Magnetic Effects of Current JEE Main previous year questions with solutions

5 solved JEE Main questions on Magnetic Effects of Current, free to read — no sign-in needed. The full chapter has 263 questions; sign in to attempt the remaining 258 in the exam simulator.

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1. Q1JEE Main 2026 (08 Apr, Shift 2)Force and Torque on Current Carrying Conductor
   A current carrying circular loop of radius $2$ cm with unit normal $\hat{n} = \dfrac{\hat{k}+\hat{i}}{\sqrt{2}}$ is placed in a magnetic field, $\vec{B} = B_0(3\hat{i}+2\hat{k})$. If $B_0 = 4\times 10^{-3}$ T and current $I=100\sqrt{2}$ A, the torque experienced by the loop is ________ Wb·A. ($\pi=3.14$)
   1. A.$16\times 10^{-5}\,\hat{k}$
   2. B.$5024\times 10^{-7}\,\hat{k}$
   3. C.$5024\times 10^{-7}\,\hat{i}$
   4. D.$5024\times 10^{-7}\,\hat{j}$
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   Answer: (D)

   The magnetic dipole moment of the current carrying loop is given by $\vec{M} = I A \hat{n}$. Given radius $r = 2$ cm $= 2 \times 10^{-2}$ m, the area of the loop is: $A = \pi r^2 = \pi (2 \times 10^{-2})^2 = 4\pi \times 10^{-4}$ m$^2$ Substituting the given values $I = 100\sqrt{2}$ A and $\hat{n} = \dfrac{\hat{i}+\hat{k}}{\sqrt{2}}$: $\vec{M} = (100\sqrt{2}) \times (4\pi \times 10^{-4}) \times \left(\dfrac{\hat{i}+\hat{k}}{\sqrt{2}}\right)$ $\vec{M} = 4\pi \times 10^{-2} (\hat{i}+\hat{k})$ A$\cdot$m$^2$ The torque experienced by the loop in the magnetic field is $\vec{\tau} = \vec{M} \times \vec{B}$. Given $\vec{B} = 4 \times 10^{-3} (3\hat{i}+2\hat{k})$ T, we have: $\vec{\tau} = [4\pi \times 10^{-2} (\hat{i}+\hat{k})] \times [4 \times 10^{-3} (3\hat{i}+2\hat{k})]$ $\vec{\tau} = 16\pi \times 10^{-5} [(\hat{i}+\hat{k}) \times (3\hat{i}+2\hat{k})]$ Evaluating the cross product: $(\hat{i}+\hat{k}) \times (3\hat{i}+2\hat{k}) = (\hat{i} \times 3\hat{i}) + (\hat{i} \times 2\hat{k}) + (\hat{k} \times 3\hat{i}) + (\hat{k} \times 2\hat{k})$ $= 0 + 2(-\hat{j}) + 3(\hat{j}) + 0 = \hat{j}$ Therefore, the torque is: $\vec{\tau} = 16\pi \times 10^{-5} \hat{j}$ Substituting $\pi = 3.14$: $\vec{\tau} = 16 \times 3.14 \times 10^{-5} \hat{j} = 50.24 \times 10^{-5} \hat{j} = 5024 \times 10^{-7} \hat{j}$ Wb$\cdot$A
2. Q2JEE Main 2025 (03 Apr, Shift 2)Magnetic Dipole
   A magnetic dipole experiences a torque of $80 \sqrt{3} \mathrm{~N} \mathrm{~m}$ 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 :
   1. A.$80 \mathrm{~J}$
   2. B.$-40 \sqrt{3} \mathrm{~J}$
   3. C.$-60 \mathrm{~J}$
   4. D.$-80 \mathrm{~J}$
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   Answer: (D)

