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 :

Solution & Explanation

### 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Magnetism and Matter

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Q46 jee_main_2026_24_january_morning Magnetic Dipole in a Magnetic Field
A short bar magnet placed with its axis at 30^circ 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 alpha times 10^-3 J. The value of alpha is ____.
Numerical Answer. Answer: 64 to 64

Solution

### Related Formula tau = M B sin theta W = Delta U = M B (cos theta_1 - cos theta_2) ### Core Logic First, find the magnetic moment M using the given torque condition: 0.016 = M B sin 30^circ 0.016 = M times B times frac12 M = frac0.032B ### Step 1: Calculate Work Done Work done moving from the most stable position (theta_1 = 0^circ) to the most unstable position (theta_2 = 180^circ) is: W = U_f - U_i = (+MB) - (-MB) = 2MB Substitute M = frac0.032B: W = 2 left(frac0.032Bright) B = 0.064 text J W = 64 times 10^-3 text J Comparing with alpha times 10^-3 text J, we get alpha = 64. ### Pattern Recognition Flipping a dipole from strictly aligned (0^circ) to anti-aligned (180^circ) always requires exactly 2MB of work. Use the initial torque to isolate the MB product, rendering the explicit B-field value (800 Gauss) computationally irrelevant. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Magnetism and Matter
Q1 jee_main_2025_03_april_evening Magnetic Dipole in a Uniform Magnetic Field
A magnetic dipole experiences a torque of 80sqrt3mathrm~N~m when placed in uniform magnetic field in such a way that dipole moment makes angle of 60^circ with magnetic field. The potential energy of the dipole is :
  • A. 80mathrm~J
  • B. -40sqrt3mathrm~J
  • C. -60mathrm~J
  • D. -80mathrm~J

Solution

### Related Formula The torque vectau experienced by a magnetic dipole in a uniform magnetic field vecB is given by: vectau = vecM times vecB Rightarrow tau = MB sintheta The potential energy U of the dipole is given by: U = -vecM cdot vecB = -MB costheta ### Core Logic Given parameters: - Torque tau = 80sqrt3mathrm~N~m - Angle theta = 60^circ ### Step 1: Calculate the value of MB Substitute the given values into the torque formula: 80sqrt3 = MB sin(60^circ) 80sqrt3 = MB left(fracsqrt32right) MB = 160mathrm~J ### Step 2: Calculate Potential Energy Using the potential energy expression: U = -MB cos(60^circ) U = -160 times frac12 = -80mathrm~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 -tantheta. ### Evaluation Rubric / Model Answer null ### 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 chi_mathrmmg = 1.2times 10^-5 ) is:
  • A. frac65 times 10^-3 \%
  • B. frac56 times 10^-5\%
  • C. frac56 times 10^-4\%
  • D. frac53 times 10^-5\%

Solution

### Related Formula The magnetic field inside a filled solenoid is: B_textnew = mu n I The magnetic field in air is: B_textold = mu_0 n I Relative permeability mu_r and susceptibility chi are related by: mu_r = fracmumu_0 = 1 + chi ### Core Logic The percentage increase in magnetic field is: \% text change in B = fracB_textnew - B_textoldB_textold times 100\% \% text change in B = fracmu - mu_0mu_0 times 100\% = (mu_r - 1) times 100\% = chi times 100\% ### Step 1: Substitute and Calculate Given susceptibility chi_mathrmmg = 1.2 times 10^-5: \% text change = 1.2 times 10^-5 times 100\% = 1.2 times 10^-3\% Convert decimal to fraction: 1.2 times 10^-3\% = frac65 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 chi. Thus, the percentage change is just chi times 100. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Magnetism and Matter
Q17 jee_main_2025_29_jan_evening Magnetic Quantities and Units
Match List-I with List-II. beginarray|l|l|l|l| hline textbfList-I & & textbfList-II & \\ hline text(A) & textMagnetic induction & text(I) & textAmpere meter^2 \\ text(B) & textMagnetic intensity & text(II) & textWeber \\ text(C) & textMagnetic flux & text(III) & textGauss \\ text(D) & textMagnetic moment & text(IV) & textAmpere meter \\ hline endarray Choose the correct answer from the options given below:
  • A. text(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  • B. text(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  • C. text(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  • D. text(A)-(III), (B)-(II), (C)-(I), (D)-(IV)

Solution

### Related Formula textMagnetic Flux: Phi = B cdot A implies textWeber textMagnetic Intensity: H = fracBmu implies textAmpere/meter ### Core Logic Analyzing official units: - **(A) Magnetic induction** rightarrow Gauss (CGS unit) rightarrow **(III)** - **(B) Magnetic intensity** rightarrow Ampere/meter rightarrow (Note: List-II text erroneously says "Ampere meter" or missing solidus bar, actual standard is mathrmAcdot m^-1) rightarrow **(IV)** - **(C) Magnetic flux** rightarrow Weber rightarrow **(II)** - **(D) Magnetic moment** rightarrow textAmpere meter^2 rightarrow **(I)** Due to typo variants in option strings on the primary list sheet, this question was technically **dropped by NTA**. If picking the closest appropriate intended layout configuration, option (2) presents the best approximation. ### Pattern Recognition Note that this question was officially declared dropped by NTA due to printing formatting discrepancies in the unit labels. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Magnetism and Matter

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