Magnetism and MatterNEET Physics · Class 12 · NCERT Chapter 5

Solution

$m=NIA$, units: $A\cdot m^2$.

Solution

Aluminium is paramagnetic. Bi and Cu are diamagnetic; Fe is ferromagnetic.

Solution

Axial: $B=2{\mu }_{0}m/(4\pi r^3)$. Twice the equatorial.

Solution

$\chi =C/T$. Inversely proportional.

Solution

Soft magnets (transformer cores) need easy magnetisation/demagnetisation: low coercivity, low retentivity.

Solution

Diamagnetic: chi is small and negative.

Solution

At magnetic equator, B is purely horizontal: dip = 0°.

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$\tau =mB\sin \theta =mB\cdot 1=mB$.

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${\mu }_r=1+\chi$.

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Above $T_C$: thermal disorder destroys domain alignment, behaves paramagnetic.

Solution

$B_{eq}/B_{ax}=1/2$.

Solution

Loop area = energy dissipated per cycle as heat.

Solution

Iron's Curie temperature is about 1043 K (770 °C).

Solution

Cobalt (Co) is ferromagnetic. Cu is diamagnetic, O₂ is paramagnetic, Bi is diamagnetic.

Solution

At poles, B is purely vertical: dip = 90°.

Solution

$\chi ={\mu }_r-1=0.0001$.

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$\chi \propto 1/T$. T → T/3, χ → 3χ = $1.2\times 10^{-3}$.

Solution

$m=NIA$.

Solution

Permanent magnets must resist demagnetisation: high coercivity AND retain magnetisation: high retentivity.

Solution

Diamagnetic: small negative chi.

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$\cos \delta =B_H/B=0.5$, so $\delta =60°$.

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$U=-mB\cos \theta$ is minimum at $\theta =0$ (m parallel to B). Stable.

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$H=nI$ for a solenoid: units A/m.

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Steel and Alnico have high coercivity, so they retain magnetisation. Soft iron does not.

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Above $T_C$: $\chi =C/(T-T_C)$. Curie-Weiss extension.

Solution

Internal poles cancel, external poles add. Total moment = 2m.

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Retentivity = M when H = 0 after the material has been saturated. The leftover magnetism.

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At magnetic poles dip = 90°: needle stands vertical.

Solution

Bismuth has notable diamagnetism.

Solution

$U=-mB\cos \theta$ varies from $-mB$ (at $\theta =0$) to $+mB$ (at $\theta =180°$).

Solution

$M=\chi H=(C/T)H$. Inversely proportional to T.

Solution

$m=m_{\text{pole}}\cdot 2L=5\times 0.1=0.5{\text{A⋅m}}^2$.