A long solenoid is formed by winding 70 turns . If current flows, then the magnetic field produced inside the solenoid is___ [24-Jan-2023 Shift 2]

A. $1232 × 10^{-4} {\text{T}}$

A charge particle moving in magnetic field B, has the components of velocity along as well as perpendicular to . The path of the charge particle will be [8-Apr-2023 shift 1]

A. Helical path with the axis along magnetic field $B$

B. Straight along the direction of magnetic field $B$

C. Helical path with the axis perpendicular to the direction of magnetic field $B$

A single current carrying loop of wire carrying current flowing in anticlockwise direction seen from +ve direction and lying in plane in shown in figure. The plot of component of magnetic field ( ) at a distance (less than radius of the coil) and on plane vs coordinate look like [29-Jan-2023 Shift 1]

in plane of coil is opposite of each other in and positions.

Magnetic Effects Of Current And Magnetism – JEE Main Physics Practice Questions & PYQs – Page 5

Q81JEE Main 2023NAT

An election in a hydrogen atom revolves around its nucleus with a speed of $6.76 \times 10^6 {\text{ms}}^{-1}$ in an orbit of radius $0.52 {\text{A}}^{\circ}$ . The magnetic field produced at the nucleus of the hydrogen atom is ______ ${\text{T}}$ . [15-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Biot-Savart law, $\; {\text{B}}=\;\frac{\mu _0}{4 \pi } \times \;{ {\text{qV}}}/{ {\text{r}}^2}$ $\; {\text{B}}=10^{-7} \times \;\frac{1.6 \times 10^{-19} \times 6.76 \times 10^6}{ (0.52 \times 10^{-10}) ^2}$ $\; {\text{B}}=40 {\text{T}}$

Q82JEE Main 2023MCQ

A circular loop of radius $r$ is carrying current I A. The ratio of magnetic field at the centre of circular loop and at a distance $r$ from the center of the loop on its axis is : [24-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Magnetic field due to current carrying circular loop on its axis is given as $\;{\mu _0 {\text{ir}}^2}/{2 ( {\text{r}}^{2+} {\text{x}}^2) ^{3 / 2}}$ At centre, $x=0, B_1=\;{\mu _0 {\text{i}}}/{2 {\text{r}}}$ $\;\;\text{ At }\; x=r, B_2=\;{\mu _0 {\text{i}}}/{2 \times 2 {\sqrt{2}} r}$ $\;\;\frac{B_1}{B_2}=2 {\sqrt{2}}$

Q83JEE Main 2023MCQ

A long solenoid is formed by winding 70 turns ${\text{cm}}^{-1}$ . If $2.0 {\text{A}}$ current flows, then the magnetic field produced inside the solenoid is___ $(\mu _0=4 \pi \times 10^{-7} {\text{TmA}}^{-1})$ [24-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{B}}=\mu _0 {\text{nI}}$ $\;=4 \pi \times 10^{-7} \times 70 \times 10^2 \times 2$ $\;=56 \pi \times 10^{-4} {\text{T}}$ $\;=176 \times 10^{-4} {\text{T}}$

Q84JEE Main 2023MCQ

A rod with circular cross-section area $2 {\text{cm}}^2$ and length $40 {\text{cm}}$ is wound uniformly with 400 turns of an insulated wire. If a current of $0.4 {\text{A}}$ flows in the wire windings, the total magnetic flux produced inside windings is $4 \pi \times 10^{-6} {\text{Wb}}$ . The relative permeability of the rod is(Given : Permeability of vacuum $\mu _0=4 \pi \times 10^{-7} {\text{NA}}^{-2} \;\text{ ) }\;$ [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\varphi =\mu _{ {\text{r}}} \mu _0 \;{ {\text{N}}}/{ℓ} {\text{I}} {\text{xA}}$ $\;\mu _{ {\text{r}}}=125$ Option 3.

Q85JEE Main 2023MCQ

The source of time varying magnetic field may be(A) A permanent magnet(B) An electric field changing linearly with time(C) Direct current(D) A decelerating charge particle(E) An antenna fed with a digital signalChoose the correct answer from the options given below: [13-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Accelerated charge particle produces EMW which has time varying ${\text{E}}$ and ${\text{B}}$ . If ${\text{E}}$ is linear function of time then ${\text{B}}$ will be constant.

Q86JEE Main 2023MCQ

A charge particle moving in magnetic field B, has the components of velocity along ${\text{B}}$ as well as perpendicular to ${\text{B}}$ . The path of the charge particle will be [8-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Path will be helical with axis along uniform ${ {\text{B}}}^{\to }-$ .

