
Taking the ratio: $\frac{T_1}{T_2} = \sqrt{\frac{H+F}{H}}$.
Substituting the values: $\frac{2}{1} = \sqrt{1 + \frac{F}{H}}$.
Squaring both sides: $4 = 1 + \frac{F}{H} \Rightarrow \frac{F}{H} = 3$.
Therefore,the ratio $\frac{H}{F} = \frac{1}{3}$.

(C) Let $M_1$ and $M_2$ be the magnetic moments of the magnets and $H$ be the horizontal component of the Earth's magnetic field.
For a magnet in equilibrium,the restoring torque of the wire equals the magnetic torque: $\tau = C\phi = MH \sin \theta$,where $C$ is the restoring couple per unit twist,$\phi$ is the twist angle of the wire,and $\theta$ is the deflection angle.
For the first magnet: $\phi_1 = 180^o - 30^o = 150^o$. Thus,$C(150^o) = M_1 H \sin 30^o$.
For the second magnet: $\phi_2 = 270^o - 30^o = 240^o$. Thus,$C(240^o) = M_2 H \sin 30^o$.
Dividing the two equations: $\frac{M_1}{M_2} = \frac{150^o}{240^o} = \frac{15}{24} = \frac{5}{8}$.
Therefore,the ratio of the magnetic moments is $5:8$.

(C) In the vertical plane perpendicular to the magnetic meridian,the effective magnetic field acting on the needle is the vertical component of the Earth's magnetic field,$B_V$. The time period is given by $T_1 = 2\pi \sqrt{\frac{I}{MB_V}} = 2 \text{ s}$.
In the horizontal plane,the effective magnetic field acting on the needle is the horizontal component of the Earth's magnetic field,$B_H$. The time period is given by $T_2 = 2\pi \sqrt{\frac{I}{MB_H}} = 2 \text{ s}$.
Since $T_1 = T_2$,we have $B_V = B_H$.
The angle of dip $\phi$ is defined by the relation $\tan \phi = \frac{B_V}{B_H}$.
Substituting $B_V = B_H$,we get $\tan \phi = 1$,which implies $\phi = 45^o$.

(C) The time period of an oscillating magnet is given by $T = 2\pi \sqrt{\frac{I}{m B_H}}$,where $I$ is the moment of inertia,$m$ is the magnetic moment,and $B_H$ is the horizontal component of the Earth's magnetic field.
Initially,$T = 0.1 \, s$ and $B_H = 24 \, \mu T$.
The magnetic field $B$ produced by a long vertical wire carrying current $i$ at a distance $a$ is $B = \frac{\mu_0}{4\pi} \cdot \frac{2i}{a}$.
Given $i = 18 \, A$ and $a = 20 \, cm = 0.2 \, m$,we have $B = 10^{-7} \cdot \frac{2 \times 18}{0.2} = 18 \times 10^{-6} \, T = 18 \, \mu T$.
Since the wire is to the east and the current is downward,by the right-hand rule,the magnetic field $B$ at the magnet's position points north,in the same direction as $B_H$.
Thus,the new total magnetic field is $B_{net} = B_H + B = 24 \, \mu T + 18 \, \mu T = 42 \, \mu T$.
The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{I}{m(B_H + B)}}$.
Taking the ratio: $\frac{T'}{T} = \sqrt{\frac{B_H}{B_H + B}} = \sqrt{\frac{24}{24 + 18}} = \sqrt{\frac{24}{42}} = \sqrt{\frac{4}{7}} \approx 0.7559$.
Therefore,$T' = 0.1 \times 0.7559 \approx 0.0756 \, s \approx 0.076 \, s$.

(B) In a tangent galvanometer,the current $i$ is related to the deflection $\theta$ by the formula: $i = K \tan \theta$,where $K$ is the reduction factor of the galvanometer.
This implies that $\theta = \arctan(i/K)$.
As $i$ increases,$\tan \theta$ increases. For small values of $\theta$,$\tan \theta \approx \theta$,making the graph nearly linear. As $\theta$ approaches $\pi/2$,$\tan \theta$ approaches infinity,meaning the slope of the $\theta$ vs $i$ curve must decrease as $\theta$ increases (the curve bends towards the $i$-axis).
Looking at the provided graph,curve $b$ shows this characteristic behavior where the rate of increase of $\theta$ with respect to $i$ decreases as $\theta$ increases,correctly representing the inverse tangent function shape.
Therefore,the correct option is $b$.

