Q.A paramagnetic sample shows a net magnetisation of $8\ \text{Am}^{-1}$ when placed in an external magnetic field of $0.6\ \text{T}$ at a temperature of $4\ \text{K}$. When the same sample is placed in an external magnetic field of $0.2\ \text{T}$ at a temperature of $16\ \text{K}$, the magnetisation will be

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\chi$ is the magnetic susceptibility, $\mu$, $\mu_0$ and $\mu_r$ are absolute, free space and relative permeabilities respectively, then (A) $\mu = \mu_r(1+\chi)$ (B) $\mu = \mu_0(1+\chi)$ (C) $\mu_r = 1-\chi$ (D) $\mu = \dfrac{\mu_r}{\mu_0}(1-\chi)$

   ›Reveal solutionSolution

   The standard magnetism relations $\mu_r=1+\chi$ and $\mu=\mu_0\mu_r$ combine directly to give $\mu=\mu_0(1+\chi)$.

   Concept and Intuition

   Magnetic susceptibility $\chi$ measures how strongly a material magnetizes in response to a field; relative permeability $\mu_r$ measures how much the material's absolute permeability exceeds that of free space. These two descriptions of the same magnetic response are linked by the identity $\mu_r = 1+\chi$, and absolute permeability is simply $\mu_r$ scaled by the free-space value $\mu_0$.

   Step-by-Step Solution

   1. Definition of relative permeability: $\mu_r = \dfrac{\mu}{\mu_0}$.
   2. Standard relation between susceptibility and relative permeability: $\mu_r = 1+\chi$.
   3. Combine: $\dfrac{\mu}{\mu_0} = 1+\chi \implies \mu = \mu_0(1+\chi)$. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.A material satisfies the relation $\mu_0(H + M) = 0$, where H and M are magnetic intensity and magnetization, respectively; then the material is (A) Nonmagnetic (B) Paramagnetic (C) Ferromagnetic (D) Diamagnetic

   ›Reveal solutionSolution

   This tests recognizing that $B=\mu_0(H+M)=0$ with $M=-H$ corresponds to diamagnetic behaviour (magnetization opposing the field). Answer: Diamagnetic.

   Concept and Intuition

   Inside any magnetic material, $B = \mu_0(H+M)$. The given condition forces $B=0$, meaning the material's own magnetization $M$ exactly cancels the applied field's contribution ($M=-H$). Materials whose induced magnetization opposes the external field (negative susceptibility, $\chi < 0$) are diamagnetic; here that opposition is complete, which is the hallmark (idealised limit) of diamagnetism, distinguishing it from paramagnetic ($M$ aligns with $H$, $\chi>0$, small positive) and ferromagnetic ($M$ strongly aligns with $H$, large positive $\chi$) materials.

   Step-by-Step Solution

   1. Start from $B = \mu_0(H+M)$, the general relation between $B$, $H$, and $M$.
   2. Given: $\mu_0(H+M) = 0 \Rightarrow H + M = 0 \Rightarrow M = -H$.
   3. Since $M$ is opposite in sign to $H$ (magnetization opposes the applied field), the susceptibility $\chi = M/H = -1$, i.e. negative. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.A magnetic material placed in a magnetic field of intensity $H = 1000\ \mathrm{Am^{-1}}$ has magnetization $M = 2\ \mathrm{Am^{-1}}$. The magnetic susceptibility, and type of material is (A) $2 \times 10^{-3}$, Diamagnetic (B) $2 \times 10^{-3}$, paramagnetic (C) $4 \times 10^{-3}$, Ferromagnetic (D) 2000, Ferromagnetic

   ›Reveal solutionSolution

   Magnetic susceptibility is $\chi = M/H$. A small positive value like $2\times10^{-3}$ identifies the material as paramagnetic (diamagnetic would be small and negative; ferromagnetic would be very large, often $\sim10^2$–$10^5$).

   Concept and Intuition

   Susceptibility measures how readily a material magnetizes in response to an applied field. Diamagnetic materials weakly oppose the field ($\chi$ small and negative). Paramagnetic materials weakly align with it ($\chi$ small and positive, typically $10^{-5}$ to $10^{-3}$). Ferromagnetic materials show enormous positive $\chi$ (hundreds to thousands).

