JEE Main · Physics

Magnetic Effects Of Current And Magnetism – JEE Main PhysicsPractice Questions & PYQs

Generate JEE Main level questions on Magnetic Effects of Current and Magnetism. Focus on Biot-Savart law, Ampere's law, and Lorentz force.

240 questions · 20 PYQs · 0 AI practice · JEE Main 2027

Q101JEE Main 2023MCQ

A long conducting wire having a current I flowing through it, is bent into a circular coil of ${\text{N}}$ turns. Then it is bent into a circular coil of ${\text{n}}$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is : [31-Jan-2023 Shift 2]

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📖 Explanation

$I\;=(2 \pi r) n$ $r\;\propto (\;\frac{I}{n})$ $B\;=n (\;\frac{\mu _0 i}{2 r}) \propto (\;\frac{\mu _0 i}{2 L}) n^2$ $\;\frac{B_1}{B_2}\;= (\;\frac{N^2}{n^2})$

Q102JEE Main 2023MCQ

The magnitude of magnetic induction at mid-point ${\text{O}}$ due to current arrangement as shown in Fig will be : [29-Jan-2023 Shift 1]

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📖 Explanation

Magnetic field due to current in BC and ET are outward at point ' ${\text{O}}$ ' ${\text{B}}_0=\;{\mu _0 {\text{i}}}/{4 \pi {\text{r}}}+\;{\mu _0 {\text{i}}}/{4 \pi {\text{r}}}=\;{\mu _0 {\text{i}}}/{2 \pi {\text{r}}}=\;{\mu _0 {\text{i}}}/{\pi {\text{a}}}$

Q103JEE Main 2022MCQ

A long solenoid carrying a current produces a magnetic field ${\text{B}}$ along its axis. If the current is doubled and the number of turns per ${\text{cm}}$ is halved, the new value of magnetic field will be equal to [25-Jun-2022-Shift-2]

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📖 Explanation

$B=\mu _0 n i$ Now $i \to 2 i$ And $n \to \;\frac{n}{2}$ $B^{'}=\mu _0 \;\frac{n}{2} \times 2 i=\mu _0 n i=B$

Q104JEE Main 2022MCQ

A charge particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charge particle increases to 4 times its initial value. What will be the ratio of new radius to the original radius of circular path of the charge particle : [29-Jun-2022-Shift-1]

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📖 Explanation

$R=\;\frac{m v}{B q}=\;{{\sqrt{2 m K}}}/{B q}$ $\Rightarrow R \propto {\sqrt{K}}$ $\Rightarrow$ ratio $=2: 1$

Q105JEE Main 2022MCQ

Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is $6: 5$ and their respective masses ratio is $9: 4$ . Then, the ratio of their charges will be : [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

We know that $R=\;\frac{m v}{B q}=\sqrt{\;\frac{2 m K}{B q}}$ $\Rightarrow$ Ratio of radii $=\;\frac{R_{1}}{R_{2}}=\sqrt{\;\frac{m_{1}}{m_{2}}} \;\frac{q_{2}}{q_{1}}$ $\Rightarrow \;\frac{6}{5}=\sqrt{\;\frac{9}{4}} \;\frac{q_{2}}{q_{1}}$ $\Rightarrow \;\frac{q_{1}}{q_{2}}=\;\frac{3}{2} \times \;\frac{5}{6}=\;\frac{5}{4}$

Q106JEE Main 2022MCQ

A cyclotron is working at a frequency of $10 {\text{MHz}}$ . If the radius of its dees is $60 {\text{cm}}$ . The maximum kinetic energy of accelerated proton will be : $(.$ Take $.: {\text{e}}=1.6 \times 10^{-19} {\text{C}}, {\text{m}}_{ {\text{p}}}=1.67 \times 10^{-27} {\text{kg}})$ [30-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Given, $f=10 \times 10^6 {\text{Hz}}$ $r=0.6 {\text{m}}$ Charge on proton $(q)=e$ We know,Radius $(r)=\;\frac{m v}{q B}$ $=\;{{\sqrt{2 m k}}}/{e B}$ Also, we know,Cyclotron oscillation frequency should be equal to the pendulum evolution frequency. $\therefore f=\;\frac{e B}{2 \pi m}$ $\Rightarrow e B=2 \pi m f$ Putting this value of eB in equation (1) we get $r=\;{{\sqrt{2 m k}}}/{2 \pi m f}$ $\Rightarrow r^2=\;\frac{2 m k}{4 \pi ^2 m^2 f^2}$ $\Rightarrow k=2 \pi ^2 m f^2 r^2$ $=2 \times (\;\frac{22}{7}) ^2 \times 1.67 \times 10^{-27} \times (10 \times 10^6) ^2 \times (0.6)^2 {\text{J}}$ $=1.2 \times 10^{-12} {\text{J}}$ $=\;\frac{1.2 \times 10^{-12}}{1.6 \times 10^{-19}} {\text{eV}}$ $=\;\frac{12}{16} \times 10^7 {\text{eV}}$ $=0.75 \times 10^7 {\text{eV}}$ $=7.5 \times 10^6 {\text{eV}}$ $=7.5 {\text{MeV}}$