(C) The neutral point for a magnet placed with its south pole pointing towards the geographic north lies on its axial line. The magnetic field due to the magnet at an axial point is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{2Mr}{(r^2 - l^2)^2}$. For a short magnet,this simplifies to $B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}$.
At the neutral point,the magnetic field of the magnet equals the horizontal component of the Earth's magnetic field $(B_H)$:
$\frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3} = B_H$
Given: $r = 20 \, cm = 0.2 \, m$,$B_H = 0.3 \times 10^{-4} \, T$,and $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$.
Substituting the values:
$10^{-7} \cdot \frac{2M}{(0.2)^3} = 0.3 \times 10^{-4}$
$10^{-7} \cdot \frac{2M}{0.008} = 0.3 \times 10^{-4}$
$2M = 0.3 \times 10^{-4} \times 0.008 \times 10^7$
$2M = 0.3 \times 8 = 2.4$
$M = 1.2 \, A \cdot m^2$.
Since $1 \, A \cdot m^2 = 1000 \, ab-amp \cdot cm^2$,we have:
$M = 1.2 \times 1000 = 1200 \, ab-amp \cdot cm^2$.

(D) The time period of oscillation for a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MH}}$.
For the sum of magnetic moments $(M_1 + M_2)$,the time period is $T_s = \frac{60}{12} = 5 \ s$.
For the difference of magnetic moments $(M_1 - M_2)$,the time period is $T_d = \frac{60}{4} = 15 \ s$.
The ratio of magnetic moments is given by $\frac{M_1}{M_2} = \frac{T_d^2 + T_s^2}{T_d^2 - T_s^2}$.
Substituting the values: $\frac{M_1}{M_2} = \frac{15^2 + 5^2}{15^2 - 5^2} = \frac{225 + 25}{225 - 25} = \frac{250}{200} = \frac{5}{4}$.

(B) The principle of a tangent galvanometer is given by the formula: $B = B_H \tan \theta$,where $B$ is the magnetic field produced by the coil,$B_H$ is the horizontal component of the Earth's magnetic field,and $\theta$ is the deflection angle.
Given:
$B_H = 0.34 \times 10^{-4} \, T$
$\theta = 30^{\circ}$
Substituting the values:
$B = (0.34 \times 10^{-4}) \times \tan(30^{\circ})$
Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}} \approx 0.57735$
$B = 0.34 \times 10^{-4} \times 0.57735$
$B \approx 0.1963 \times 10^{-4} \, T$
$B \approx 1.963 \times 10^{-5} \, T$
Thus,the magnetic field of the coil is $1.96 \times 10^{-5} \, T$.

(B) The time period $T$ of a freely suspended magnet in a magnetic field $B_H$ is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.
From the formula,$T \propto \frac{1}{\sqrt{M}}$.
Let the initial magnetic moment be $M_1 = 100$ and the final magnetic moment be $M_2 = 100 - 19 = 81$.
The ratio of time periods is $\frac{T_2}{T_1} = \sqrt{\frac{M_1}{M_2}} = \sqrt{\frac{100}{81}} = \frac{10}{9}$.
Calculating the percentage change: $\frac{T_2 - T_1}{T_1} \times 100 = (\frac{10}{9} - 1) \times 100 = \frac{1}{9} \times 100 \approx 11.11\%$.
Thus,the time period increases by approximately $11\%$.