   Step-by-Step Solution

   1. $\chi = \dfrac{M}{H} = \dfrac{2\ \text{Am}^{-1}}{1000\ \text{Am}^{-1}} = 2\times10^{-3}$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The temperature at which the susceptibility of nickel becomes 0.5 times its susceptibility at a temperature of $460^\circ C$ is (Curie temperature of nickel is $360^\circ C$) (A) 560 K (B) 833 K (C) 733 K (D) 760 K

   ›Reveal solutionSolution

   The Curie–Weiss law for a ferromagnet above its Curie point gives $\chi\propto 1/(T-T_C)$; solving for the new temperature gives 833 K.

   Concept and Intuition

   Above the Curie temperature $T_C$, a ferromagnetic material behaves like a paramagnet, and its susceptibility obeys the Curie–Weiss law: $\chi=\dfrac{C}{T-T_C}$, where $C$ is the material's Curie constant. Susceptibility keeps falling as temperature rises further above $T_C$.

   Step-by-Step Solution

   1. Convert to Kelvin: $T_C=360+273=633$ K; $T_1=460+273=733$ K.
   2. $\chi_1=\dfrac{C}{T_1-T_C}=\dfrac{C}{733-633}=\dfrac{C}{100}$.
   3. We want $T_2$ such that $\chi_2=0.5\chi_1=\dfrac{C}{200}$.
   4. So $T_2-T_C=200 \Rightarrow T_2=200+633=833$ K.

   Common Mistakes …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A bulk magnetic material has volume of $2\ m^3$ and its magnetization is found to be $2\ A\ m^{-1}$, then the magnetic moment of the bulk material is (A) $1\ Am^2$ (B) $4\ Am^2$ (C) $2\ Am^2$ (D) $8\ Am^2$

   ›Reveal solutionSolution

   Magnetization is magnetic moment per unit volume: $M = m/V$, so $m = MV$. Answer: $4\ Am^2$.

   Concept and Intuition

   Magnetization $\vec M$ describes how strongly a bulk material is magnetized, defined as the net magnetic dipole moment per unit volume, $M = m/V$. It's directly analogous to how polarization is dipole moment per unit volume in dielectrics. So the total magnetic moment of the whole sample is simply magnetization times volume.

   Step-by-Step Solution

   1. Definition: $M = \dfrac{m}{V}$, so $m = M\times V$.
   2. Given $M = 2\ A\,m^{-1}$, $V = 2\ m^3$.
   3. $m = 2\times2 = 4\ A\,m^2$.

   Common Mistakes …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.Materials suitable for permanent magnets should have (A) low retentivity and low coercivity (B) low retentivity and high coercivity (C) high retentivity and low coercivity (D) high retentivity and high coercivity

   ›Reveal solutionSolution

   Permanent magnet materials need both high retentivity (to stay strongly magnetized) and high coercivity (to resist demagnetization) — steel is the classic example, unlike soft iron used for electromagnets.

   Concept and Intuition

   Retentivity is the magnetization a material keeps after the external field is removed; coercivity is the reverse field needed to bring that magnetization to zero. A permanent magnet should hold a strong field indefinitely (high retentivity) and not be easily demagnetized by small stray fields, vibration, or temperature changes (high coercivity). This corresponds to a hysteresis loop that is tall (high $B_r$) and wide (high $H_c$) — materials like steel and alnico. In contrast, electromagnet cores (soft iron) want high retentivity is not needed; instead they need low retentivity/low coercivity so they can be easily magnetized and demagnetized as the current is switched.

   Step-by-Step Solution

   1. Identify what is needed for a permanent magnet: it must stay magnetized after the magnetizing field is removed → requires high retentivity.
   2. It must also resist losing its magnetism due to small opposing fields, mechanical shocks, or temperature → requires high coercivity. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.If the magnetisation of a material is M and the magnetic field in the material is B, then the magnetic intensity is ($\mu_0$ - permeability of free space) (A) $\dfrac{B}{\mu_0}$ (B) $\dfrac{B-\mu_0 M}{\mu_0}$ (C) $\dfrac{BM}{\mu_0}$ (D) $\dfrac{\mu_0 M}{B}$

   ›Reveal solutionSolution

   A direct rearrangement of the fundamental relation between magnetic field, magnetisation and magnetic intensity inside matter.

