A magnet has total length 2l = 20 units and the field point P is at a distance d = 10 units from the centre of the…

Year	2026	2025	2024	Total
Questions	2	6	1	9

Q45 jee_main_2026_23_january_morning Magnetic Properties of Materials

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Consider a ferromagnetic material: Assertion (A) : The individual atoms in a ferromagnetic material possess a magnetic dipole moment and interact with one another in such a way that they spontaneously align themselves forming domains. Reason (R) : At high enough temperature, the domain structure of ferromagnetic material disintegrates. Thus, magnetization will disappear at high enough temperature known as Curie temperature. In the light of the above statements, choose the correct answer from the options given below :

A. (A) is true but (R) is false
B. Both (A) and (R) are true but (R) is not the correct explanation of (A)
C. Both (A) and (R) are true and (R) is the correct explanation of (A)
D. (A) is false but (R) is true

Solution

Core Logic

Assertion (A) describes the fundamental microscopic origin of ferromagnetism, which is the spontaneous alignment of atomic dipoles into magnetic domains due to quantum exchange interactions. This is a factually correct statement.

Reason (R) describes the macroscopic thermal breakdown of this alignment. At the Curie temperature, thermal agitation overcomes the exchange coupling, causing the domains to scramble and the material transitions from ferromagnetic to paramagnetic. This is also a factually correct statement.

Step 1: Final Conclusion

While both statements are true descriptions of ferromagnetic behavior, Reason (R) (thermal destruction of domains) is not the cause of Assertion (A) (the formation of domains). They describe opposite phenomena (ordering vs. disordering). Therefore, (R) does not explain (A).

Pattern Recognition

Sees: Assertion about "cause of property X" and Reason about "destruction of property X" → They are disconnected causally. Both true, but R is not the correct explanation for A.

Chapter Mix

Class 12 Physics: Magnetism and Matter

Q46 jee_main_2026_24_january_morning Magnetic Dipole in a Magnetic Field

A short bar magnet placed with its axis at

30°

with an external field of 800 Gauss, experiences a torque of 0.016 N.m. The work done in moving it from most stable to most unstable position is

α \times 10⁻³

J. The value of

α

is ____.

Numerical Answer. Answer: 64 to 64

Solution

Related Formula

τ = M B θ

W = Δ U = M B ( θ₁ - θ₂)

Core Logic

First, find the magnetic moment $M$ $M$ using the given torque condition:

0.016 = M B 30°

0.016 = M \times B \times (1)/(2)

M = (0.032)/(B)

Step 1: Calculate Work Done

Work done moving from the most stable position ( $θ₁ = 0°$ $\theta_1 = 0^{\circ}$) to the most unstable position ( $θ₂ = 180°$ $\theta_2 = 180^{\circ}$) is:

W = Uf - Uᵢ = (+MB) - (-MB) = 2MB

Substitute $M = (0.032)/(B)$ $M = \frac{0.032}{B}$:

W = 2 ((0.032)/(B)) B = 0.064 J

W = 64 \times 10⁻³ J

Comparing with $α \times 10⁻³ J$ $\alpha \times 10^{-3} \text{ J}$, we get $α = 64$ $\alpha = 64$.

Pattern Recognition

Flipping a dipole from strictly aligned ( $0^°$ $0^\circ$) to anti-aligned ( $180^°$ $180^\circ$) always requires exactly $2MB$ $2MB$ of work. Use the initial torque to isolate the $MB$ $MB$ product, rendering the explicit $B$ $B$-field value (800 Gauss) computationally irrelevant.

Chapter Mix

Class 12 Physics: Magnetism and Matter

Q10 jee_main_2025_02_april_morning Magnetic Intensity and Susceptibility

The relationship between the magnetic susceptibility

(χ)

and the magnetic permeability

(μ)

is given by: (

μ₀

is the permeability of free space and

μᵣ

is relative permeability)

A. $χ = (μ)/(μ₀) - 1$ $\chi = \frac{\mu}{\mu_0} - 1$
B. $χ = μᵣμ₀ + 1$
C. $χ = μᵣ + 1$ $\chi = \mu_{\mathrm{r}} + 1$
D. $χ = 1 - (μ)/(μ₀)$ $\chi = 1 - \frac{\mu}{\mu_0}$

Solution

Related Formula

μᵣ = 1 + χ

μ = μ₀ μᵣ

Core Logic

The relative magnetic permeability $μᵣ$ $\mu_{\mathrm{r}}$ of a magnetic material is related to its susceptibility $χ$ $\chi$ by the basic relation:

μᵣ = 1 + χ χ = μᵣ - 1

Since relative permeability $μᵣ$ $\mu_{\mathrm{r}}$ is the ratio of absolute permeability $μ$ $\mu$ to the permeability of free space $μ₀$ $\mu_0$:

μᵣ = (μ)/(μ₀)

Substituting this into the susceptibility relation yields:

χ = (μ)/(μ₀) - 1

Step 1: Final Conclusion

The correct option is $χ = (μ)/(μ₀) - 1$ $\chi = \frac{\mu}{\mu_0} - 1$.