Q87JEE Main 2023MCQ

Find the magnetic field at the point ${\text{P}}$ in figure. The curved portion is a semicircle connected to two long straight wires. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$B_{ {\text{P}}}= (\;\frac{\mu _0 i}{4 r}+\;\frac{\mu _0 i}{4 \pi r}) =\;\frac{\mu _0 i}{2 r} (\;\frac{1}{2}+\;\frac{1}{2 \pi })$

Q88JEE Main 2023MCQ

As shown in the figure, a long straight conductor with semicircular arc of radius $\;\frac{\pi }{10} {\text{m}}$ is carrying current ${\text{I}}=3 {\text{A}}$ . The magnitude of the magnetic field. at the center ${\text{O}}$ of the arc is:(The permeability of the vacuum $=4 \pi \times 10^{-7} {\text{NA}}^{-2}$ ) [1-Feb-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{B}}_{ {\text{C}}}\;=\;{\mu _0 {\text{I}}}/{4 \pi {\text{R}}}(\pi )( {\text{B}} \;\text{ at centre of circular arc) }\;$ $\;=\;{\mu _0 {\text{I}}}/{4 {\text{R}}}=\;\frac{4 \pi \times 10^{-7} \times 3}{4 \times \;\frac{\pi }{10}}$ $\;=3 \times 10^{-6} {\text{T}}=3 \mu {\text{T}}$

Q89JEE Main 2023MCQ

An electron is allowed to move with constant velocity along the axis of current carrying straight solenoid.A. The electron will experience magnetic force along the axis of the solenoid.B. The electron will not experience magnetic force.C. The electron will continue to move along the axis of the solenoid.D. The electron will be accelerated along the axis of the solenoid.E. The electron will follow parabolic path-inside the solenoid.Choose the correct answer from the options given below: [11-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

We know thatIn the solenoid magnetic field along the axis of solenoid.When charge particle moving inside solenoid along the axis ${\text{F}}=0$ $\;{ {\text{F}}_{ {\text{m}}}^{\to }}= {\text{q}}({ {\text{V}}}^{\to } \times { {\text{B}}}^{\to })$ $\;\;\text{ So, }\; { {\text{F}}_{ {\text{m}}}^{\to }}=0$ And it moves with constant velocity.

Q90JEE Main 2023MCQ

The electric current in a circular coil of four turns produces a magnetic induction $32 {\text{T}}$ at its centre. The coil is unwound and is rewound into a circular coil of single turn, the magnetic induction at the centre of the coil by the same current will be : [29-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{B}}=\;{\mu _0 {\text{i}}}/{2 {\text{R}}} \times 4$ $\; {\text{B}}^{'}=\;{\mu _0 {\text{i}}}/{2 {\text{R}}^{'}}$ $\; {\text{R}}^{'}=4 {\text{R}}$ $\; {\text{B}}^{'}=\;{\mu _0 {\text{i}}}/{8 {\text{R}}}$ $\;\;{ {\text{B}}^{'}}/{ {\text{B}}}=\;\frac{1}{16}$ $\; {\text{B}}^{'}=2 {\text{T}}$

Q91JEE Main 2023NAT

A single turn current loop in the shape of a right angle triangle with sides $5 {\text{cm}}, 12 {\text{cm}}, 13 {\text{cm}}$ is carrying a current of $2 {\text{A}}$ . The loop is in a uniform magnetic field of magnitude $0.75 {\text{T}}$ whose direction is parallel to the current in the $13 {\text{cm}}$ side of the loop. The magnitude of the magnetic force on the $5 {\text{cm}}$ side will be $\;{ {\text{x}}}/{130} {\text{N}}$ . The value of ${\text{x}}$ is____ [24-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Force on $5 {\text{cm}}$ side is $\;|{ \vec{\text{F}}|= {\text{ILB}} sin \;\theta$ $\;=(2) (5 \times 10^{-2}) \times \;\frac{3}{4} \times \;\frac{12}{13}=\;\frac{9}{130} {\text{N}}$ So, $x=9$

Q92JEE Main 2023NAT

The ratio of magnetic field at the centre of a current carrying coil of radius ${\text{r}}$ to the magnetic field at distance ${\text{r}}$ from the centre of coil on its axis is ${\sqrt{x}}: 1$ . The value of $x$ is _______. [8-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{B}}_{\;\text{axis }\;}=\;{\mu _0 {\text{i}} {\text{R}}^2}/{2 ( {\text{R}}^{2+} {\text{x}}^2) ^{3 / 2}}$ $\;\;{ ( {\text{B}}_{\;\text{axis }\;}) {\text{x}}= {\text{R}}}/{ ( {\text{B}}_{\;\text{axis }\;}) {\text{x}}=0}=\;{\;{\mu _0 {\text{iR}}^2}/{2 ( {\text{R}}^{2+} {\text{R}}^2) ^{3 / 2}}}/{{\mu _0 {\text{iR}}^2}/{2 (R^2) ^{3 / 2}}}=\;{R^3}/{2^{3 / 2} {\text{R}}^3}=\;{1}/{{\sqrt{8}}}$ $\;\;{(B)_{ {\text{At}} \;\text{ centrs }\;}}/{(B)_{A t} x=R}=\;{{\sqrt{8}}}/{1}$ $\;x=8$

