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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

Q181JEE Main 2019MCQ

A rigid square loop of side 'a' and carrying current $I_2$ is lying on a horizontal surface near a long current $I_1$ carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be:

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📖 Explanation

$\ {F_{A B}}^ {---- }+\ {F_{D C}}^{----}=0$ $F_{A D}=B I_{2} a$ $=\ \frac{\mu_{0} I}{2 \pi (\alpha )} I_{2}(\alpha )$ (right wards) $=\ \frac{\mu I_{1} I_{2}}{2 \pi }$ $F_{B C}=B^{ ,} I_{2} a$ $=\ \frac{\mu_{0} I_{1} I_{2}}{4 \pi }$ (up + wards) $F_{n e t}=\ \frac{\mu_{0} I_{1} I_{2}}{2 \pi }-\ \frac{\mu_{0} I_{1} I_{2}}{4 \pi }$ $=\ \frac{\mu _{0} I_{1} I_{2}}{4 \pi}$ (right wards ev (repulsive)

Q182JEE Main 2019MCQ

Two very long, straight, and insulated wires are kept at 90° angle from each other in xy-plane as shown in the figure. These wires carry currents of equal magnitude I, whose directions are shown in the figure. The net magnetic field at point P will be:

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📖 Explanation

$B_{1}= ( \frac{\mu_{0} I}{2 \pi d} )(- \vec{K})$ $\Rightarrow B_{2}= ( \frac{\mu_{0} I}{2 \pi d} )( \vec{K}) \Rightarrow B_{p}=0$

Q183JEE Main 2019MCQ

A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 $\pi \text{ rad s}^{-1}$ about its axis, perpendicular to its plane. If the magnetic field at its centre is $3.8 \times 10^{-9} \text{ T}$ , then the charge carried by the ringis close to $(\mu_0 = 4\pi \times 10^{-7} \text{ N/A}^2)$

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📖 Explanation

$B= \frac{\mu_{0} q \omega }{r 2 \pi }$ $3.8\times 10^{-9}= \frac{4 \pi \times 10^{-7}\times \rho \times 40 \pi }{0.1\times 2 \pi }$ $\rho =2\times 10^{-6} C$

Q184JEE Main 2019MCQ

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10A is: (take $\mu_{0}=4\pi \times 10^{-7} \text{N A}^{-2}$ )

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📖 Explanation

$B^{1}=3 B$ $=3 \times {\mu_{0} I}/{4 \pi \times \frac{l}{2 \\sqrt{3}}}\ (\sin 60^{\0}+\sin 60^{\0}\ )$ $=18 \\times 10^{-6}$ Tesla $=18 \muT$

Explanation Figure 1

Q185JEE Main 2019MCQ

A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field $B \hat{i}$ . The torque on the coil due to the magnetic field is:

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📖 Explanation

${\tau ^{\to }}= \vec{\mu}\times \vec{B}=N (\pi r^{2}\ ) I(B)$ $\vec{\tau }=B N \pi r^{2} I$

Q186JEE Main 2019MCQ

A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be :

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📖 Explanation

$m=l^{2} I$ $m^{1}=\pi r^{2} I, B u t\ 2 \pi r=4 l$ $r=\ \frac{4 l}{2 \pi }= \frac{2 l}{\pi }$ So, $\ \frac{m^{1}}{m}=\ \frac{\pi r^{2} I}{l^{2} I}$ $=\ {\pi \times \frac{4 l^{2}}{\pi ^{2}}}/{l^{2}}=\ \frac{4}{\pi }$ $\Rightarrowm^{1}=\ \frac{4 m}{\pi }$

Explanation Figure 1

Explanation Figure 2

Q187JEE Main 2019MCQ

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magneticfield in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are $T_h$ and $T_c$ respectively, then :

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📖 Explanation

T = 2π $\sqrt{I/{\mu B}}$ $T_h$ = 2π $\sqrt{\frac{mR^2}{(2\mu )B}}$ $T_c$ = 2π $\sqrt{\frac{1/2mR^2}{\mu B}}$

Q188JEE Main 2019MCQ

A moving coil galvanometer has a coil with 175 turns and area 1 cm $^{2}$ . It uses a torsion band of torsion constant 10 $^{-6}$ N - m / rad . The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 1o for a current of 1 mA. The value of B (in Tesla) is approximately:

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📖 Explanation

$\tau =K \theta =N t A B$ $\Rightarrow 10^{-6} (1 .\frac{\pi }{180}\ ) \;\;\;=175\times 10^{-3} \\times 10^{-4} \times B$ $\Rightarrow B \approx 10^{-3} T$

Q189JEE Main 2019MCQ

A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given chargeof electron = 1.6 × $10^{-19}$ C)