   Concept and Intuition

   Inside a magnetized material, the total magnetic field B has two contributions: the field due to free currents (captured by H) and the field due to the material's own magnetisation M. The relation $B=\mu_0(H+M)$ ties all three together.

   Step-by-Step Solution

   1. Start from $B=\mu_0(H+M)$.
   2. Rearranging: $H+M=\dfrac{B}{\mu_0}$.
   3. $H=\dfrac{B}{\mu_0}-M=\dfrac{B-\mu_0 M}{\mu_0}$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.A sample of paramagnetic salt contains $2 \times 10^{24}$ atomic dipoles each of dipole moment $1.5 \times 10^{-23} \mathrm{~J~T^{-1}}$. The sample is placed under homogeneous magnetic field of 0.6 T and cooled to a temperature 4.2 K. The degree of magnetic saturation achived is 20%. Then total dipole moment of the sample for a magnetic field of 0.9 T and a temperature of 2.8 K is (A) $4.5 \mathrm{~J~T^{-1}}$ (B) $13.5 \mathrm{~J~T^{-1}}$ (C) $0.64 \mathrm{~J~T^{-1}}$ (D) $7 \mathrm{~J~T^{-1}}$

   ›Reveal solutionSolution

   Curie's law ($M\propto B/T$) scales the dipole moment between the two states; the result is $13.5\ \mathrm{J\,T^{-1}}$.

   Concept and Intuition

   In a paramagnetic sample well below full saturation, thermal agitation competes with the aligning magnetic field, and Curie's law says the net magnetic dipole moment is proportional to $B/T$ — stronger field aligns more dipoles, higher temperature randomizes them more. The 'degree of saturation' tells us what fraction of the theoretical maximum (all dipoles perfectly aligned) is actually achieved at a given $(B,T)$.

   Step-by-Step Solution

   1. Maximum possible (fully saturated) dipole moment: $M_{sat} = N p = (2\times10^{24})(1.5\times10^{-23}) = 30\ \mathrm{J/T}$.
   2. At $B_1=0.6\,\mathrm T$, $T_1=4.2\,\mathrm K$, the sample achieves 20% saturation: $M_1 = 0.20\times30 = 6\ \mathrm{J/T}$.
   3. Curie's law: $M \propto \dfrac{B}{T}$, so $\dfrac{M_2}{M_1} = \dfrac{B_2/T_2}{B_1/T_1} = \dfrac{0.9/2.8}{0.6/4.2}$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.A sample of a ferromagnetic iron in the shape of a cube of side $1.0\ \mu\mathrm{m}$ contains $8.7 \times 10^{28}$ atoms per cubic metre and the magnetic dipole moment of each iron atom is $9.3 \times 10^{-24}\ \mathrm{A\,m^2}$. Then the maximum possible magnetic dipole moment (in $\mathrm{A\,m^2}$) of the sample is nearly (A) $8.1 \times 10^{-12}$ (B) $8.1 \times 10^{-14}$ (C) $81 \times 10^{-14}$ (D) $81 \times 10^{-16}$

   ›Reveal solutionSolution

   The maximum possible dipole moment of a ferromagnetic sample occurs when every atomic dipole is perfectly aligned (saturation); here it works out to $81\times10^{-14}\ \text{A m}^2$.

   Concept and Intuition

   In a ferromagnetic material, atomic magnetic moments can be aligned in the same direction by an external field (magnetic saturation). The maximum possible dipole moment of the whole sample is simply the number of atoms times each atom's dipole moment, since in that ideal aligned state nothing cancels.