(B) The time period of a magnetic needle oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $B_H = B \cos \phi$ is the horizontal component of the Earth's magnetic field.
Thus,$T = 2\pi \sqrt{\frac{I}{MB \cos \phi}}$,which implies $T \propto \frac{1}{\sqrt{B \cos \phi}}$.
Given $n_1 = 20$ oscillations in $60 \ s$,so $T_1 = \frac{60}{20} = 3 \ s$ at $\phi_1 = 30^{\circ}$.
Given $n_2 = 15$ oscillations in $60 \ s$,so $T_2 = \frac{60}{15} = 4 \ s$ at $\phi_2 = 60^{\circ}$.
Taking the ratio: $\frac{T_1}{T_2} = \sqrt{\frac{B_2 \cos \phi_2}{B_1 \cos \phi_1}} \Rightarrow \frac{3}{4} = \sqrt{\frac{B_2 \cos 60^{\circ}}{B_1 \cos 30^{\circ}}}$.
Squaring both sides: $\frac{9}{16} = \frac{B_2}{B_1} \times \frac{1/2}{\sqrt{3}/2} = \frac{B_2}{B_1} \times \frac{1}{\sqrt{3}}$.
Therefore,$\frac{B_1}{B_2} = \frac{16}{9\sqrt{3}}$.

(B) The time period of an oscillating magnet in a magnetic field $B$ is given by $T = 2\pi \sqrt{\frac{I}{MB}}$.
Initially,the magnetic field is $H$,so $T = 2\pi \sqrt{\frac{I}{MH}} = 2 \, s$.
When an external field $F$ is applied in the same direction,the net magnetic field becomes $B_{net} = H + F$.
The new time period is $T' = 2\pi \sqrt{\frac{I}{M(H+F)}} = 1 \, s$.
Taking the ratio: $\frac{T}{T'} = \sqrt{\frac{H+F}{H}} = \sqrt{1 + \frac{F}{H}}$.
Substituting the values: $\frac{2}{1} = \sqrt{1 + \frac{F}{H}} \Rightarrow 4 = 1 + \frac{F}{H} \Rightarrow \frac{F}{H} = 3$.
Therefore,the ratio $\frac{H}{F} = \frac{1}{3}$.

(D) The time period of a magnet oscillating in a horizontal magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the Earth's magnetic field.
From the formula,we see that $T \propto \sqrt{I}$.
Initially,the moment of inertia is $I$. When a wooden block of the same moment of inertia $I$ is attached to the magnet,the total moment of inertia of the system becomes $I' = I + I = 2I$.
The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{I'}{MB_H}} = 2\pi \sqrt{\frac{2I}{MB_H}}$.
Therefore,$T' = \sqrt{2} \times (2\pi \sqrt{\frac{I}{MB_H}}) = \sqrt{2}T$.

(C) For a short bar magnet in $\tan A$ position,the magnetic field $B$ is given by $B = \frac{\mu_0}{4\pi} \frac{2M}{d^3}$.
In a deflection galvanometer,$B = H \tan \theta$,where $H$ is the horizontal component of Earth's magnetic field.
Thus,$\frac{\mu_0}{4\pi} \frac{2M}{d^3} = H \tan \theta$ ..... $(i)$
When the distance is doubled $(d' = 2d)$,the new deflection $\theta'$ is given by:
$\frac{\mu_0}{4\pi} \frac{2M}{(2d)^3} = H \tan \theta'$ ..... $(ii)$
Dividing equation $(ii)$ by $(i)$:
$\frac{\tan \theta'}{\tan \theta} = \frac{1}{2^3} = \frac{1}{8}$
$\tan \theta' = \frac{\tan 60^\circ}{8} = \frac{\sqrt{3}}{8}$
$\theta' = \tan^{-1}\left(\frac{\sqrt{3}}{8}\right)$

(C) The formula for a tangent galvanometer is $I = K \tan \theta$,where $K = \frac{2r B_H}{\mu_0 n}$.
Given: $I = 10 \, mA = 10 \times 10^{-3} \, A$,$\theta = 45^{\circ}$,$B_H = 3.6 \times 10^{-5} \, T$,$r = 10 \, cm = 0.1 \, m$.
Since $\tan 45^{\circ} = 1$,we have $I = K$,so $n = \frac{2r B_H}{\mu_0 I}$.
Substituting the values: $n = \frac{2 \times 0.1 \times 3.6 \times 10^{-5}}{4\pi \times 10^{-7} \times 10 \times 10^{-3}}$.
$n = \frac{0.72 \times 10^{-5}}{4 \times 3.14159 \times 10^{-10}} = \frac{0.72 \times 10^5}{12.566} \approx 5729.7 / 10 = 573$ (approx $570$ based on standard approximation $\pi \approx 3.14$ and $4\pi \approx 12.56$).
Using $\pi = 3.14$,$n = \frac{0.72 \times 10^{-5}}{12.56 \times 10^{-10}} = \frac{72000}{125.6} \approx 573$. Given the options,$570$ is the correct choice.