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📖 Explanation

${mv^2}/R$ = qvB mv = qBR ... (i) Path is straight line it qE = qvB E = vB ... (ii) From equation (i) & (ii) m = ${qB^2R}/E$ m = 2.0 × $10^{-24}$ kg

Q190JEE Main 2019MCQ

A rectangular coil (Dimension $5 cm\times 2.5cm$ ) with 100 turns, carrying a current of 3A in the clock-wise direction, is kept centered at the origin and in the X-Z plane. Amagnetic field of 1 T is applied along X-axis. If the coil is tilted through 45° aboutZ-axis, then the torque on the coil is

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📖 Explanation

Torque $=M \\times B$ $=N x i \\times A \\times B \\times \sin 4 J$ $=100 \\times 3 \\times 12.5 \times 10^{-4} \\times 1 \\times 1 \frac{1}{\sqrt{2}}$ $\tau =0.265$

Q191JEE Main 2019MCQ

An infinitely long current carrying wire and asmall current carrying loop are in the plane ofthe paper as shown. The radius of the loop is aand distance of its centre from the wire is d (d ≫ a).If the loop applies a force F on the wire then : [9-Jan-2019 Shift 1]

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📖 Explanation

inifinitely long wire Eqvilent dipole of given loop F = m . $\frac{dB}{dr}$ Now $\frac{dB}{dx}$ = $d/{dx} (\frac{\mu _0I}{2\pi x})$ α $1/x^2$ \Rightarrow So F α $M/x^2$ [Since M = NIA] ∴ F α $a^2/d^2$

Explanation Figure 2

Q192JEE Main 2019MCQ

In an experiment electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. $[Charge of the electron = 1.6 \times $10^{-19}$ C Mass of the electron = 9.1 \times $10^{-31}$ kg]$

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📖 Explanation

r = $\sqrt\frac{2mk}{eB}$ = $\sqrt\frac{2me\Delta v}{eB}$ r = $\sqrt{{2m}/e . \Delta v}/B$ = $\sqrt{\frac{2\times 9.1\times 10^{-31}}{1.6\times 10^{-19}} (500)} /{100\times 10^{-3}}$ r = $\sqrt\frac{\frac{9.1}{0.16} \times 10^{-10}}{10^{-1}}$ = $3/4 \times 10^{-4}$ = 7.5 × $10^{-4}$

Q193JEE Main 2019MCQ

A proton,an electron, and a Helium nucleus, have the same energy. they are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let $r_{p}, r_{e} \text{and} r_{H e}$ be their respective radii, then,

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📖 Explanation

The radius of a charged particle in a magnetic field is $r = \frac{mv}{qB}$ . Using the kinetic energy relation $K = \frac{1}{2}mv^2$ , we substitute $v = \sqrt{\frac{2K}{m}}$ to get $r = \frac{\sqrt{2mK}}{qB}$ . Since $K$ and $B$ are constant, $r \propto \frac{\sqrt{m}}{q}$ .
For the proton, electron, and Helium nucleus, the ratios are:
$r_p \propto \frac{\sqrt{m_p}}{e}$ , $r_e \propto \frac{\sqrt{m_e}}{e}$ , and $r_{He} \propto \frac{\sqrt{m_{He}}}{2e}$ .
Using $m_{He} \approx 4m_p$ , the ratio for Helium is $r_{He} \propto \frac{\sqrt{4m_p}}{2e} = \frac{2\sqrt{m_p}}{2e} = \frac{\sqrt{m_p}}{e}$ .
Since $m_e < m_p$ , it follows that $r_e < r_p$ .
Because $\frac{\sqrt{m_p}}{e} = \frac{\sqrt{m_p}}{e}$ , we have $r_p = r_{He}$ .
Thus, $r_e < r_p = r_{He}$ .

Q194JEE Main 2019MCQ

The region between y = 0 and y = d contains a magnetic field $\vec{B} = B\hat{k}$ . A particle of mass m and charge q enters the region with a velocity $\vec{v} = v\hat{i}$ . If d = $\frac{mv}{2qB}$ , the acceleration of thecharged particle at the point of its emergenceat the other side is :-

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📖 Explanation

The magnetic force on the particle is given by the Lorentz force law:
$\vec{F} = q(\vec{v} \times \vec{B})$
Given the velocity $\vec{v} = v \hat{i}$ and the magnetic field $\vec{B} = B \hat{z}$ (where $\hat{z} = \hat{k}$ ), the force at the point of entry is:
$\vec{F} = q(v \hat{i} \times B \hat{k}) = -qvB \hat{j}$
The acceleration $\vec{a}$ of the particle is:
$\vec{a} = \frac{\vec{F}}{m} = -\frac{qvB}{m} \hat{j}$
Because the force has a negative $y$ -component, the particle immediately exits the region $0 \le y \le d$ at the point of entry. As $-\frac{qvB}{m} \hat{j}$ does not match any of the provided options (A-D), the correct choice is E.

