A uniform magnetic field of acts along positive Y-direction. A rectangular loop of sides 20 and with current of is plane. The current is in anticlockwise sense with reference to negative axis. Magnitude and direction of the torque is : [31-Jan-2024 Shift 2]

B. $2 × 10^{-4} {\text{N}}- {\text{m}}$ along negative Z-direction

A. $2 × 10^{-4} {\text{N}}- {\text{m}}$ along positive ${\text{Z}}$-direction

B. $2 × 10^{-4} {\text{N}}- {\text{m}}$ along negative Z-direction

C. $2 × 10^{-4} {\text{N}}- {\text{m}}$ along positive ${\text{X}}$-direction

D. $2 × 10^{-4} {\text{N}}- {\text{m}}$ along positive ${\text{Y}}$-direction

A proton and a deutron having same kinetic energies enter a region of uniform magnetic field , moving perpendicular to . The ratio of the radius of deutron path to the radius of the proton path is :

In uniform magnetic field, Since same K.E

An element is placed at the origin and carries a large current . The magnetic field on the -axis at a distance of from the elements of length is :

C. $12 \times 10^{-8} \text{T}$

D. $10 \times 10^{-8} \text{T}$

A straight magnetic strip has a magnetic moment of . If the strip is bent in a semicircular shape, its magnetic moment will be ______ (Given )

Magnetic moment of straight wire Magnetic moment of arc

The electrostatic force and magnetic force acting on a charge moving with velocity can be written :

C. $\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{V} × \vec{B})$

A. $\vec{F}_1=q \vec{V} ⋅ \vec{E}, \vec{F}_2=q(\vec{B} ⋅ \vec{V})$

B. $\vec{F}_1=q \vec{B}, \vec{F}_2=q(\vec{B} × \vec{V})$

C. $\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{V} × \vec{B})$

D. $\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{B} × \vec{V})$

Magnetic Effects Of Current And Magnetism – JEE Main Physics Practice Questions & PYQs – Page 3

Q41JEE Main 2025NAT

A 4.0 cm long straight wire carrying a current of 8 A is placed perpendicular to a uniform magnetic field of strength 0.15 T . The magnetic force on the wire is $\_\_\_\_$ mN .

🚀 Solve in Practice Mode

📖 Explanation

$F\;=i / B$ $\;=8 \times 4 \times 10^{-2} \times 0.15 \;\text{ newton }\;$ $\;=8 \times 4 \times 10^{-2} \times 0.15 \times 10^3 {\text{mN}}$ $\;=32 \times 1.5 {\text{mN}}$ $\;=48 {\text{mN}}$

Q42JEE Main 2024NAT

A $2 {\text{A}}$ current carrying straight metal wire of resistance $1 Ω$ , resistivity $2 \times 10^{-6} Ω {\text{m}}$ , area of cross-section $10 {\text{mm}}^2$ and mass $500 {\text{g}}$ is suspended horizontally in mid air by applying a uniform magnetic field $\vec{B}$ . The magnitude of $B$ is ______ $\times 10^{-1} {\text{T}}$ (given, ${\text{g}}=10 {\text{m}} / {\ s}^2$ )

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{R}}=\;{\rho ℓ}/{ {\text{A}}} \Rightarrow \;\frac{2 \times 10^{-6} \times ℓ}{110^{-5}}=1 \Rightarrow ℓ=5$ $\; {\text{mg}}= {\text{Bi}} ℓ$ $\; {\text{B}}=\;{ {\text{mg}}}/{ {\text{i}} ℓ}=\;\frac{5}{2 \times 5}=0.5=5 \times 10^{-1} \;\text{ Tesla }\;$

Q43JEE Main 2024MCQ

An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then :

🚀 Solve in Practice Mode

📖 Explanation

Since $\vec{v} \| \vec{B}$ so force on electron due to magnetic field is zero. So it will move along axis with uniform velocity.