Q107JEE Main 2022MCQ

Two parallel, long wires are kept $0.20 {\text{m}}$ apart in vacuum, each carrying current of $x {\text{A}}$ in the same direction. If the force of attraction per meter of each wire is $2 \times 10^{-6} {\text{N}}$ , then the value of ${\text{x}}$ is approximately : [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{d F}{d l}=2 \times 10^{-6} {\text{N}} / {\text{m}}=\;\frac{\mu _0 i_1 i_2}{2 \pi d}$ $2 \times 10^{-6}=\;\frac{2 \times 10^{-7} \times x^2}{0.2}$ $x={\sqrt{2}} ≃ 1.4$

Q108JEE Main 2022MCQ

A wire $X$ of length $50 {\text{cm}}$ carrying a current of $2 {\text{A}}$ is placed parallel to a long wire $Y$ of length $5 {\text{m}}$ . The wire ${\text{Y}}$ carries a current of $3 {\text{A}}$ . The distance between two wires is $5 {\text{cm}}$ and currents flow in the same direction. The force acting on the wire $Y$ is [29-Jul-2022-Shift-2]

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📖 Explanation

Length of wire $Y$ is very high compared to length of wire $X$ . So we can assume length of wire $Y$ is infinite compare to wire $X$ .Magnetic field due to wire $y$ from $5 {\text{cm}}$ from wire towards wire $X$ is, $B=\;\frac{\mu _0 I_1}{2 \pi d}=\;\frac{\mu _0 \times 3}{2 \pi \times 0.05}$ And direction of $B$ is outward.Now force on wire $X$ due to wire $Y$ , $F_{Y X}=I_2 B l$ $=2 \times \;\frac{\mu _0 \times 3}{2 \pi \times 0.05} \times 0.5$ $=1.2 \times 10^{-5} {\text{N}}$ Current flowing in both the wire in same direction so both wire will attract each other with equal force. $\therefore F_{Y X}=F_{X Y}$ $\therefore F_{X Y}=1.2 \times 10^{-5} N$ , attractive force towards wire $X$ .

Explanation Figure 2

Q109JEE Main 2022MCQ

The magnetic field at the center of current carrying circular loop is $B_1$ . The magnetic field at a distance of $\sqrt{3}$ times radius of the given circular loop from the center on its axis is $B_2$ . The value of $B_1 / B_2$ will be [28-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$B_1=\;\frac{\mu _0 i}{2 R}$ $B_2=\;{\mu _0 i R^2}/{2(R^{2+}x^2)^{\;\frac{3}{2}}}$ $\Rightarrow \;\frac{B_1}{B_2}=\;\frac{1}{R^3}(R^{2+}x^2)^{\;\frac{3}{2}}$ $=\;\frac{1}{R^3}(8 R^3)$ $=8$

Q110JEE Main 2022MCQ

The magnetic field at the centre of a circular coil of radius $r$ , due to current I flowing through it, is $B$ . The magnetic field at a point along the axis at a distance $\;\frac{r}{2}$ from the centre is : [24-Jun-2022-Shift-1]

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📖 Explanation

$B=\;\frac{\mu _0 I}{2 r}$ $B_a=\;\frac{\mu _0 I r^2}{2(r^{2+}\;\frac{r^2}{4})}$ $\Rightarrow \;\frac{B_a}{B}=(\;\frac{2}{\sqrt{5}})^3$ $\Rightarrow B_a=(\;\frac{2}{\sqrt{5}})^3 B$

Q111JEE Main 2022MCQ

The current sensitivity of a galvanometer can be increased by : (A) decreasing the number of turns (B) increasing the magnetic freld (C) decreasing the area of the coil (D) decreasing the torsional constant of the spring Choose the most appropriate answer from the options given below : [28-Jul-2022-Shift-1]

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📖 Explanation

$N i A B=k \theta$ $\Rightarrow \;\frac{\theta }{i}=\;\frac{N A B}{k}$ $\Rightarrow$ Sensitivity increases if $B ↑$ and $k ↓$

Q112JEE Main 2022MCQ

Given below are two statements :Statement I : The electric force changes the speed of the charged particle and hence changes its kinetic energy; whereas the magnetic force does not change the kinetic energy of the charged particle.Statement II : The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field.In the light of the above statements, choose the most appropriate answer from the options given below : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Electric field accelerates the particle in the direction of field $(\vec{F}=q \vec{E}=m \vec{a})$ and magnetic field accelerates the particle perpendicular to the field $(\vec{F}=\vec{q v} \times \vec{B}=m \vec{a})$ .