Q93JEE Main 2023MCQ

A current carrying rectangular loop PQRS is made of uniform wire. The length ${\text{PR}}= {\text{QS}}=5 {\text{cm}}$ and ${\text{PQ}}= {\text{RS}}=100 {\text{cm}}$ . If ammeter current reading changes from I to 2I, the ratio of magnetic forces per unit length on the wire ${\text{PQ}}$ due to wire ${\text{RS}}$ in the two cases respectively $f_{ {\text{PQ}}}^{ {\text{I}}}: f_{ {\text{PQ}}}^{2 {\text{I}}}$ is : [30-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;F \propto I_1 I_2$ $\;F_1: F_{21}=1: 4$

Q94JEE Main 2023MCQ

A single current carrying loop of wire carrying current $I$ flowing in anticlockwise direction seen from +ve $z$ direction and lying in $xy$ plane in shown in figure. The plot of $\hat{j}$ component of magnetic field ( $B_y$ ) at a distance $a$ (less than radius of the coil) and on $yz$ plane vs $z$ coordinate look like [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$B_y=0$ in plane of coil ${\text{B}}_{ {\text{y}}}$ is opposite of each other in $- {\text{z}}$ and $+ {\text{z}}$ positions.

Q95JEE Main 2023MCQ

An electron is moving along the positive ${\text{x}}$ -axis. If the uniform magnetic field is applied parallel to the negative z-axis, thenA. The electron will experience magnetic force along positive y-axisB. The electron will experience magnetic force along negative y-axisC. The electron will not experience any force in magnetic fieldD. The electron will continue to move along the positive ${\text{x}}$ -axisE. The electron will move along circular path in magnetic field Choose the correct answer from the options given below : [13-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;{ {\text{F}}}^{\to }= {\text{q}}({ {\text{V}}}^{\to } \times { {\text{B}}}^{\to })$ $\;{ {\text{F}}}^{\to }=- {\text{e}}({ {\text{V}}}^{\to } \times { {\text{B}}}^{\to })$ Force will be along -ve yaxis As magnetic force is $⟂_r$ to velocity, path of electron must be circle.

Q96JEE Main 2023NAT

A proton with a kinetic energy of $2.0 {\text{eV}}$ moves into a region of uniform magnetic field of magnitude $\;\frac{\pi }{2} \times 10^{-3} {\text{T}}$ . The angle between the direction of magnetic field and velocity of proton is $60^{\circ}$ . The pitch of the helical path taken by the proton is _______ ${\text{cm}}$ .(Take, mass of proton $=1.6 \times 10^{-27} {\text{kg}}$ and Charge on proton $=1.6 \times 10^{-19} {\text{C}}$ ) . [6-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{B}}=\;\frac{\pi }{2} \times 10^{-3}$ $\;\;\text{ K.E. }\;=\;\frac{1}{2} {\text{mV}}^2$ $\;\Rightarrow {\text{V}}=\sqrt{\;{2 {\text{KE}}}/{ {\text{m}}}}$ $\;\;\text{ PItch }\;= {\text{v}} \cos 60^{\circ} \times \;\text{ time period of one rotation }\;$ $\;= {\text{v}} \cos 60^{\circ} \times \;{2 \pi {\text{m}}}/{ {\text{eB}}}$ $\;=\sqrt{\;\frac{2 \times 2 \times 1.6 \times 10^{-9}}{1.6 \times 10^{-27}}} \times \cos 60^{\circ} \times \;\frac{2 \pi \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times \;\frac{\pi }{2} \times 10^{-3}}$ $\;=2 \times 10^4 \times \;\frac{1}{2} \times 4 \times 10^{-5}$ $\;=4 \times 10^{-1} {\text{m}}=40 {\text{cm}}$

Q97JEE Main 2023NAT

A charge particle of $2 \mu {\text{C}}$ accelerated by a potential difference of $100 {\text{V}}$ enters a region of uniform magnetic field of magnitude $4 {\text{mT}}$ at right angle to the direction of field. The charge particle completes semicircle of radius $3 {\text{cm}}$ inside magnetic field. The mass of the charge particle is ______ $\;\; \times 10^{-18} {\text{kg}}$ . [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$r=\;\frac{m v}{q B}=\;{\sqrt{2 {\text{km}}}}/{q B}, m=\;\frac{r^2 q^2 B^2}{2 k}$ ${\text{m}}=\;\frac{\;\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 10^{-3} \times 4 \times 10^{-3} \times 10^{-12}}{2 \times (100) \times 10^{-6}}$ $\;=144 \times 10^{-18} {\text{kg}}$