Q44JEE Main 2024NAT

An electron with kinetic energy $5 {\text{eV}}$ enters a region of uniform magnetic field of $3 \mu {\text{T}}$ perpendicular to its direction. An electric field ${\text{E}}$ is applied perpendicular to the direction of velocity and magnetic field. The value of ${\text{E}}$ , so that electron moves along the same path, is ______ ${\text{NC}}^{-1}$ .(Given, mass of electron $=9 \times 10^{-31} {\text{kg}}$ , electric charge $=1.6 \times 10^{-19} {\text{C}}$ )

🚀 Solve in Practice Mode

📖 Explanation

For the given condition of moving undeflected, net force should be zero. ${\text{qE}}\;= {\text{qVB}}$ ${\text{E}}\;= {\text{VB}}$ $\;=\sqrt{\;{2 \times {\text{KE}}}/{ {\text{m}}}} \times {\text{B}}$ $\;=\sqrt{\;\frac{2 \times 5 \times 1.6 \times 10^{-19}}{9 \times 10^{-31}}} \times 3 \times 10^{-6}$ $\;=4 {\text{N}} / {\text{C}}$

Q45JEE Main 2024MCQ

A uniform magnetic field of $2 \times 10^{-3} {\text{T}}$ acts along positive Y-direction. A rectangular loop of sides 20 ${\text{cm}}$ and $10 {\text{cm}}$ with current of $5 {\text{A}}$ is ${\text{Y}}- {\text{Z}}$ plane. The current is in anticlockwise sense with reference to negative ${\text{X}}$ axis. Magnitude and direction of the torque is : [31-Jan-2024 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;{ {\text{M}}}^{\to }= {\text{i}} { {\text{A}}}^{\to }$ $\;=5 \times (0.2) \times (0.1)(-{ {\text{i}}^{\^}})$ $\;=0.1(-{ {\text{i}}^{\^}})$ $\;\vec{\tau }={ {\text{M}}}^{\to } \times { {\text{B}}}^{\to }=0.1(-{ {\text{i}}^{\^}})$ $\times (2 \times 10^{-3}) ({ {\text{j}}^{\^}})$ $\;=2 \times 10^{-4}(-{ {\text{k}}^{\^}}) {\text{N}}- {\text{m}}$

Q46JEE Main 2024NAT

A square loop of edge length $2 \text{m}$ carrying current of $2 \text{A}$ is placed with its edges parallel to the $x-y$ axis. A magnetic field is passing through the $x-y$ plane and expressed as $\vec{B}=B_0(1+4 x) \hat{k}$ , where $B_0=5 \text{T}$ . The net magnetic force experienced by the loop is ______ $\text{N}$ .

🚀 Solve in Practice Mode

📖 Explanation

$\;B(x=0)=B_0, \;\; B(x=2)=9 B_0$ $\;\;\text{ Also, }\; F=i ℓ B$ $\;\Rightarrow F_1=i ℓ B_0 \;\text{ & }\; F_2=9 i ℓ B_0$ $\;F=F_2-F_1=8 i ℓ B_0=8 \times 2 \times 2 \times 5$ $\;F=160 N$

Q47JEE Main 2024MCQ

Given below are two statements:Statement (I) : When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic field is taken into account.Statement (II) : Ampere's circuital law does not depend on Biot-Savart's law.In| the light of the above statements, choose the correct answer from the options given below:

