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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

Q161JEE Main 2020MCQ

A charged particle of mass 'm' and charge 'q' moving under the influence of uniform electric field $E\vec{i}$ and a uniform magnetic field $B\vec{k}$ follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, $v\vec{i}$ and $-2v\vec{j}$ . Then which of the following statements (A, B, C, D) are the correct ? (Trajectory shown is schematic and not to scale) : (A) $E=3/4(\frac{mv^2}{qa})$ (B) Rate of work done by the electric field at P is $3/4({mv^2}/a)$ (C) Rate of work done by both the fields at Q is zero(D) The difference between the magnitude of angular momentum of the particle at P and Q is 2 mav.

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📖 Explanation

Option (A) $W = k_f - k_j$ $qE(2a-0)=1/2m(2V)^{2-}1/2mV^2$ $qE2a=3/2mV^2$ $E=3/4\frac{mv^2}{qa}$ Option (B) Rate of work done P $= \vec{F}.\vec{V} = FV\cos\theta = FV$ Power = qEV $= q(3/4\frac{mV^2}{qa})V=q3/4\frac{mV^3}{qa}$ $=3/4{mV^3}/a$ Option (C)Angle between electric force and velocity is 90°, hence rate of work done will be zero at Q.Option (D)Initial angular momentum $L_i = mVa$ Final angular momentum $L_f = m(2V) (2a)$ Change in angular momentum $L_f - L_i = 3mVa$ (Note : angular momentum is calculated about O)

Q162JEE Main 2020NAT

A galvanometer coil has 500 turns and each turnhas an average area of $3 \times 10^{-4} \text{ m}^{2}$ . If a torqueof $1.5 \text{ Nm}$ is required to keep this coil parallelto magnetic field when a current of $0.5 \text{ A}$ is flowing through it, the strength of the field(in T) is ______.

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📖 Explanation

${\tau } ^{\to }={{m} ^{\to }} \times {{B} ^{\to }}$ $\tau = {NI}\times {A} \times {B}$ $105=500 \\times 3 \\times 10^{-4} \$ $\times \frac{1}{2}\times {B}$ $\ {B}=20$

Q163JEE Main 2020MCQ

An electron is moving along $+x$ direction with a velocity of $6 \times 10^{6} {\text{ms}}^{-1} .$ It enters a region of uniform electric field of $300 {\text{V}} / {\text{cm}}$ pointing along $+y$ direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be:

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📖 Explanation

$\ov {{\text{B}}}$ must be in $+{\text{z}}$ axis. $\overline{{\text{V}}}=6 \times 10^{6} \hat{i}$ $\overline{{E}}=300 \hat{j} {\text{V}} / {\text{cm}}=3 \times 10^{4} {\text{V}} / {\text{m}}$ ${q} \overline{{\text{E}}}+{\text{q}} \overline{{\text{V}}}\times \overline{{\text{B}}}=0$ ${\text{E}}={\text{VB}}$ ${\text{B}}={{\text{E}}}/{{\text{V}}}$ $=\frac{3 \times 10^{4}}{6 \times 10^{6}}=5 \times 10^{-3} {\text{T}}$

Explanation Figure 1

Q164JEE Main 2020MCQ

Magnitude of magnetic field (in SI units) at thecentre of a hexagonal shape coil of side $10 \text{ cm}$ ,50 turns and carrying current $I$ (Ampere) inunits of $\frac{\mu_0 I}{\pi}$ is :

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📖 Explanation

$\ {B}= \frac{6 \mu_{0} {I}}{4 \pi {a} \cos 30°} \\times 2 \sin 30° \\times 50$ $= \frac{\mu_{0} {I}}{\pi } \frac{150}{\\sqrt{3} {a}}= \frac{50 \\sqrt{3}}{0.1} \frac{\mu_{0} {I}}{\pi }$ $=500 \\sqrt{3} \frac{\mu_{0} {I}}{\pi }$

Explanation Figure 1

Q165JEE Main 2020MCQ

A particle of mass m and charge q has an initial velocity $\vec{\upsilon}=\upsilon_0\hat{j}$ . If an electric field $\vec{E}=E_0\hat{i}$ and magnetic field $\vec{B}=B_0\hat{i}$ act on the particle, its speed will double after a time:

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📖 Explanation

$(2V_0)^2=v_0^{2+}υ_x^2$ $v_x=\sqrt3v_0$ $\sqrt3v_0=0+{qE_0}/mt$ $t= \frac{\sqrt3mυ_0}{qE_0}$