Pattern Recognition

Relative permeability $μᵣ$ $\mu_r$ represents total magnetic response (vacuum background + material response), while susceptibility $χ$ $\chi$ represents the material's specific response alone. Hence they always differ by exactly 1: $μᵣ = 1 + χ$ $\mu_r = 1 + \chi$.

Chapter Mix

Class 12 Physics: Magnetic Properties of Matter

Q1 jee_main_2025_03_april_evening Magnetic Dipole in a Uniform Magnetic Field

A magnetic dipole experiences a torque of

80\sqrt(3)~N~m

when placed in uniform magnetic field in such a way that dipole moment makes angle of

60°

with magnetic field. The potential energy of the dipole is :

A. $80~J$ $80\mathrm{~J}$
B. $-40\sqrt(3)~J$ $-40\sqrt{3}\mathrm{~J}$
C. $-60~J$ $-60\mathrm{~J}$
D. $-80~J$ $-80\mathrm{~J}$

Solution

Related Formula

The torque $τ$ $\vec{\tau}$ experienced by a magnetic dipole in a uniform magnetic field $B$ $\vec{B}$ is given by:

τ = M \times B \Rightarrow τ = MB θ

The potential energy $U$ $U$ of the dipole is given by:

U = - M \cdot B = -MB θ

Core Logic

Given parameters:

Torque $τ = 80\sqrt(3)~N~m$ $\tau = 80\sqrt{3}\mathrm{~N~m}$
Angle $θ = 60°$ $\theta = 60^{\circ}$

Step 1: Calculate the value of $MB$ $MB$

Substitute the given values into the torque formula:

80\sqrt(3) = MB (60°)

80\sqrt(3) = MB ( \sqrt(3)2)

MB = 160~J

Step 2: Calculate Potential Energy

Using the potential energy expression:

U = -MB (60°)

U = -160 \times (1)/(2) = -80~J

Pattern Recognition

A standard dipole problem checking the relation between torque and potential energy. Note that torque goes with the sine of the angle, whereas potential energy goes with the negative cosine. Dividing torque by potential energy gives $- θ$ $-\tan\theta$.

Chapter Mix

Class 12 Physics: Magnetism and Matter

Q7 jee_main_2025_07_april_morning Magnetic Materials

The percentage increase in magnetic field (B) when space within a current carrying solenoid is filled with magnesium (magnetic susceptibility

χmg = 1.2\times 10⁻⁵

) is:

A. $(6)/(5) \times 10⁻³ %$ $\frac{6}{5} \times 10^{-3} \%$
B. $(5)/(6) \times 10⁻⁵%$ $\frac{5}{6} \times 10^{-5}\%$
C. $(5)/(6) \times 10⁻⁴%$ $\frac{5}{6} \times 10^{-4}\%$
D. $(5)/(3) \times 10⁻⁵%$ $\frac{5}{3} \times 10^{-5}\%$

Solution

Related Formula

The magnetic field inside a filled solenoid is:

Bnew = μ n I

The magnetic field in air is:

Bold = μ₀ n I

Relative permeability $μᵣ$ $\mu_r$ and susceptibility $χ$ $\chi$ are related by:

μᵣ = (μ)/(μ₀) = 1 + χ

Core Logic

The percentage increase in magnetic field is:

% change in B = Bnew - BoldBold \times 100%

% change in B = (μ - μ₀)/(μ₀) \times 100% = (μᵣ - 1) \times 100% = χ \times 100%

Step 1: Substitute and Calculate

Given susceptibility $χmg = 1.2 \times 10⁻⁵$ :

% change = 1.2 \times 10⁻⁵ \times 100% = 1.2 \times 10⁻³%

Convert decimal to fraction:

1.2 \times 10⁻³% = (6)/(5) \times 10⁻³%

Pattern Recognition

Sees: Susceptibility and solenoid field percentage change. Shortcut: The fractional change in magnetic field when inserting a core is exactly equal to its magnetic susceptibility $χ$ $\chi$. Thus, the percentage change is just $χ \times 100$ $\chi \times 100$.

Chapter Mix

Class 12 Physics: Magnetism and Matter

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Magnetism and Matter Previous-Year Questions

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Work Done

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the value of MB$MB$

Step 2: Calculate Potential Energy

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitute and Calculate

Pattern Recognition

Chapter Mix

More Magnetism and Matter Questions — jee_main_2025_28_jan_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

Step 1: Calculate the value of $MB$ $MB$