Q98JEE Main 2023NAT

Two long parallel wires carrying currents $8 {\text{A}}$ and $15 {\text{A}}$ in opposite directions are placed at a distance of $7 {\text{cm}}$ from each other. A point $P$ is at equidistant from both the wires such that the lines joining the point $P$ to the wires are perpendicular to each other. The magnitude of magnetic field at $P$ is ______ $\times 10^{-6} {\text{T}}$ . (Given : ${\sqrt{2}}=1.4)$ [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Magnetic fields due to both wires will be perpendicular to each other. $\; {\text{B}}_1=\;{\mu _0 {\text{i}}_1}/{2 \pi {\text{d}}} \;\; {\text{B}}_2=\;{\mu _0 {\text{i}}_2}/{2 \pi {\text{d}}}$ $\; {\text{B}}_{\;\text{net }\;}=\sqrt{ {\text{B}}_1^{2+} {\text{B}}_2^2} \Rightarrow \;{\mu _0}/{2 \pi {\text{d}}} \sqrt{ {\text{i}}_1^{2+} {\text{i}}_2^2}$ $\;\Rightarrow \;{4 \pi \times 10^{-7}}/{2 \pi \times (7 / {\sqrt{2}}) \times 10^{-2}} \times {\sqrt{8^{2+}15^2}} ( {\text{d}}=\;{7}/{{\sqrt{2}}} {\text{cm}})$ $\;\Rightarrow 68 \times 10^{-6} {\text{T}}$

Q99JEE Main 2023NAT

A straight wire $A B$ of mass $40 {\text{g}}$ and length $50 {\text{cm}}$ is suspended by a pair of flexible leads in uniform magnetic field of magnitude $0.40 {\text{T}}$ as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is _______ A. $(.$ Take $g=10 {\text{ms}}^{-2})$ [13-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

For equilibrium : $\; {\text{Mg}}= {\text{IlB}}$ $\; {\text{I}}=\;{ {\text{Mg}}}/{ {\text{IB}}}=\;\frac{40 \times 10^{-3} \times 10}{50 \times 10^{-2} \times 0.4}=2 {\text{A}}$

Q100JEE Main 2023MCQ

Math List I with List IIChoose the correct answer from the option given below: [25-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

(A) $\; {\text{B}}_{ {\text{ab}}}=\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}} \;\text{ (out of the plane) }\;$ $\; {\text{B}}_{ {\text{bcd}}}=\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}(2 \pi ) \;\text{ (\in the plane) }\;$ $\; {\text{B}}_{ {\text{de}}}=\;\frac{\mu _0}{4\pi } \;{ {\text{I}}}/{ {\text{r}}} \;\text{ (out of the plane) }\;$ Hence magnetic field at ${\text{O}}$ is $\; {\text{B}}_0=-\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}+\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}(2 \pi )-\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}$ $\; {\text{B}}_0=\;\frac{\mu _0}{2 \pi } \;{ {\text{I}}}/{{\text{r}}}(\pi -1) . . . . . . . . \;\text{ (III) }\;$ (B) $\; {\text{B}}_{ {\text{ab}}}=\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}} \;\text{ (out of the plane) }\;$ $\; {\text{B}}_{ {\text{bcd}}}=\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}(\pi ) \;\text{ (out of the plane) }\;$ $\; {\text{B}}_{ {\text{de}}}=\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}} \;\text{ (out of theplane) }\;$ Hence magnetic field at ${\text{O}}$ is $\; {\text{B}}_0=\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}+\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}(\pi )+\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}$ $\; {\text{B}}_0=\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}(\pi +2) . . . \;\text{. (I) }\;$ (C) ${\text{B}}_{ {\text{ab}}}=\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}$ (in the plane) ${\text{B}}_{ {\text{bcd}}}=\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}(\pi )$ (in the plane) ${\text{B}}_{ {\text{de}}}=0$ (at the axis)Hence magnetic field at ${\text{O}}$ is ${\text{B}}_0=\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}(1+\pi ) . . .$ (D) ${\text{B}}_{ {\text{ab}}}=0$ (at the axis) ${\text{B}}_{ {\text{bcd}}}=\;\frac{\mu _0}{4 \pi } \;{ {\text{I}}}/{ {\text{r}}}(\pi )$ (out of the plane) ${\text{B}}_{ {\text{de}}}=0$ (at the axis)Hence magnetic field at ${\text{O}}$ is ${\text{B}}_0=\;{\mu _0 {\text{I}}}/{4 {\text{r}}} . . .$

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

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