Q166JEE Main 2020MCQ

A charged particle going around in a circle can be considered to be a current loop. A particle of mass m carrying charge q is moving in a plane with speed $v$ under the influence of magnetic field $\vec{B}$ . The magnetic moment of this moving particle :

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📖 Explanation

Magnetic moment, ${\text{M}}={\text{iA}}$ $M=(\frac{q}{T})\times \pi r^{2}$ $=\frac{q \pi r^{2}}{(\frac{2 \pi r}{v})}=\frac{q v r}{2}$ $M=\frac{q v}{2} \times \frac{v m}{q B}=\frac{mv^2}{2\text{B}}$ As we can see from the figure, direction of magnetic moment (M) is opposite to magnetic field. ${\vec{M}}=-\frac{{mv}^{2}}{2 {B}} \hat{B}$ $=-{{mv}^{2} {{\\vec{B}}}}/{2{\text{B}}^{2}}$

Explanation Figure 1

Q167JEE Main 2020MCQ

A wire carrying current I is bent in the shape ABCDEFA as shown, where rectangle ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of lengths $a$ and $b$ , then the magnitude and direction of magnetic moment of the loop ABCDEFA is :

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📖 Explanation

${M}= {NIA}$ ${N}=1$ For ${ABCD}$ ${\vec{M}}_{1}= {abI} \hat{K}$ For DEFA ${\vec{M}}_{2}= {abI} \hat{j}$ ${\vec{M}} ={\vec{M}}_{1}+{\vec{M}}_{2}$ $=\ {ab} {I}(\hat{k}+\hat{j})$ = abI, $\sqrt{2}(\frac{\hat{j}}{\sqrt{2}}+ \frac{\hat{k}}{\\sqrt{2}} )$

Q168JEE Main 2020MCQ

An iron rod of volume $10^{-3} \text{m}^3$ and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be :

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📖 Explanation

${M}=\mu_{ {r}} {NiA}$ Here $\mu_{ {r}}=$ Relative permeability ${N}= {\text{No}} .$ of turns ${i}= {\text{Current}}$ ${A}=$ Aea of cross section ${M}=\mu_{ {r}} {NiA}=\mu_{ {r}} {n} l{i} {A}$ ${M}=\mu_{ {r}} {niV}$ $=1000(1000) 0.5 (10^{-3} )$ $=500=5\times 10^{2} \ {Am}^{2}$

Q169JEE Main 2020MCQ

A charged particle carrying charge $1 \mu\text{C}$ is moving with velocity $(2\hat{i}+3\hat{j}+4\hat{k}) \text{ ms}^{-1}$ . If an external magnetic field of $(5\hat{i}+3\hat{j}-6\hat{k}) \times 10^{-3} \text{ T}$ exists in the region where the particle is moving then the force on the particle is $\vec{F} \times 10^{-9} \text{ N}$ . The vector $\vec{F}$ is :

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📖 Explanation

$\vec{F}=9(\vec{V} \times \vec{B})$ (Force on charge particle moving in magnetic field) $\vec{V}\times \vec{B}=(2\hat{i}+3 \hat{j}+4 \hat{k})$ $\times (5 {\hat{i}}+3 {\hat{j}}-6 {\hat{k}}) \\times 10^{-3}$ $(\text{table} \hat{i},\hat{j},\hat{k};2,3,4;5,3,-6)\times 10^{-3}$ $=[\hat{i}[-18-12]- \hat{j}[-12-20]$ $+\hat{k}[6-15]] \\times 10^{-3}$ $=[\hat{i}[-30]+\hat{j}[32]+\hat{k}[-9]]$ $\times 10^{-3}$ Force $=10^{-6}[-30 \hat{i}+32 \hat{j}-9 \hat{k}] \\times 10^{-3}$ $=10^{-9}[-30 \hat{i}+32\hat{j}-9 \hat{k}]$

Q170JEE Main 2020MCQ

A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is :

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📖 Explanation

We say we have 3 parts (A, B, C) $B =B_A + B_B + B_C$ $=\frac{\mu _0I}{4\pi R}(\sin90°-\sin45°) ⊗$ $+ \frac{\mu _0I}{2R} ⊚$ $+\frac{\mu _0I}{4\pi R}(\sin45°+\sin90°)⊚$ $=\frac{\mu _0I}{2\pi R}(\pi +1/\sqrt2)$

Explanation Figure 2

Q171JEE Main 2020MCQ

A circular coil has moment of inertia $0.8 \text{ kg m}^2$ around any diameter and is carrying current to produce a magnetic moment of $20 \text{ Am}^2.$ The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4 \text{ T}$ is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by $60^{\circ}$ will be:

