Magnetism and Matter

The magnetic potential energy $ U $ of a magnetic dipole moment $ \vec{m} $ in a magnetic field $ \vec{B} $ is given by the formula:

$ U = -\vec{m} \cdot \vec{B} $

Here, the dot product $ \vec{m} \cdot \vec{B} $ represents the scalar product of the vectors. When the magnetic moment $ \vec{m} $ is placed perpendicular to the magnetic field $ \vec{B} $, the angle $ \theta $ between these two vectors is 90 degrees. The dot product in this case can be written as:

$ \vec{m} \cdot \vec{B} = m B \cos \theta $

Since $ \theta = 90^\circ $, we have $ \cos 90^\circ = 0 $. Therefore, the magnetic potential energy $ U $ becomes:

$ U = -m B \cos 90^\circ $

$ U = -m B \cdot 0 $

$ U = 0 $

So, the magnetic potential energy when a magnetic bar of magnetic moment $ \vec{m} $ is placed perpendicular to the magnetic field $ \vec{B} $ is zero.

The correct answer is: Option B "Zero".

For diamagnetic material,

$\begin{aligned} & 0 \leq \mu_r<1 \\ & \chi=\mu_r-1 \\ & \Rightarrow-1 \leq \chi<0 \\ & \mu_r=\frac{\mu}{\mu_0} \\ & \Rightarrow \mu<\mu_0 \quad\left(\because \mu_r<1\right) \end{aligned}$

Time period of oscillation of magnet inside the magnetic field $T=2 \pi \sqrt{\frac{I}{M B}}$

$\begin{aligned} & T=\frac{t}{n}=\frac{10}{10}=1 \mathrm{~s} \\ & 1=2 \pi \sqrt{\frac{10^{-6}}{\pi^2 \times 1.0 \times 10^{-2} \times B}} \\ & \frac{1}{4}=\frac{10^{-4}}{B} \Rightarrow B=0.4 \mathrm{~mT} \end{aligned}$

$M=m \cdot L$

$\begin{aligned} & R \theta=L \\ & \frac{R \pi}{3}=L \\ & R=\frac{3 L}{\pi} \\ & M^{\prime}=m(2 R) \sin 30^{\circ} \\ & =m(2) \frac{3 L}{\pi} \times \frac{1}{2}=\frac{3}{\pi} m L=\frac{3 M}{\pi} \end{aligned}$

Time period of Oscillation, $T=2 \pi \sqrt{\frac{I}{M B}}$

$\begin{aligned} & \Rightarrow \frac{1}{4}=2 \pi \sqrt{\frac{9.8 \times 10^{-6}}{M \times 0.049}} \\ & \Rightarrow \quad \frac{1}{16}=4 \pi^2 \times \frac{9.8 \times 10^{-6}}{M \times 49 \times 10^{-3}} \\ & \Rightarrow \quad M=\frac{4 \pi^2 \times 9.8 \times 10^{-6}}{49 \times 10^{-3}} \times 16 \\ & \quad=\frac{4 \pi^2 \times 9.8 \times 16 \times 10^{-3}}{49} \\ & \quad=12.8 \pi^2 \times 10^{-3} \times 10^{-2} \times 10^2 \\ & =1280 \pi^2 \times 10^{-5} \mathrm{Am}^2 \end{aligned}$

North of magnet is moving away from solenoid 1 so end $B$ of solenoid 1 is South and as south of magnet is approaching solenoid 2 so end $C$ of solenoid 2 is South.

To match the magnetic materials with their respective magnetic susceptibilities, we must understand the characteristics of each type of material regarding its magnetic behavior. Here are the implications of each type:

Diamagnetic Materials: These materials are repelled by magnetic fields. The susceptibility ($\chi$) of diamagnetic materials is negative but very small, near zero. Therefore, $\chi > -1$ and closer to zero, reflects this property.

Ferromagnetic Materials: They have a very high, positive susceptibility because these materials can be permanently magnetized. Their susceptibility ($\chi$) is much greater than one ($\chi >> 1$).

Paramagnetic Materials: These materials are weakly attracted by a magnetic field, and their susceptibility is positive. However, the value of $\chi$ for paramagnetic materials is small but greater than zero ($0 < \chi < \varepsilon$), where $\varepsilon$ represents a small positive value.

Non-magnetic Materials: For materials that are considered non-magnetic, the susceptibility ($\chi$) is zero ($\chi = 0$), as they are neither attracted nor repelled by magnetic fields.

Let's match these descriptions with the given List II in your question:

• Diamagnetic (A) matches with $0 > \chi \ge -1$ (II).
• Ferromagnetic (B) corresponds to $\chi >> 1$ (III).
• Paramagnetic (C) should be aligned with $0 < \chi < \varepsilon$ (IV).
• Non-magnetic (D) clearly fits $\chi = 0$ (I).

Therefore, the correct answer matches:

• A-II, B-III, C-IV, D-I.

This aligns with Option A.