🚀 Solve in Practice Mode

Q48JEE Main 2024NAT

Two circular coils P and Q of 100 turns each have same radius of $\pi {\text{cm}}$ . The currents in ${\text{P}}$ and ${\text{R}}$ are $1 {\text{A}}$ and 2 A respectively. P and Q are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is $\sqrt{ {\text{x}}} {\text{mT}}$ , where ${\text{x}}=$ ____ $[.$ Use $\mu _0=4 \pi \times 10^{-7} {\text{TmA}}^{-1}$ ] [31-Jan-2024 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{B}}_{ {\text{P}}}=\;{\mu _0 {\text{Ni}}_1}/{2 {\text{r}}}=\;\frac{\mu _0 \times 1 \times 100}{2 \pi }$ $=2 \times 10^{-3} {\text{T}}$ $\; {\text{B}}_{ {\text{Q}}}=\;{\mu _0 {\text{Ni}}_2}/{2 {\text{r}}}=\;\frac{\mu _0 \times 2 \times 100}{2 \pi }$ $=4 \times 10^{-3} {\text{T}}$ $\; {\text{B}}_{\;\text{net }\;}=\sqrt{ {\text{B}}_{ {\text{P}}}^{2+} {\text{B}}_{ {\text{Q}}}^2}$ $\;={\sqrt{20}} {\text{mT}}$ $\; {\text{x}}=20$

Q49JEE Main 2024MCQ

Two particles ${\text{X}}$ and ${\text{Y}}$ having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of ${\text{X}}$ and ${\text{Y}}$ is : [29-Jan-2024 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{R}}=\;{ {\ mv}}/{ {\text{qB}}}=\;{ {\text{p}}}/{ {\text{qB}}}=\;{\sqrt{2 {\text{m}}( {\text{KE}})}}/{ {\text{qB}}}$ $=\;{\sqrt{2 {\text{mqV}}}}/{ {\text{qB}}}$ $\; {\text{R}} \propto \sqrt{ {\text{m}}}$ $\; {\text{m}} \propto {\text{R}}^2$ $\;\;{ {\text{m}}_1}/{ {\text{m}}_2}= (\;{ {\text{R}}_1}/{ {\text{R}}_2}) ^2$

Q50JEE Main 2024NAT

A solenoid of length $0.5 {\text{m}}$ has a radius of $1 {\text{cm}}$ and is made up of ' $m$ ' number of turns. It carries a current of $5 {\text{A}}$ . If the magnitude of the magnetic field inside the solenoid is $6.28 \times 10^{-3} {\text{T}}$ , then the value of ${\text{m}}$ is :

🚀 Solve in Practice Mode

📖 Explanation

$\;\mu _0 {\ni}= {\text{B}} \;\; {\text{n}}=\;\text{ number of turns per unit length }\;$ $\;\mu _0 (\;{ {\text{m}}}/{ℓ}) {\text{i}}= {\text{B}}$ $\; {\text{m}}=\;{ {\text{B}} . ℓ}/{\mu _0 {\text{i}}}=\;\frac{6.28 \times 10^{-3} \times 0.5}{12.56 \times 10^{-7} \times 5}$ $\; {\text{m}}=500$

Q51JEE Main 2024NAT

A charge of $4.0 \, \mu\text{C}$ is moving with a velocity of $4.0 \times 10^6 \, \text{ms}^{-1}$ along the positive y-axis under a magnetic field $\vec{B}$ of strength $(2 \hat{k}) \, \text{T}$ . The force acting on the charge is $x \hat{i} \, \text{N}$ . The value of $x$ is__ [29-Jan-2024 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{q}}\;=4 \mu {\text{C}}, { {\text{v}}}^{\to }=4 \times 10^6 { {\text{j}}^{\^} m} / {\ s}$ ${ {\text{B}}}^{\to }\;=2 { {\text{k}}^{\^}} {\text{T}}$ ${ {\text{F}}}^{\to }\;= {\text{q}}({ {\text{v}}}^{\to } \times { {\text{B}}}^{\to })$ $\;=4 \times 10^{-6} (4 \times 10^6 { {\text{j}}^{\^}} \times 2 { {\text{k}}^{\^}})$ $\;=4 \times 10^{-6} \times 8 \times 10^6 { {\text{i}}^{\^}}$ ${ {\text{F}}}^{\to }\;=32 { {\text{i}}^{\^}} {\text{N}}$ ${\text{x}}\;=32$

