Magnetism And Matter – JEE Main PhysicsPractice Questions & PYQs

According to lenz's law, the rate of charge of flum produced by bar magnet will be approused by the conducting loops.

Diamagnetism do not depends on temperature. So statement-1 is wrongStatement-1 is given as: Diamagnetic property depends on temperature. WrongStatement-2 : in diamagnetic material induced B is apposite to applied magnetic field.

Magnetic field at centre inside the solenoid is given by${\text{B}}=\mu _0 {\text{nI}}$So magnetic intensity at centre$\; {\text{H}}=\;{ {\text{B}}}/{\mu _0}= {\text{nI}}= (\;\frac{1200}{2}) (2)$ $\; {\text{H}}=1.2 \times 10^3 {\text{Am}}^{-1}$

${\text{H}}=1.6 \times 10^3 {\text{A}} / {\text{m}}, {\text{n}}=8$ per ${\text{cm}}=800$ per ${\text{m}}$ ${\text{H}}= {\text{nI}} \Rightarrow {\text{I}}=\;{ {\text{H}}}/{ {\text{n}}}$ ${\text{I}}=\;\frac{1.6 \times 10^3}{8 \times 10^2} \Rightarrow {\text{I}}=2 {\text{A}}$

$\; {\text{X}}=2 \times 10^{-2}$ $\;\mu {\text{r}}=1+ {\text{x}}=1+0.02=1.02$Bo $\to$ magnetic field due to magnetic material${\text{B}}_{ {\text{m}}} \to$ magnetic field due to magnetic material${\text{B}}_{ {\text{m}}}=\mu _{ {\text{r}}} {\text{B}}_0$ $\Delta {\text{B}}=\;{ {\text{B}}_{ {\text{m}}}- {\text{B}}_0}/{ {\text{B}}_0} \times 100=\;{\mu _{ {\text{r}}} {\text{B}}_0- {\text{B}}_0}/{ {\text{B}}_0} \times 100$ $\Delta {\text{B}} \%=\;{( {\text{X}}+1)-1}/{1} \times 100= {\text{X}} \times 100$ $\Delta {\text{B}} \%=2 \times 10^{-2} \times 100=2 \%$Ans. Option (3)

Diamagnetic substances have the property due to which they tends to move away from stronger magnetic field to weaker magnetic field, as their magnetic susceptibility is negative.Therefore both statements are correct.

$\therefore sin \;37^{\circ}=\;\frac{B_v}{B_{n e t}}$ $\Rightarrow B_v=B_{n e t} sin \;37^{\circ}$ $\Rightarrow B_{n e t}=\;\frac{B_v}{ sin \;37^{\circ}}$ $=\;\frac{6 \times 10^{-5}}{\frac{3}{5}}$ $=10 \times 10^{-5} {\text{T}}$ $=1 \times 10^{-4} {\text{T}}$

Statement $I$ is true as susceptibility of ferromagnetic and paramagnetic materials is inversely related to temperature.Statement II is true as because of orbital motion of electrons the diamagnetic material is able to oppose external magnetic field.

{B^\vec{'}=\mu _0(1+X) n i (in the material){B^\vec{'}=\mu _0 n i (without material)So fractional increase is$\;\frac{B^{'}-B}{B}=X=1.2 \times 10^{-5}$

$\mu _{ {\text{r}}}= {\text{x}}+1$ $=99+1=100$ $\Rightarrow \mu =\mu _{ {\text{r}}} \mu _0=100 \times 4 \mu \times 10^{-7} {\text{Wb}} / {\text{Am}}$ $=4 \mu \times 10^{-5} {\text{Wb}} / {\text{Am}}$

Electromagnet requires high permeability and Iow retentivity.