(B) The time period $T$ of a vibration magnetometer is given by the formula $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the Earth's magnetic field.
From the formula,we see that $T \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the time periods is $\frac{T_1}{T_2} = \sqrt{\frac{M_2}{M_1}}$.
Given that the magnetic moment is reduced by $36\%$,if the initial magnetic moment $M_1 = 100$,then the final magnetic moment $M_2 = 100 - 36 = 64$.
Substituting these values,we get $\frac{T_1}{T_2} = \sqrt{\frac{64}{100}} = \frac{8}{10} = 0.8$.
This implies $T_2 = \frac{T_1}{0.8} = 1.25 T_1$.
The percentage increase in the time period is $\frac{T_2 - T_1}{T_1} \times 100\% = (1.25 - 1) \times 100\% = 25\%$.

(A) Let the magnetic moments be $M_1$ and $M_2$ such that $\frac{M_1}{M_2} = \frac{13}{5}$.
When like poles are together,the effective magnetic moment is $M_s = M_1 + M_2$ and the frequency is $\nu_s = 15 \text{ oscillations/min}$.
When unlike poles are together,the effective magnetic moment is $M_d = M_1 - M_2$ and the frequency is $\nu_d$.
The frequency of a vibration magnetometer is given by $\nu \propto \sqrt{M}$.
Therefore,$\frac{\nu_s}{\nu_d} = \sqrt{\frac{M_1 + M_2}{M_1 - M_2}}$.
Squaring both sides: $\frac{\nu_s^2}{\nu_d^2} = \frac{M_1 + M_2}{M_1 - M_2} = \frac{13 + 5}{13 - 5} = \frac{18}{8} = \frac{9}{4}$.
Taking the square root: $\frac{\nu_s}{\nu_d} = \frac{3}{2}$.
Given $\nu_s = 15$,we have $\frac{15}{\nu_d} = \frac{3}{2}$.
Solving for $\nu_d$: $\nu_d = \frac{15 \times 2}{3} = 10 \text{ oscillations/min}$.

(D) The time period of a magnetic dipole oscillating in a uniform magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the earth's magnetic field.
Initially,the moment of inertia is $I$. So,$T = 2\pi \sqrt{\frac{I}{MB_H}}$.
When a piece of wood with the same moment of inertia $I$ is attached to the magnet,the new moment of inertia of the system becomes $I' = I + I = 2I$.
The magnetic moment $M$ remains unchanged.
The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{I'}{MB_H}} = 2\pi \sqrt{\frac{2I}{MB_H}}$.
Comparing this with the initial time period,we get $T' = \sqrt{2} \times (2\pi \sqrt{\frac{I}{MB_H}}) = \sqrt{2}T$.

(C) The time period of a magnet in a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MH}}$.
When a magnet of length $L$,mass $m$,and magnetic moment $M$ is cut into $6$ equal parts along its length,each part has a mass $m' = m/6$,length $l' = L/6$,and magnetic moment $M' = M/6$.
The moment of inertia of each part about the center of the original magnet is $I' = \frac{1}{12} m' (l')^2 + m' d^2$. However,considering the standard approximation for such problems where the moment of inertia of each piece is $I_{part} = \frac{I}{6^3}$,the total moment of inertia of the system is $I_{net} = 6 \times \frac{I}{216} = \frac{I}{36}$.
From the figure,there are $2$ magnets oriented in one direction and $4$ magnets in the opposite direction. The net magnetic moment is $M_{net} = \frac{2M}{6} - \frac{4M}{6} = -\frac{2M}{6} = -\frac{M}{3}$. The magnitude is $|M_{net}| = \frac{M}{3}$.
The new time period $T'$ is $T' = 2\pi \sqrt{\frac{I_{net}}{|M_{net}|H}} = 2\pi \sqrt{\frac{I/36}{(M/3)H}} = 2\pi \sqrt{\frac{I}{12MH}} = \frac{1}{\sqrt{12}} \times 2\pi \sqrt{\frac{I}{MH}} = \frac{T}{2\sqrt{3}}$.

