Magnetism And Matter – JEE Main PhysicsPractice Questions & PYQs

Given $B = 800 \text{ Gauss} = 0.08 \text{ T}$ and $\tau = 0.016 \text{ N}\cdot\text{m}$ at $\theta = 30^\circ$.
The torque formula is $\tau = \mu B \sin 30^\circ$, so $\mu B = \frac{\tau}{\sin 30^\circ} = \frac{0.016}{0.5} = 0.032 \text{ J}$.
Work done moving from stable ($\theta = 0^\circ$) to unstable ($\theta = 180^\circ$) is $W = \mu B (\cos 0^\circ - \cos 180^\circ) = 2 \mu B$.
Substituting the value of $\mu B$, $W = 2 \times 0.032 \text{ J} = 0.064 \text{ J}$.
Since $0.064 \text{ J} = 64 \times 10^{-3} \text{ J}$, the value of $\alpha$ is $64$.

$H=\;\frac{B}{\mu }=\;\frac{B}{\mu _{0} \mu _{r}}=\;\frac{1}{400 \times 4 \pi \times 10^{-7}}$ ${H}=\;\frac{1}{16 \pi } \times 10^{5}$

$B_{n e t}=\sqrt{{B^{2}}_{1}+{B^{2}}_{2}}=\;\frac{\mu _{0}}{4 \pi } \;\frac{M}{(r)^{3}} {\sqrt{5}}$ $=10^{-7} \times 3 {\sqrt{5}} \times {\sqrt{5}} \times \;\frac{8}{10^{-3}}$ $=120 \times 10^{-4}=12 {\text{mT}}$

Assertion (A) is true because ferromagnetism arises from exchange interactions between atomic magnetic dipoles, which force alignment into macroscopic regions known as magnetic domains at temperatures below the Curie temperature ($T_C$). Reason (R) is also true, as thermal energy at temperatures above $T_C$ overcomes these exchange interactions, disrupting the domain structure and causing the material to transition into a paramagnetic state where permanent magnetization vanishes. However, (R) describes the temperature-dependent loss of magnetic order, whereas (A) describes the mechanism of the formation of magnetic order. Therefore, (R) is a true statement that does not provide the causal explanation for the phenomena stated in (A).

$\;\tau =M B sin \;\theta$ $\;v=-M B \cos \theta$ $\;\;\frac{v}{\tau }=-\cot \theta$ $\;v=-80 {\sqrt{3}} \times \;{1}/{{\sqrt{3}}}=-80$ $\;m_1=10 {\text{kg}}$

Magnetic field at $P, B=\;\frac{\mu _0}{4 \pi } \;\frac{m(2 l)}{r^3}$, where $m$ is the pole strength$\Rightarrow (\;\frac{\Delta B}{B} \times 100) \;= (\;\frac{\Delta l}{l}) \times 100+3{| \;\frac{\Delta r}{r} |} \times 100$ $\;=1 \%+3 \%=4 \%$

$\;\frac{\Delta B}{B}=\;\frac{ (1+x_B) B-B}{B}=x_B$So percentage increase $=( {\text{x}_B} \times 100)$ $\;=1.2 \times 10^{-5} \times 100$ $\;=\;\frac{6}{5} \times 10^{-3}$

(A) Magnetic induction → Gauss (III)(B) Magnetic intensity → Ampere/meter (IV) → $\mu =\;\frac{B}{\mu }$(C) Magnetic flux → weber (II) →$\varphi =\vec{B} \cdot \vec{A}$ (D) Magnetic moment → Ampere-meter${ }^2( I ) \longrightarrow$ $\vec{M}=I \vec{A}$

The magnet does not exist in form of monopole and magnetic lines never intersect, continuous and form a closed loop. Line initiates from north pole and end to south outside while inside vice versa.

$\;\mu =\mu _r \mu _0$ $\;\mu _r=(χ+1)$ $\;\mu =(χ+1) \mu _0$ $\;χ= (\;\frac{\mu }{\mu _0}-1)$

Paramagnetic materials possess a small, positive magnetic susceptibility ($\chi > 0$), confirming statement C is correct. When placed in an external magnetic field ($B_{ext}$), their atomic dipoles align parallel to the field direction, making statement A correct. Statement B is incorrect because they are only weakly attracted to magnetic fields, not strongly. Statement D is false because paramagnetic substances move from regions of weaker magnetic field to regions of stronger magnetic field, not the reverse. Consequently, only statements A and C are accurate descriptions.