Q195JEE Main 2019MCQ

One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop( $B_L$ ) to that at the centre of the coil ( $B_C$ ), i.e. R $B_L/B_C$ will be :

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📖 Explanation

L = 2πR , L = N × 2π R = Nr $B_L$ = $\frac{\mu _0i}{2R}$ , $B_C$ = $\frac{\mu _0Ni}{2r}$ $B_C$ = $\frac{\mu _0N^2i}{2R}$ $B_L/B_C$ = $1/N^2$

Explanation Figure 1

Q196JEE Main 2019MCQ

An insulating thin rod of length l has a x linear charge density p(x) = $\rho _0$ $x/l$ on it. The rod is rotated about an axis passing through the origin (x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magneticmoment of the rod is :

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📖 Explanation

Since M = NIA dq = λdx & A = $\pi x^2$ ∫ dm = ∫ (x) ${\rho _0x}/l$ dx . $\pi x^2$ M = ${n\rho _0\pi }/l$ . $∫^{l}_{0}x^3$ . dx = ${n\rho _0\pi }/l . [L^4/4]$ M = ${n\rho _0\pi l^3}/4$ or $\pi /4 n\rho l^3$

Q197JEE Main 2019MCQ

A particle of mass m and charge q is in an electric and magnetic field given by $\vec{E} = 2\hat{i} + 3\hat{j}$ ; $\vec{B} = 4\hat{j} + 6\hat{k}$ . The charged particle is shifted from the origin to the point P(x = 1 ; y = 1) along a straight path. The magnitude of the total work done is :-

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📖 Explanation

$F_\vec{\text{net}}$ = $q\vec{E}$ + $q(\vec{v}\times \vec{B})$ = $(2q\hat{i}+3q\hat{j})$ + q $(\vec{v}\times \vec{B})$ W = $F_\vec{\text{net}} . \vec{S}$ = 2q + 3q = 5q

Q198JEE Main 2018MCQ

A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity ω with respect to normal axis then the magnetic moment of the loop is :[Main 16 April 2018 S1]

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📖 Explanation

The magnetic moment $\mu$ of a current loop is defined as $\mu = I A$ , where $I$ is the current and $A$ is the area of the loop.
The current $I$ produced by the rotating charge is $I = \frac{q}{T}$ , where $T$ is the time period of one revolution.
Given the angular velocity $\omega = \frac{2\pi}{T}$ , we find $T = \frac{2\pi}{\omega}$ , so $I = \frac{q\omega}{2\pi}$ .
The area $A$ of the loop with radius $r$ is $A = \pi r^2$ .
Substituting $I$ and $A$ into the magnetic moment formula:
$\mu = \left( \frac{q\omega}{2\pi} \right) (\pi r^2) = \frac{1}{2} q \omega r^2$

Q199JEE Main 2018MCQ

A Helmholtz coil has a pair of loops, each with N turns and radius R. They are placed coaxially at distance R and the same current I flows through the loops in the same direction. The magnitude of magnetic field at P, midway between the centres A and C, is given by [Refer to figure given below] :[Main 15 April 2018 S1]

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📖 Explanation

The magnetic field $B$ at a distance $x$ along the axis of a circular coil with radius $R$ and $N$ turns is given by $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$ .
Since point $P$ is midway between two coaxial loops separated by distance $R$ , the distance from each coil center is $x = R/2$ .
The total field at $P$ is the sum of the fields from both loops: $B_{total} = 2 \times \frac{\mu_0 N I R^2}{2(R^2 + (R/2)^2)^{3/2}} = \frac{\mu_0 N I R^2}{(5R^2/4)^{3/2}}$ .
Simplifying the denominator: $(5R^2/4)^{3/2} = \frac{5^{3/2} R^3}{8}$ .
Substituting this back gives $B_{total} = \frac{\mu_0 N I R^2 \cdot 8}{5^{3/2} R^3} = \frac{8 \mu_0 N I}{5^{3/2} R}$ .

Q200JEE Main 2018MCQ

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii $r_e,r_p,r_\alpha$ respectively in a uniform magnetic field B. The relation between $r_e,r_p,r_\alpha$ is[Main 8 April 2018]

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📖 Explanation

$r=\frac{mv}{qB}=\sqrt\frac{2mk}{qB}\alpha \sqrt{k/q^2}$ $r_c < r_p=r_a$

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