Moving Charges and Magnetism JEE Main PYQs (2020–2026)

Solved Moving Charges and Magnetism PYQs

3 recent questions with answers. The full set of 171 is free to practise.

1. JEE Main 2026MediumMagnetic Field due to Current (Biot-Savart Law)
   Q1

   A small cube of side 1 mm is placed at the centre of a circular loop of radius 10 cm carrying a current of 2 A . The magnetic energy stored inside the cube is $\times 10^{-14}J$ . The value of $\alpha$ is $\_\_\_\_$ .
   
   $$\left({\mu }_o=4\pi \times 10^{-7}Tm/A,\pi =3.14\right)$$

   • A6.28
   • B$6.28\times 10^{-6}$
   • C628
   • D$6.28\times 10^{-4}$

   Show answer & solution

   Correct answer: (A)

   The magnetic field produced at the center of a circular wire of radius $r$ carrying current $I$ is:
   
   $$B=\frac{{\mu }_{0}I}{2r}$$
   
   The electric current in the circular loop is $I=2A$ , the radius of the loop is $r=10cm=0.1m$ .
   
   $$B=\frac{4\pi \times 10^{-7}\times 2}{2\times 0.1}=4\pi \times 10^{-6}T$$
   
   Magnetic energy is stored in the space where a magnetic field exists. The energy per unit volume is:
   
   $$u_B=\frac{B^2}{2{\mu }_0}$$
   
   For a very small volume V where the field B is approximately uniform:
   
   $$U=u_B\times V=\frac{B^2}{2{\mu }_0}\times V$$
   
   The volume of the cube is,
   
   $$V=(1mm{)}^3={\left(10^{-3}m\right)}^3=10^{-9}{m}^3$$
   
   So, the total energy stored in the cube is,
   
   $$U=\frac{{\left(4\pi \times 10^{-6}\right)}^2}{2\times 4\pi \times 10^{-7}}\times 10^{-9}$$
   
   $$\Rightarrow U=\frac{16{\pi }^2\times 10^{-12}}{8\pi \times 10^{-7}}\times 10^{-9}=2\pi \times 10^{-5}\times 10^{-9}$$
   
   $$\Rightarrow U=2\times 3.14\times 10^{-14}=6.28\times 10^{-14}J=\alpha \times 10^{-14}J$$
   
   $$\Rightarrow \alpha =6.28$$
   
   Therefore, the correct option is (A).
2. JEE Main 2026EasyMotion of Charged Particles in Electric and Magnetic Fields
   Q2

   The charged particle moving in a uniform magnetic field of $(3\hat{i}+2\hat{j})T$ has an acceleration $\left(4\hat{i}-\frac{x}{2}\hat{j}\right)m/s^2$ . The value of $x$ is

   Numerical answer type

   Show answer & solution

   Correct answer: 12

   The force ( $\overrightarrow{F}$ ) acting on a particle with charge $q$ moving with velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$ is given by the Lorentz force law:
   
   $$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$$
   
   According to Newton’s second law, $\overrightarrow{F}=m\overrightarrow{a}$ .
   
   Therefore,
   
   $$m\overrightarrow{a}=q(\overrightarrow{v}\times \overrightarrow{B})$$
   
   The resulting vector is always perpendicular to both $\overrightarrow{v}$ and $\overrightarrow{B}$ .
   
   Since $\overrightarrow{a}$ is in the same direction as $\overrightarrow{F}$ , the acceleration vector $\overrightarrow{a}$ must be perpendicular to the magnetic field vector $\overrightarrow{B}$ .
   
   $$\overrightarrow{a}\perp \overrightarrow{B}$$
   
   If two vectors are perpendicular, their dot product must be zero:
   
   $$\overrightarrow{a}\cdot \overrightarrow{B}=0$$
   
   The magnetic field is given as $\overrightarrow{B}=3\hat{ı}+2\hat{ȷ}$ and $\overrightarrow{a}=4\hat{ı}-\frac{x}{2}\hat{ȷ}$ :
   
   $$\left(4\hat{ı}-\frac{x}{2}\hat{ȷ}\right)\cdot (3\hat{ı}+2\hat{ȷ})=0$$
   
   $$\Rightarrow (4\times 3)+\left(-\frac{x}{2}\times 2\right)=0$$
   
   $$\Rightarrow 12-x=0$$
   
   $$\Rightarrow x=12$$
   
   Therefore, the value of x is 12 .
3. JEE Main 2025MediumAmpere's Circuital Law and Solenoids
   Q3

   A solenoid having area A and length ’ $l$ ’ is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is :

   • A$\frac{B^{2}Al}{4{\mu }_0}$
   • B$\frac{B^{2}Al}{2{\mu }_0}$
   • C$\frac{B^{2}Al}{{\mu }_0}$
   • D$B^{2}Al$

   Show answer & solution

   Correct answer: (A)

   To determine the magnetic energy stored in a solenoid filled with a material of relative permeability 2, we start with the expression for energy density in a magnetic field:
   
   $$\frac{U}{V}=\frac{B^2}{2{\mu }_r{\mu }_0}$$
   
   Given that the relative permeability ${\mu }_r$ is 2, this becomes:
   
   $$\frac{U}{V}=\frac{B^2}{2\times 2{\mu }_0}=\frac{B^2}{4{\mu }_0}$$
   
   The energy $U$ stored in the solenoid can be expressed as:
   
   $$U=\frac{B^2}{4{\mu }_0}\times V$$
   
   Here, $V$ is the volume of the solenoid, calculated as $A\times l$ (where $A$ is the cross-sectional area and $l$ is the length). Therefore, substituting for $V$ , we get:
   
   $$U=\frac{B^2}{4{\mu }_0}\times Al$$
   
   This formula gives us the magnetic energy stored in the solenoid.

Topic-wise question count

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

• Motion of Charged Particles in Electric and Magnetic Fields54
• Magnetic Field due to Current (Biot-Savart Law)52
• Ampere's Circuital Law and Solenoids24
• Magnetic Dipole Moment and Torque on a Current Loop18
• Force on a Current-Carrying Conductor in a Magnetic Field12
• Force between Parallel Current-Carrying Conductors6
• Moving Coil Galvanometer5

Year-wise Moving Charges and Magnetism questions in JEE Main

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