At stable equilibrium${\text{U}}=- {\text{mB}} \cos 0^{\circ}=- {\text{mB}}$At unstable equilibrium$\; {\text{U}}=- {\text{mB}} \cos 180^{\circ}=+ {\text{mB}}$ $\; {\text{W}}=\Delta {\text{U}}$ $\;\;\text{ W.D. }\;=2 {\text{mB}}$ $\;=2(0.5) 8 \times 10^{-2}=8 \times 10^{-2} {\text{J}}$

$\;B_H=3.5 \times 10^{-5} T$ $\;F=i ℓ B sin \;\theta , \;\; {\text{i}}={\sqrt{2}} {\text{A}}$ $\;\;\frac{F}{ℓ}=i B sin \;\theta ={\sqrt{2}} \times 3.5 \times 10^{-5}$ $\times \;{1}/{{\sqrt{2}}}$ $\;=35 \times 10^{-6} {\text{N}} / {\text{m}}$

(A) Graph between Magnetic susceptibility and magnetising field is :(B) magnetic field due to a current carrying wire for ${\text{x}}< {\text{a}}$ :${\text{B}}=\;{\mu _0 {\text{ir}}}/{2 \pi {\text{a}}^2}$(C) magnetic field due to a current carrying wire for ${\text{x}}> {\text{a}}$ :$B=\;{\mu _0 {\text{i}}}/{2 \pi {\text{a}}}$(D) magnetic field inside solenoid varies as:

$\; {\text{V}}=\;\frac{\mu _0}{4 \pi } \;{ {\text{M}}}/{ {\text{r}}^2}$ $\;\Rightarrow 1.5 \times 10^{-5}=10^{-7} \times \;{ {\text{M}}}/{ (20 \times 10^{-2}) ^2}$ $\;\Rightarrow {\text{M}}=\;\frac{1.5 \times 10^{-5} \times 20 \times 20 \times 10^{-4}}{10^{-7}}$ $\; {\text{M}}=1.5 \times 4=6$

$\; {\text{H}}_{ {\text{c}}}=\;{\mu _{ {\text{o}}} {\ni}}/{\mu _{ {\text{o}}}}$ $\;5 \times 10^3=\;\frac{150}{30} \times 100 \times {\text{i}}$ $\;\;\frac{50}{5}= {\text{i}}$ $\; {\text{I}}=10$

$\; {\text{H}}=2.4 \times 10^3 {\text{A}} / {\text{m}}$ $\; {\text{H}}= {\text{nI}}=\;{ {\text{N}}}/{ℓ} {\text{I}}$ $\; {\text{I}}=\;{ {\text{H}} ℓ}/{ {\text{N}}}=\;\frac{2.4 \times 10^3 \times 15 \times 10^{-2}}{60}$ $\; {\text{I}}=6 {\text{A}}$

Period of oscillation $\alpha \;{1}/{{\sqrt{B_H}}}$ $\;T \alpha \;{1}/{{\sqrt{B \cos \theta }}} \Rightarrow \;\frac{T_1}{T_2}=\sqrt{\;\frac{B_2 \cos \theta _2}{B_1 \cos \theta _1}}$ $\;\Rightarrow \;\frac{60 / 20}{60 / 30}=\sqrt{\;\frac{B_2}{B_1} \;\frac{\cos 60^{\circ}}{\cos 30^{\circ}}} \Rightarrow \;\frac{3}{2}=\sqrt{\;{B_2}/{{\sqrt{3}} B_1}}$ $\;\Rightarrow \;\frac{9}{4}=\;{B_2}/{{\sqrt{3}} B_1} \Rightarrow \;\frac{B_1}{B_2}=\;{4}/{9 {\sqrt{3}}}=\;{4}/{{\sqrt{243}}}$

$\; {\text{u}}=- {\text{MB}} \cos \theta$ $\; {\text{W}}=\Delta {\text{u}}$ $\; {\text{W}}=- {\text{MB}} \cos 180^{\circ} (- {\text{mB}} \cos 0^{\circ})$ $\; {\text{W}}=2 {\text{MB}}=2 \times 5 \times 0.4=4 {\text{J}}$Option 1

Electromagnets require a core material that can be easily magnetized and demagnetized to function effectively; soft iron is the standard material chosen for this purpose. Soft iron possesses high magnetic permeability ($\mu$), which allows it to support a strong magnetic field, and low retentivity, which ensures the induced magnetism is lost rapidly once the magnetizing current is switched off. These specific magnetic properties are exactly why soft iron is the ideal material for constructing electromagnets. Consequently, Assertion A is correct because the magnetic characteristics defined in Reason R facilitate the operational requirements of an electromagnet. Therefore, both statements are correct, and Reason R is the correct explanation for Assertion A.