Magnetism and Matter JEE Main PYQs (2020–2026)

Solved Magnetism and Matter PYQs

3 recent questions with answers. The full set of 52 is free to practise.

1. JEE Main 2026EasyMagnetization, Magnetic Intensity, and Susceptibility
   Q1

   A solenoid has a core made of material with relative permeability 400 . The magnetic field produced in the interior of solenoid is 1.0 T . The magnetic intensity in SI units is $\times 10^5$ . The value of $\alpha$ is $\_\_\_\_$ .
   
   (Free space permeability ${\mu }_0=4\pi \times 10^{-7}$ SI units.)

   • A$\frac{25}{\pi }$
   • B$\frac{1}{16\pi }$
   • C$\frac{1}{\pi }$
   • D$\frac{1}{4\pi }$

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   Correct answer: (B)

   The relative permeability of the core is ${\mu }_r=400$ .
   
   The magnetic field applied is, $B=1.0T$ .
   
   The permeability of free space is ${\mu }_0=4\pi \times 10^{-7}T\cdot m/A$ .
   
   The magnetic induction $B$ in a material is related to the magnetic intensity $H$ by the total permeability of the medium ( $\mu$ ) :
   
   $$B=\mu H\dots \text{ (i) }$$
   
   The total permeability $\mu$ is the product of the permeability of free space and the relative permeability of the material :
   
   $$\mu ={\mu }_0{\mu }_r$$
   
   Substituting this into the first equation :
   
   $$B={\mu }_0{\mu }_{r}H$$
   
   $$\Rightarrow H=\frac{B}{{\mu }_0{\mu }_r}$$
   
   Substituting the given values,
   
   $$H=\frac{1.0}{\left(4\pi \times 10^{-7}\right)\times 400}$$
   
   $$\Rightarrow H=\frac{1}{1600\pi \times 10^{-7}}$$
   
   $$\Rightarrow H=\frac{1}{16\pi }\times 10^5=a\times 10^5\Rightarrow \alpha =\frac{1}{16\pi }$$
   
   Therefore, the value of $\alpha$ is $\frac{1}{16\pi }$ .
   
   Thus, the correct option is (B).
2. JEE Main 2025EasyMagnetic Dipole, Torque, and Oscillations in a Magnetic Field
   Q2

   Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
   
   Assertion (A): Magnetic monopoles do not exist.
   
   Reason (R): Magnetic field lines are continuous and form closed loops.
   
   In the light of the above statements, choose the most appropriate answer from the options given below:

   • ABoth (A) and (R) are correct and (R) is the correct explanation of (A)
   • Bis not correct but (R) is correct
   • Cis correct but (R) is not correct
   • DBoth (A) and (R) are correct but (R) is not the correct explanation of (A)

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   Correct answer: (A)

   The assertion states that magnetic monopoles do not exist. This is consistent with current experimental findings and the standard formulation of Maxwell’s equations.
   
   The reason provided is that magnetic field lines are continuous and form closed loops. This is mathematically expressed by the Maxwell equation:
   
   $$\nabla \cdot B=0$$
   
   This equation tells us that there are no “sources” or “sinks” of magnetic fields (unlike electric fields, which can originate or terminate at charges). In other words, magnetic field lines neither begin nor end; they loop continuously.
   
   Since the fact that the magnetic field lines form closed loops is a direct consequence of the absence of isolated magnetic charges (monopoles), the reason correctly explains the assertion.
   
   Thus, both (A) and (R) are correct, and (R) is the correct explanation of (A).
   
   The most appropriate answer is Option A.
3. JEE Main 2024MediumHysteresis, Electromagnets, and Permanent Magnets
   Q3

   The coercivity of a magnet is $5\times 10^{3}A/m$ . The amount of current required to be passed in a solenoid of length $30cm$ and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is ________ A.

   Numerical answer type

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   Correct answer: 10

   Coercivity is a measure of the resistance of a ferromagnetic material to becoming demagnetized. It is defined as the intensity of the applied magnetic field required to reduce the magnetization of a material to zero after the magnetization of the sample has been driven to saturation. In this case, coercivity $H_c$ is given to be $5\times 10^{3}A/m$ .
   
   The relationship between the magnetic field $H$ inside a solenoid and the current $I$ passed through it is given by the formula:
   
   $$H=\frac{N\cdot I}{L}$$
   
   where:
   
   $H$ is the magnetic field strength inside the solenoid in amperes per meter (A/m),
   
   $N$ is the total number of turns of wire,
   
   $I$ is the current in amperes (A), and
   
   $L$ is the length of the solenoid in meters (m).
   
   Given that the number of turns of the solenoid $N=150$ and the length of the solenoid $L=30\text{cm}=0.3\text{m}$ , we can rearrange the formula to solve for $I$ :
   
   $$I=\frac{H\cdot L}{N}$$
   
   Substituting the given values:
   
   $$I=\frac{(5\times 10^3)\times 0.3}{150}$$
   
   Simplifying, we get:
   
   $$I=\frac{1500}{150}$$
   
   $$I=10\text{A}$$
   
   Therefore, the amount of current required to be passed in the solenoid for demagnetizing the magnet when inside the solenoid is 10 A.

Topic-wise question count

Where JEE Main has asked from Magnetism and Matter, most frequent first.

• Magnetic Dipole, Torque, and Oscillations in a Magnetic Field14
• Magnetization, Magnetic Intensity, and Susceptibility12
• Magnetic Materials: Dia, Para, and Ferromagnetism11
• Earth's Magnetism9
• Hysteresis, Electromagnets, and Permanent Magnets6

Year-wise Magnetism and Matter questions in JEE Main

Number of questions from this chapter in each year's paper(s).