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📖 Explanation

${I}_{ {\text{dia}}}=0.8 {kg} / {m}^{2}$ $\ {M}=20 \ {Am}^{2}$ ${U}_{ {i}}+ {K}_{ {i}}= {U}_{ {f}}+ {K}_{ {f}}$ $0+0=- {MB} \cos 30°+ \frac{1}{2} {I} \omega ^{2}$ $20 \\times 4\times \frac{\\sqrt{3}}{2}= \frac{1}{2}(0.8) \omega ^{2}$ $\omega =\\sqrt{100 \\sqrt{3}}=10(3)^{1 / 4}$

Explanation Figure 1

Q172JEE Main 2020MCQ

A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $a/3$ and 2a, respectively from the axis of the wire is :

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📖 Explanation

Let current density be J.∴ Applying Ampere's law. $∮\vec{B}.d\vec{l}=\mu _0i$ $\Rightarrow B_A2\pi a/3=\mu _0J\pi (a/3)^2$ $\therefore B_A={\mu _0Ja}/6$ Similarly, $\therefore B_B={\mu _0Ja}/4$ $\therefore B_A/B_B=2/3$

Explanation Figure 1

Q173JEE Main 2020MCQ

The figure shows a region of length ' $l$ with a uniform magnetic field of $0.3 \text{ T}$ in it and a proton entering the region with velocity $4 \times 10^{5} \text{ ms}^{-1}$ making an angle $60^\circ$ with the field. If the proton completes 10 revolution by thetime it cross the region shown, ' $l'$ is close to (mass of proton $= 1.67 \times 10^{-27} \text{ kg},$ charge of the proton $= 1.6 \times 10^{-19} \text{ C}$ )

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📖 Explanation

${T}= \frac{2 \pi {m}}{ {qB}}$ total time ${t}=10 {T}$ Kinematics $l=\frac{{V}}{2} {t}$ $l=\frac{{V}}{2} 10 \\times \frac{2 \pi {m}}{ {qB}}$ $=4 \\times 10^{5} \\times 10 \$ $\times \frac{3.14 \\times 1.67 \\times 10^{-27}}{1.6 \\times 10^{-19} \\times 0.3}$ $=0.439$

Explanation Figure 2

Q174JEE Main 2019MCQ

An electron, moving along the $x$ -axis with an initial energy of $100\text{eV}$ , enters a region of magnetic field $\vec{B}=(1.5 \times 10^{-3}\text{T})\hat{k}$ at S (See figure). The field extends between $x = 0$ and $x = 2\text{cm}$ . The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is: (electron's charge $1.6 \times 10^{-19}\text{C}$ , mass of electron $9.1 \times 10^{-31}\text{kg}$ )

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📖 Explanation

$\ \frac{1}{2} m v^{2}=K=100 e v \ \frac{1}{2} \\times 9.1 \\times 10^{-31}$ $v^{2}=100 \\times 1.6 \times 10^{-19}$ $v^{2}=\ {200 \times 1.6}{9.1 \times 10^{-12}}$ $v=0.59 \\times 10^{7}$ $v=5.9 \\times 10^{6} \ {m} / {s}$ $x_{1}=v t_{1}$ $y_{1}=\ \frac{1}{2} a t_{1}^{2}=\ \frac{1}{2}\ (\ \frac{B q v}{m}\ )\ (\ \frac{x}{v}\ )^{2}$ $y_{1}=\ \frac{1}{2} \\times (\ \frac{1.5 \times 10^{-3} \\times 1.6 \\times 10^{-19} \\times 5.9 \times 10^{6}}{9.1 \\times 15^{31}}\ )\ (\ \frac{2 \times 10^{-2}}{5.9 \\times 10^{6}}\ )^{2}$ $=0.00894 m=0.894 \ {cm}$ $r= \frac{m v}{B q}$ $=\ \frac{9.1 \\times 10^{-31} \\times 5.9 \times 10^{6}}{1.5 \\times 10^{-3} \\times 1.6 \\times 10^{-19}}=2.24 \ {cm}$ $\sin \theta = \frac{d}{r}=\ \frac{2}{2.24} \Rightarrow \theta =63.23^{0}$ $y_{2}=u \sin \theta t_{1}- \frac{1}{2} g t_{1}^{2}$ $y_{2}= {Tan} \theta x_{2}-\ \frac{g}{2 x \cos ^{2} \theta } x_{2}^{2}$ $x_{2}=6 c m=0.119 m=11.9 \ {cm}$ $d=y_{1}+y_{2}=12.87 \ {cm}$