Q52JEE Main 2024NAT

A moving coil galvanometer has 100 turns and each turn has an area of $2.0 {\text{cm}}^2$ . The magnetic field produced by the magnet is $0.01 {\text{T}}$ and the deflection in the coil is 0.05 radian when a current of $10 {\text{mA}}$ is passed through it. The torsional constant of the suspension wire is $x \times 10^{-5} {\text{N}}- {\text{m}} / {\text{rad}}$ . The value of $x$ is_____ [1-Feb-2024 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\tau =\;\text{ BINAsin }\;\phi$ $\; {\text{C}} \theta =\;\text{ BINAsin }\; 90^{\circ}$ $\; {\text{C}}=\;\frac{\;\text{ BINA }\;}{\theta }$ $=\;\frac{0.01 \times 10 \times 10^{-3} \times 100 \times 2 \times 10^{-4}}{0.05}$ $\;=4 \times 10^{-5} {\text{N}}- {\text{m}} / {\text{rad}} .$ $\; {\text{x}}=4$

Q53JEE Main 2024MCQ

A proton and a deutron $( {\text{q}}=+ {\text{e}}, m=2.0 {\text{u}})$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$ , moving perpendicular to $\vec{B}$ . The ratio of the radius $r_d$ of deutron path to the radius $r_p$ of the proton path is :

🚀 Solve in Practice Mode

📖 Explanation

In uniform magnetic field, ${\text{R}}=\;{ {\text{m}} v}/{ {\text{qB}}}=\;{\sqrt{2 {\text{m}}( {\text{K}} \cdot {\text{E}})}}/{ {\text{qB}}}$ Since same K.E $\; {\text{R}} \propto \;{\sqrt{ {\text{m}}}}/{ {\text{q}}}$ $\; \therefore \;{ {\text{R}}_{\;\text{deutron }\;}}/{ {\text{R}}_{\;\text{proton }\;}}=\sqrt{\;{ {\text{m}}_{ {\text{d}}}}/{ {\text{m}}_p}} \times \;{ {\text{q}}_{ {\text{p}}}}/{ {\text{q}}_d}$ $\;={\sqrt{2}} \times 1$ $\; \therefore \gamma _d: \gamma _{ {\text{p}}}={\sqrt{2}}: 1$

Q54JEE Main 2024NAT

The magnetic field at the centre of a wire loop formed by two semicircular wires of radii $R_1=2 \pi {\text{m}}$ and $R_2=4 \pi {\text{m}}$ carrying current $I=4 {\text{A}}$ as per figure given below is $\alpha \times 10^{-7} {\text{T}}$ . The value of $\alpha$ is ____. (Centre ${\text{O}}$ is common for all segments) [27-Jan-2024 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;{\mu _0 {\text{i}}}/{2 {\text{R}}_2} (\;\frac{\pi }{2 \pi }) ⊗+\;{\mu _0 {\text{i}}}/{2 {\text{R}}_1} (\;\frac{\pi }{2 \pi }) ⊗$ $\; (\;{\mu _0 {\text{i}}}/{4 {\text{R}}_2}+\;{\mu _0 {\text{i}}}/{4 {\text{R}}_1}) ⊗$ $\;\;\frac{4 \pi \times 10^{-7} \times 4}{4 \times 4 \pi }+\;\frac{4 \pi \times 10^{-7} \times 4}{4 \times 2 \pi }$ $\;=3 \times 10^{-7}=\alpha \times 10^{-7}$ $\;\alpha =3$

Q55JEE Main 2024MCQ

An element $\Delta l = \Delta x \hat{i}$ is placed at the origin and carries a large current $I = 10 \text{A}$ . The magnetic field on the $y$ -axis at a distance of $0.5 \text{m}$ from the elements $\Delta x$ of $1 \text{cm}$ length is :