$U=-\vec{M} \cdot \vec{B}$So $U_f-U_i=-M B(1-\cos \theta )$ $=-14 J$So $W=-\Delta U=14 J$

$B_H=B \cos \theta$ $\therefore B=\;\frac{B_H}{\cos \theta }=\;{0.5 {\text{G}}}/{\cos 30^{\circ}} \Rightarrow \;{G}/{{\sqrt{3}}}$

$\tan \theta ^{'}=\;\frac{\tan \theta }{\cos \alpha }$ $\theta ^{'}=60^{\circ}$ $\alpha =45^{\circ}$ $\sqrt{3}=\;{\tan \theta }/{\frac{1}{\sqrt{2}}}$ $\tan \theta =\sqrt{\;\frac{3}{2}}$ $\theta =\tan ^{-1} \sqrt{\;\frac{3}{2}}$

${\text{T}}=2 \pi \sqrt{\;{ {\text{I}}}/{ {\text{MB}}_{ {\text{H}}}}}$ $\;{ {\text{T}}_1}/{ {\text{T}}_2}=\;{2 \pi \sqrt{{ {\text{I}}_1}/{ {\text{M}}_1 {\text{B}}_{ {\text{H}}}}}}/{2 \pi \sqrt{\;{ {\text{I}}_2}/{ {\text{M}}_2 {\text{B}}_{ {\text{H}}}}}}=\;\frac{3}{4}$ $\sqrt{\;{ {\text{I}}_1}/{ {\text{I}}_2} \times \;{ {\text{M}}_2}/{ {\text{M}}_1}}=\;\frac{3}{4}$ $\sqrt{\;{ {\text{I}}_1}/{ {\text{I}}_2}} \times \sqrt{\;{ {\text{M}}_2}/{ {\text{M}}_1}}=\;\frac{3}{4}$ $\sqrt{\;\frac{3}{2}} \times \sqrt{\;{ {\text{M}}_2}/{ {\text{M}}_1}}=\;\frac{3}{4}$ $\;\frac{3}{2} \times \;{ {\text{M}}_2}/{ {\text{M}}_1}=\;\frac{9}{16}$ $\;{ {\text{M}}_1}/{ {\text{M}}_2}=\;\frac{8}{3}$

$T=\;\frac{2 \pi m}{B q}$ $\Rightarrow$ Frequency $f=\;\frac{B q}{2 \pi m}$ $=\;\frac{10^{-4} \times 1.6 \times 10^{-19}}{2 \pi \times 9 \times 10^{-31}}$ $≃ 2.8 \times 10^{6} {Hz}$

$T=2 \pi \sqrt{\;{ {\text{I}}}/{ {\text{B}}_{ {\text{H}}} {\text{M}}}}$ $T_1=3 \ sec =2 \pi \sqrt{\;\frac{I}{(B_P \cos 30^{\circ}) M}}$ ${\text{T}}_2=6 {\ sec}=2 \pi \sqrt{\;{ {\text{I}}}/{( {\text{B}}_{ {\text{Q}}} \cos 60^{\circ}) {\text{M}}}}$ $\;\frac{3}{6}=\sqrt{\;{1}/{( {\text{B}}_{ {\text{P}}} \;\frac{\sqrt{3}}{2})} \times ( {\text{B}}_{ {\text{Q}}} / 2)}$ $\;\frac{3}{6}=\sqrt{(\;{B_{ {\text{Q}}}}/{\sqrt{3} B_P})}$ $\;\frac{\sqrt{3}}{4}=\;{B_{ {\text{Q}}}}/{ {\text{B}}_{ {\text{P}}}}$ $B_Q: B_P=\sqrt{3}: 4$

A $\tan \delta=\tan \delta' \cos \theta$ $=\tan 45°\cos 30°$ $\tan \delta=1 \\times \frac{\\sqrt{3}}{2}$ $\delta=\tan ^{-1}(\frac{\\sqrt{3}}{2})$

$\mu=\mu_{0}(1+{x}_{{m}})$ $=4 \pi \\times 10^{-7} \\times 500$ $=2 \pi \\times 10^{-4} {H} / {m}$

The magnetic susceptibility $(χ)$ of ferromagnetic material is in the range of $10^{3}$ to $10^{4}$, so it gets easily magnetised and demagnetised in the presence of external field region. Therefore, when a soft ferromagnetic material is placed in an external magnetic field region, the magnetic domains may increase or decrease in size and change its orientation.