To understand which statement(s) is/are correct, we must explore the interactions between the magnetic field and different types of materials (magnetic, non-magnetic, conducting, non-conducting).

Statement A: If a sheet is magnetic and placed near a strong magnetic pole, it will experience a magnetic force due to the magnetic field. This force could either attract or repel the sheet depending on the polarity of the magnetic pole and the induced or inherent poles in the sheet. A force is needed to hold the sheet stationary against this magnetic force. Hence, statement A is true.

Statement B: If the sheet is non-magnetic, it will not experience any magnetic force because non-magnetic materials do not respond to magnetic fields in a manner where a force would be exerted on them. Thus, no external force is needed to hold a non-magnetic sheet in place near a magnetic pole. Statement B is false.

Statement C: If the sheet is conducting and you move it away from the magnetic pole, it will experience a change in the magnetic flux through it. According to Faraday's Law of Electromagnetic Induction, a change in magnetic flux induces an electromotive force (EMF) and consequently, currents known as eddy currents in the conductor. These induced currents produce their own magnetic fields, which interact with the original magnetic field, leading to a force (Lenz's Law). To move the sheet away with uniform velocity, a force must counteract this magnetic interaction. Therefore, statement C is true.

Statement D: If the sheet is non-conducting and non-polar, it will neither induce currents (since it's non-conducting) nor experience magnetic forces (since it's non-magnetic). Thus, moving such a sheet away from a magnetic pole with uniform velocity does not require overcoming any electromagnetic forces. Statement D is false.

Given these explanations, the correct statements are A and C only.

The correct choice is: Option B: A and C only.

$ \begin{aligned} & \Delta l=2 \frac{I}{2} \sin 30^{\circ} \\ & =\frac{I}{2} \\ & M^{\prime}=m I / 2 \\ & =M / 2 \end{aligned} $

By magnetic property

$\chi \propto \frac{1}{T}$

$\chi$ vs $\frac{1}{T}$ graph will be straight line.

${\phi _B} = NBA\cos \omega t$

$\varepsilon = {{ - d{\phi _B}} \over {dt}} = - NBA\omega ( - \sin \omega t)$

$\varepsilon = NBA\omega \sin \omega t$

${i_{\max }} = {{{\varepsilon _{\max }}} \over R} - {{NBA\omega } \over R}$

$ = {{1000 \times 2 \times {{10}^{ - 5}} \times \pi {{(10)}^2} \times 2} \over {12.56}}$

= 1 A

Charge on $\alpha$-particle, $q=2 e$

Time period of $\alpha$-particle

$T=\frac{2 \pi r}{v}$

$\therefore$ Current associated by $\alpha$-particle,

$I=\frac{q}{T}=\frac{2 e}{\frac{2 \pi r}{v}} \Rightarrow I=\frac{e v}{\pi r}$ ..... (i)

Magnetic dipole moment,

$\begin{aligned} & \mu=I \cdot A=\frac{e v}{\pi r} \cdot \pi r^2 \\ & \mu=e v r \end{aligned}$

When paramagnetic substances are kept in a external magnetic field, then it develops feeble magnetisation in the direction of external applied magnetic field i.e., magnetic dipoles are aligned along external magnetic field weakly. Hence, both assertion and reason are true and reason is the correct explanation of assertion.

Vertical component of earth’s magnetic field,

$\begin{aligned} B_V & =B_H \tan \delta \\ & =0.36 \times 10^{-4} \times \tan 60^{\circ} \\ & =0.36 \times 10^{-4} \times \sqrt{3} \\ & =0.622 \times 10^{-4} \mathrm{~T} \\ & =0.622 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^2 \end{aligned}$

$T=2 \pi \sqrt{\frac{I}{M B}}$

or $T^{\prime}=2 \pi \sqrt{\frac{I}{m V^{\prime}}}$

$B=$ Total earth's field intensity

$V=$ Vertical component of $B$

$\begin{array}{rlrl} \frac{T^{\prime}}{T} & =\sqrt{\frac{B}{V}} \\ \Rightarrow \quad & \frac{\left(T^{\prime}\right)^2}{T^2} =\frac{B}{V} \\ & \text { but } \quad V=B \sin \theta \\ \Rightarrow \quad & \frac{\left(T^{\prime}\right)^2}{T^2} =\frac{1}{\sin \theta} \\ \Rightarrow \quad & \sin \theta =\frac{T^2}{\left(T^{\prime}\right)^2} \\ T & =\frac{60}{40}=\frac{3}{2} \mathrm{~s} \end{array}$

Also, and $T^{\prime}=\frac{60}{30}=2 \mathrm{~s}$

$\begin{aligned} & \therefore \quad \sin \theta=0.5625 \\ & \Rightarrow \quad \theta=\sin ^{-1}(0.5625) \\ \end{aligned}$

We know the magnetism of the magnet is achieved due to the spin motion of electrons. Also the spinning electrons possess magnetic dipole moment. This dipole moment is much greater than that due to orbital moment of electrons around the nucleus.