Q113JEE Main 2022MCQ

A long straight wire with a circular cross-section having radius ${\text{R}}$ , is carrying a steady current ${\text{I}}$ . The current ${\text{I}}$ is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance $r(r< R)$ from its centre will be : [25-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$∫ \overline{B} \cdot \overline{d l}=\mu _0 I_{i n}$ $\Rightarrow B \times 2 \pi r=\;\frac{\mu _0 I}{\pi R^2} \times \pi r^2$ $\Rightarrow B \propto r$

Q114JEE Main 2022NAT

A singly ionized magnesium atom $( {\text{A}}=24)$ ion is accelerated to kinetic energy $5 {\text{keV}}$ , and is projected perpendicularly into a magnetic field ${\text{B}}$ of the magnitude $0.5 {\text{T}}$ . The radius of path formed will be _____ ${\text{cm}}$ . [28-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$R=\;\frac{m v}{q B}$ $R=\;{{\sqrt{2 m K E}}}/{q B}$ $=\;{\sqrt{2 \times 24 \times 1.67 \times 10^{-27} \times 5 \times 1.6 \times 10^{-16}}}/{1.6 \times 10^{-19} \times 0.5}$ $=10.009 {\text{cm}}=10 {\text{cm}}$

Q115JEE Main 2022MCQ

A charge particle is moving in a uniform magnetic field $(2\hat{i} + 3\hat{j}) \text{T}$ . If it has an acceleration of $(\alpha \hat{i} - 4\hat{j}) \text{m/s}^2$ , then the value of $\alpha$ will be : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

As magnetic force is perpendicular to magnetic fieldSo, $\vec{F} \cdot \vec{B}$ must be 0So, $2 \alpha -12=0$ $\alpha =6$

Q116JEE Main 2022MCQ

A proton, a deutron and an $\alpha$ -particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is : [24-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\therefore r=\;\frac{m v}{q B}=\;\frac{\sqrt{2 m(K E)}}{q B}$ $\Rightarrow r_1: r_2: r_3=\;\frac{\sqrt{m_1}}{q_1}: \;\frac{\sqrt{m_2}}{q_2}: \;\frac{\sqrt{m_3}}{q_3}$ $=\;\frac{\sqrt{1}}{1}: \;\frac{\sqrt{2}}{1}: \;\frac{\sqrt{4}}{2}$ $=1: \sqrt{2}: 1$

Q117JEE Main 2022MCQ

A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$R=\;\frac{m v}{q B}$ $\;{ {\text{R}}_\alpha }/{ {\text{R}}_{ {\text{P}}}}=\;{ {\text{M}}_\alpha }/{ {\text{M}}_{ {\text{P}}}} \times \;{ {\text{q}}_{ {\text{P}}}}/{ {\text{q}}_\alpha }$ $\;{ {\text{R}}_\alpha }/{ {\text{R}}_{ {\text{P}}}}=\;\frac{4}{1} \times \;\frac{1}{2}=2$

Q118JEE Main 2022MCQ

Two concentric circular loops of radii $r_1=30 \text{cm}$ and $r_2=50 \text{cm}$ are placed in $\text{X}-\text{Y}$ plane as shown in the figure. A current $I=7 \text{A}$ is flowing through them in the direction as shown in figure. The net magnetic moment of this system of two circular loops is approximately : [26-Jul-2022-Shift-2]

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Q119JEE Main 2022NAT

A closely wounded circular coil of radius $5 {\text{cm}}$ produces a magnetic field of $37.68 \times 10^{-4} {\text{T}}$ at its center. The current through the coil is _______ A. [Given, number of turns in the coil is 100 and $\pi =3.14$ ] [29-Jul-2022-Shift-1]

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📖 Explanation

$B=\;\frac{\mu _0 n I}{2 R}$ $37.68 \times 10^{-4}=\;\frac{4 \pi \times 10^{-7} 100 I}{2 \times 5 \times 10^{-2}}$ $I=\;\frac{300 A}{100}$ $=3 {\text{A}}$

Q120JEE Main 2022NAT

A deuteron and a proton moving with equal kinetic energy enter into a uniform magnetic field at right angle to the field. If $r_d$ and ${\text{r}}_p$ are the radii of their circular paths respectively, then the ratio $\;\frac{r_d}{r_p}$ will be $\sqrt{x}: 1$ where ${\text{x}}$ is_____ [27-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$R=\;{{\sqrt{2 m K}}}/{q B}$ So, $\;\frac{r_d}{r_p}=\;{{\sqrt{m_d}} / q_d}/{{\sqrt{m_p}} / q_p}$ $={\sqrt{2}}$ So $x=2$

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