   Step-by-Step Solution

   1. Volume of the cube: $V=(1.0\times10^{-6}\,\text{m})^3=1.0\times10^{-18}\,\text{m}^3$.
   2. Number of atoms in the sample: $N=n\times V=8.7\times10^{28}\times1.0\times10^{-18}=8.7\times10^{10}$.
   3. Maximum dipole moment: $M_{max}=N\mu=8.7\times10^{10}\times9.3\times10^{-24}$. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.If the given figure shows the relation between magnetic field (B-along y-axis) and magnetic intensity (H-along x-axis) of a ferromagnetic material, then the point that represents coercivity of the material is [FIGURE] (a B-H hysteresis loop: B on the vertical axis, H on the horizontal axis; the loop passes through P in the upper-right (positive B and H), Q on the positive B-axis where H=0, R on the negative H-axis where B=0 to the left of the origin O, U on the positive H-axis where B=0 to the right of the origin O, T on the negative B-axis where H=0, and S in the lower-left (negative B and H); an arrow on the upper part of the loop shows the traversal direction from S/lower branch up through Q towards P) (A) P (B) Q (C) R (D) S

    ›Reveal solutionSolution

    This tests reading coercivity and retentivity off a hysteresis ($B$–$H$) loop; the answer is (C) point R.

    Concept and Intuition

    On a $B$–$H$ hysteresis loop, two special sets of points are commonly asked about: retentivity (residual magnetism when $H=0$, found where the loop crosses the vertical $B$-axis) and coercivity (the reverse field needed to demagnetise the material completely, i.e. bring $B$ to zero, found where the loop crosses the horizontal $H$-axis). Coming down from positive saturation ($P$), the material retains a positive $B$ at $H=0$ (point $Q$, retentivity), and needs a negative $H$ to bring $B$ to zero — that crossing point is the coercivity.

    Step-by-Step Solution

    1. Starting from positive saturation at $P$ and decreasing $H$, the loop crosses the $B$-axis (at $H=0$) at point $Q$ — this gives the retentivity (residual field).
    2. Continuing to decrease $H$ into negative values, $B$ falls to zero at the point where the loop crosses the $H$-axis on the negative side — this is point $R$.
    3. By definition, coercivity is the magnitude of this reverse field required to demagnetise the sample, i.e., it is represented by point $R$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
    Q.Relative permeability ($\mu_r$) of a sample is given by $1 < \mu_r < 1+\epsilon$, where $\epsilon$ is a small positive number. Then nature of the sample is (A) Diamagnetic material (B) Paramagnetic material (C) Ferro magnetic material (D) Super conducting material

    ›Reveal solutionSolution

    A relative permeability just above 1 is characteristic of paramagnetism; diamagnets sit just below 1, ferromagnets are far above 1, and superconductors have $\mu_r=0$.

    Concept and Intuition

    Relative permeability $\mu_r = 1+\chi$, where $\chi$ is the magnetic susceptibility.

    • Diamagnetic substances: $\chi$ is small and negative, so $\mu_r$ is slightly less than 1.
    • Paramagnetic substances: $\chi$ is small and positive, so $\mu_r$ is slightly greater than 1 (a value like $1+\epsilon$ with small $\epsilon>0$).
    • Ferromagnetic substances: $\chi$ is large and positive (hundreds to thousands), so $\mu_r \gg 1$.
    • Ideal superconductors: perfect diamagnets, $\chi=-1$, so $\mu_r = 0$ (Meissner effect).

    Step-by-Step Solution

    1. Given: $1 < \mu_r < 1+\epsilon$ with $\epsilon$ small and positive.
    2. This means $\mu_r$ is only marginally above 1 — a small positive susceptibility.
    3. Small positive susceptibility with $\mu_r$ close to (but above) 1 is exactly the paramagnetic regime. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
    Q.A super conductor exhibits (A) Ferro magnetism (B) Para magnetism (C) Dia magnetism (D) Ferri magnetism

    ›Reveal solutionSolution

    The Meissner effect makes a superconductor a perfect diamagnet, expelling all internal magnetic field. Answer: (C) Dia magnetism.

    Concept and Intuition

    Below its critical temperature, a superconductor doesn't merely have zero resistance — it also actively expels any externally applied magnetic field from its interior (the Meissner effect), so that the magnetic field inside is exactly zero. This is the signature of a perfect diamagnet (magnetic susceptibility $\chi=-1$), since diamagnetism is characterized by an induced magnetization that opposes and cancels the applied field.

    Step-by-Step Solution

    1. Recall the defining property of superconductivity below $T_c$: zero electrical resistance and the Meissner effect (expulsion of magnetic flux from the bulk).
    2. A material with $\chi=-1$ (magnetic field completely cancelled inside) is, by definition, a perfect diamagnet. …

    🔒 Start your 14-day free trial to see the full solution →