(A) Given:
Magnetic moment,$M = 6.7 \times 10^{-2} \ Am^2$
Magnetic field,$B = 0.01 \ T$
Moment of inertia,$I = 7.5 \times 10^{-6} \ kgm^2$
The time period $T$ of a magnetic needle performing simple harmonic oscillations is given by:
$T = 2\pi \sqrt{\frac{I}{MB}}$
Substituting the values:
$T = 2\pi \sqrt{\frac{7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times 0.01}}$
$T = 2\pi \sqrt{\frac{7.5 \times 10^{-6}}{6.7 \times 10^{-4}}}$
$T = 2\pi \sqrt{1.1194 \times 10^{-2}}$
$T = 2\pi \times 0.1058 \approx 0.665 \ s$
Time taken for $10$ complete oscillations is:
$t = 10 \times T = 10 \times 0.665 = 6.65 \ s$

(B) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal magnetic field.
For the combination of two identical magnets (each with moment of inertia $I_0$ and magnetic moment $M_0$) placed perpendicularly and bisecting each other:
Total moment of inertia $I_1 = I_0 + I_0 = 2I_0$.
Resultant magnetic moment $M_1 = \sqrt{M_0^2 + M_0^2} = M_0\sqrt{2}$.
Given $T_1 = 4 \, s$,so $4 = 2\pi \sqrt{\frac{2I_0}{M_0\sqrt{2} B_H}} = 2\pi \sqrt{\frac{I_0\sqrt{2}}{M_0 B_H}}$.
When one magnet is removed:
New moment of inertia $I_2 = I_0$.
New magnetic moment $M_2 = M_0$.
New time period $T_2 = 2\pi \sqrt{\frac{I_0}{M_0 B_H}}$.
Taking the ratio:
$\frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{I_0\sqrt{2}}{M_0 B_H}}}{2\pi \sqrt{\frac{I_0}{M_0 B_H}}} = \sqrt{\sqrt{2}} = 2^{1/4}$.
Therefore,$T_2 = \frac{T_1}{2^{1/4}} = \frac{4}{2^{1/4}} = \frac{2^2}{2^{1/4}} = 2^{2 - 1/4} = 2^{7/4} \, s$.

(B) The period of oscillation of a magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is the magnetic dipole moment.
For the original magnet of mass $m$ and length $\ell$,$I = \frac{m\ell^2}{12}$ and $M = q_m \ell$ (where $q_m$ is pole strength).
When the magnet is cut into two equal halves along its length,the new length is $\ell' = \ell/2$ and the new mass is $m' = m/2$.
The new pole strength $q_m$ remains the same.
New magnetic moment $M' = q_m \ell' = q_m (\ell/2) = M/2$.
New moment of inertia $I' = \frac{m'(\ell')^2}{12} = \frac{(m/2)(\ell/2)^2}{12} = \frac{m\ell^2}{12 \times 8} = I/8$.
The new period of oscillation is $T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{I/8}{(M/2)B}} = 2\pi \sqrt{\frac{1}{4} \cdot \frac{I}{MB}} = \frac{1}{2} T$.
Therefore,the ratio $T'/T = 1/2$.

(A) The time period of oscillation for a combination of two magnets in a magnetic field $B_{H}$ is given by $T = 2\pi \sqrt{\frac{I_{total}}{M_{total} B_{H}}}$.
Case $1$: When similar poles are on the same side,the effective magnetic moment is $M_{eff} = M + 2M = 3M$. The moment of inertia is $I = I_{1} + I_{2}$.
Thus,$T_{1} = 2\pi \sqrt{\frac{I_{1} + I_{2}}{3M B_{H}}}$.
Case $2$: When the polarity of one magnet is reversed,the effective magnetic moment is $M_{eff} = |2M - M| = M$. The moment of inertia remains $I = I_{1} + I_{2}$.
Thus,$T_{2} = 2\pi \sqrt{\frac{I_{1} + I_{2}}{M B_{H}}}$.
Comparing $T_{1}$ and $T_{2}$,since the denominator in $T_{2}$ is smaller than in $T_{1}$,we have $T_{1} < T_{2}$.