Explanation Figure 2

Q175JEE Main 2019MCQ

As shown in the figure, two infinitely long, identical wires are bent by 90° and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If OP = OQ = 4cm, and the magnitude of the magneticfield at O is $10^{-4}$ T, and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be ( $\mu _0$ = 4π × $10^{-7} NA^{- 2}$ ) :

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📖 Explanation

Magnetic field at 'O' will be done to 'PS' and'QN' only i.e. $B_0$ = $B_{PS}$ + $B_{QN}$ → Both inwards Let current in each wire = i ∴ $B_0$ = $\frac{\mu _0i}{4\pi d}+\frac{\mu _0i}{4\pi d}$ or $10^{-4}$ = $\frac{\mu _0i}{2\pi d}$ = $\frac{2\times 10^{-7}\times i}{4\times 10^{-2}}$ ∴ i = 20 A

Q176JEE Main 2019MCQ

A current loop, having two circular arcs joinedby two radial lines is shown in the figure. Itcarries a current of 10 A. The magnetic field atpoint O will be close to : [9-Jan-2019 Shift 1]

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📖 Explanation

$\vec{B}$ = $\frac{\mu _0i}{4\pi } \theta [1/r_1-1/r_2] \hat{k}$ $r_1$ = 3 cm = 3 × $10^{-2}$ m $r_2$ = 5 cm = 5 × $10^{-2}$ m θ = $\pi /4$ , i = 10 A \Rightarrow $\vec{B}$ = $\frac{4\pi \times 10^{-7}}{16}$ × 10 $[1/{3\times 10^{-2}} - 1/{5\times 10^{-2}}]\hat{k}$ \Rightarrow $|\vec{B}|$ = $\pi /3$ × $10^{-5}$ T = 1 × $10^{-5}$ T

Explanation Figure 2

Q177JEE Main 2019MCQ

Two wires A&B are carrying currents $I_1$ and $I_2$ as shown in the figure . the separation between them is $d$ . A third wire C carrying a current $I$ is to be kept parallel to them at a distance $x$ from $A$ such that the net force acting on it is Zero . The possible values of $x$ are:

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📖 Explanation

The magnetic force per unit length between two parallel wires carrying currents $I_A$ and $I_B$ at a distance $r$ is given by $F = \frac{\mu_0 I_A I_B}{2 \pi r}$ . For the net force on wire C to be zero, the force from wire A must balance the force from wire B:
$\frac{\mu_0 I_1 I}{2 \pi x} = \frac{\mu_0 I_2 I}{2 \pi |x-d|}$
Simplifying, we obtain the condition:
$\frac{I_1}{x} = \frac{I_2}{|x-d|}$
This absolute value equation can be written as $\frac{I_1}{x} = \pm \frac{I_2}{x-d}$ , which leads to the linear relation $I_1(x-d) = \pm I_2 x$ . Solving for $x$ in these cases, we find the possible positions:
$I_1 x - I_1 d = \pm I_2 x \implies x(I_1 \mp I_2) = I_1 d \implies x = \pm \frac{I_1 d}{I_1 - I_2}$
Thus, the possible values for $x$ are $x = \pm \frac{I_1 d}{I_1 - I_2}$ .

Q178JEE Main 2019MCQ

Find the magnetic field at point P due to a straight line segment AB of length 6 cm carring a current of 5 A.(See figure) $(\mu_{0}=4\pi \times 10^{-7} N-A^{-2})$

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📖 Explanation

$B=\ \frac{\mu_{0} i}{4 \pi r} (\sin \theta _{1}+\sin \theta _{2} )$ $B=\ \frac{10^{-7} \\times 5}{4 \\times 10^{-2}} \\times 2 \\times \ \frac{3}{5}$ $=1.5 \\times 10^{-5} T$

Explanation Figure 2

Q179JEE Main 2019MCQ

A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5 Nm $^{-1}$ (see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N. If the mass of the strip is 50 grams, its resistance 10 Ω. and air drag negligible, N will be close to :

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📖 Explanation

Force on oscillating wire if it's speed v is $F= \frac{B^{2} l^{2} V}{R}=b V$ Time period T is $=25 \sqrt{ \frac{m}{k}}$ No. of oscillations is $\frac{t}{T}$ and tis given by $t= \frac{2 m}{b}$

Explanation Figure 2

Q180JEE Main 2019MCQ

A proton and an a-particle (with their masses in the ratio of 1 : 4 and charges in the ratio of 1 : 2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of theradii $r_p$ : $r_a$ of the circular paths described by them will be :

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📖 Explanation

KE = qΔV r = $\sqrt\frac{2mq\Delta V}{qB}$ r α $\sqrt{m/q}$ $r_p/r_{\infty }$ = $1/\sqrt2$

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