   $\tau=\mathrm{M} \times \mathrm{B}=\mathrm{MB} \sin 60=\frac{\sqrt{3}}{2} \mathrm{MB}=80 \sqrt{3}$ $\begin{aligned} & \mathrm{MB}=160 \\ & \mathrm{U}=-\mathrm{M} \cdot \mathrm{B}=-\mathrm{MB} \cos 60 \\ & \mathrm{U}=-160 \times 1 / 2=-80 \mathrm{~J}\end{aligned}$
3. Q3JEE Main 2024 (08 Apr, Shift 2)Magnetic Field
   A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at $\frac{a}{2}$ and $2 a$ from axis of the wire is :
   1. A.$1: 4$
   2. B.$1: 1$
   3. C.$3: 4$
   4. D.$4: 1$
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   Answer: (B)

   $\begin{aligned} & \mathrm{B}_1 2 \pi \frac{\mathrm{a}}{2}=\mu_{\mathrm{o}} \frac{\mathrm{I}}{4} \\ & \mathrm{~B}_1=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi \mathrm{a}} \\ & \mathrm{B}_2 2 \pi 2 \mathrm{a}=\mu_{\mathrm{o}} \mathrm{I} \\ & \mathrm{B}_2=\frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \pi \mathrm{a}}\end{aligned}$
4. Q4JEE Main 2023 (08 Apr, Shift 2)Biot-Savart's Law and Ampere's Circuital Law
   The ratio of magnetic field at the centre of a current carrying coil of radius $r$ to the magnetic field at distance $r$ from the centre of coil on its axis is $\sqrt{x}:1$. The value of $x$ is _____.
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   Answer: 8

   The formula to calculate the magnetic field at any axial point situated at a distance $x$ from the centre of a circular current carrying conductor of radius $r$ can be written as $B=\dfrac{{\mu }_{0}I{r}^{2}}{2{\left({r}^{2}+{x}^{2}\right)}^{3/2}}...\left(1\right)$ Substitute $0$ for $x$ in equation (1) to calculate the magnetic field $\left({B}_{c}\right)$ at the centre of the conductor. $\begin{matrix}{B}_{c} & = & \dfrac{{\mu }_{0}I{r}^{2}}{2{\left({r}^{2}+{0}^{2}\right)}^{3/2}} \\ & = & \dfrac{{\mu }_{0}I{r}^{2}}{2{r}^{3}} \\ & = & \dfrac{{\mu }_{0}I}{2r}...\left(2\right)\end{matrix}$ Substitute $r$ for $x$ in equation (1) to calculate the magnetic field $\left({B}_{r}\right)$ at the given distance from the centre of the conductor. $\begin{matrix}{B}_{r} & = & \dfrac{{\mu }_{0}I{r}^{2}}{2{\left({r}^{2}+{r}^{2}\right)}^{3/2}} \\ & = & \dfrac{{\mu }_{0}I{r}^{2}}{4\sqrt{2}{r}^{3}} \\ & = & \dfrac{{\mu }_{0}I}{4\sqrt{2}r}...\left(3\right)\end{matrix}$ Divide equation (2) by equation (3) to calculate the required ratio of the magnetic fields. $\begin{matrix}\dfrac{{B}_{c}}{{B}_{r}} & = & \dfrac{\dfrac{{\mu }_{0}I}{2r}}{\dfrac{{\mu }_{0}I}{4\sqrt{2}r}} \\ & = & 2\sqrt{2} \\ & = & \sqrt{8}\end{matrix}$
5. Q5JEE Main 2022 (25 Jul, Shift 1)Magnetic Moment
   The magnetic moment of an electron $\left(e\right)$ revolving in an orbit around nucleus with an orbital angular momentum is given by
   1. A.${\vec{\mu }}_{L}=\dfrac{eL}{2m}$
   2. B.${\vec{\mu }}_{L}=-\dfrac{eL}{2m}$
   3. C.${\vec{\mu }}_{L}=-\dfrac{e\vec{L}}{m}$
   4. D.${\vec{\mu }}_{L}=\dfrac{2e\vec{L}}{m}$
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   Answer: (B)

   When a charged body is rotating in a circle, the ratio of magnetic moment and angular momentum is given by, $\dfrac{\vec{\mu }}{\vec{L}}=\dfrac{q}{2m}$ The charge of electron $q=-e$ $\vec{\mu }=-\dfrac{e}{2m}\vec{L}$