🚀 Solve in Practice Mode

📖 Explanation

${ {\text{dB}}}^{\to }\;=\;{\mu _0 {\text{I}}}/{4 \pi } \;{({ {\text{d}}}^{\to } \times { {\text{r}}}^{\to })}/{ {\text{r}}^3} \;\text{ (Tesla) }\;$ $\;=\;{10^{-7} \times 10 \times (\frac{1}{2} \times \frac{1}{100}) (+{ {\text{k}}^{\^}})}/{ (\;\frac{1}{2}) ^3}=4 \times 10^{-8} {\text{T}}(+{ {\text{k}}^{\^}})$

Q56JEE Main 2024NAT

A straight magnetic strip has a magnetic moment of $44 {\text{Am}}^2$ . If the strip is bent in a semicircular shape, its magnetic moment will be ______ ${\text{Am}}^2$ (Given $\pi =\;\frac{22}{7}$ )

🚀 Solve in Practice Mode

📖 Explanation

Magnetic moment of straight wire $= {\text{mx}} ℓ=44$ Magnetic moment of arc $\;= {\text{m}} \times 2 {\text{r}}$ $\;= {\text{m}} \times \;\frac{2 ℓ}{\pi }$ $\;=\;\frac{44 \times 2}{\pi }=\;\frac{88}{\pi }=28$

Q57JEE Main 2024MCQ

In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero.

🚀 Solve in Practice Mode

📖 Explanation

$∮ { {\text{B}}}^{\to } \cdot {\text{d}} \vec{ℓ}=\mu _0 {\text{i}}_{ {\text{enc}}}=0$ $˙eq B=0$ outside the cable

Q58JEE Main 2024MCQ

The electrostatic force $(\vec{F}_1)$ and magnetic force $(\vec{F}_2)$ acting on a charge $q$ moving with velocity $v$ can be written :

🚀 Solve in Practice Mode

📖 Explanation

$\;{ {\text{F}}}^{\to }_1= {\text{q}} { {\text{E}}}^{\to } \;\; \;\text{ (Theory) }\;$ $\;{ {\text{F}}}^{\to }_2= {\text{q}}({ {\text{V}}}^{\to } \times { {\text{B}}}^{\to })$

Q59JEE Main 2024NAT

A square loop PQRS having 10 turns, area $3.6 \times$ $10^{-3} {\text{m}}^2$ and resistance $100 Ω$ is slowly and uniformly being pulled out of a uniform magnetic field of magnitude ${\text{B}}=0.5 {\text{T}}$ as shown. Work done in pulling the loop out of the field in $1.0 {\ s}$ is ______ $\times 10^{-6} {\text{J}}$ .

🚀 Solve in Practice Mode

📖 Explanation

$\;E=N B ℓ v$ $\;i=\;\frac{E}{R}=\;\frac{N B ℓ v}{R}$ $\;F=N(i ℓ B)=\;\frac{N^2 B^2 ℓ^2 v}{R}$ $\; {\text{W}}= {\text{F}} \times ℓ=\;{ {\text{N}}^2 {\text{B}}^2 ℓ^3}/{ {\text{R}}} (\;{ℓ}/{ {\text{t}}})$ $\; {\text{A}}=ℓ^2$ $\; {\text{W}}=\;\frac{(10 \times 10)(0.5)^2 \times (3.6 \times 10^{-3}) ^2}{100 \times 1}$ $\; {\text{W}}=3.24 \times 10^{-6} {\text{J}}$

Q60JEE Main 2024MCQ

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 $\;{ {\text{a}}}/{2}$ and 2a from axis of the wire is :

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{B}}_1 2 \pi \;{ {\text{a}}}/{2}=\mu _{ {\text{o}}} \;{ {\text{I}}}/{4}$ $\; {\text{B}}_1=\;{\mu _{ {\text{o}}} {\text{I}}}/{4 \pi {\text{a}}}$ $\; {\text{B}}_2 2 \pi 2 {\text{a}}=\mu _{ {\text{o}}} {\text{I}}$ $\; {\text{B}}_2=\;{\mu _0 {\text{I}}}/{4 \pi {\text{a}}}$

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

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