datavorous/entrance-exam-dataset huggingface dataset:
A charge 'Q' $\mu$ C is placed at the centre of a cube. The flux through one face and two
opposite faces of the cube is respectively

Options:
A $\frac{\mathrm{Q}}{6 \epsilon_{0}} \mu \mathrm{Vm}, \quad \frac{\mathrm{Q}}{3 \epsilon_{0}} \mu \mathrm{Vm}$
B $\frac{\mathrm{Q}}{12 \epsilon_{0}} \mu \mathrm{Vm}, \quad \frac{\mathrm{Q}}{\epsilon_{0}} \mu \mathrm{Vm}$
C $\frac{\mathrm{Q}}{6 \epsilon_{0}} \mu \mathrm{Vm}, \quad \frac{\mathrm{Q}}{2 \epsilon_{0}} \mu \mathrm{Vm}$
D $\frac{\mathrm{Q}}{12 \epsilon_{0}} \mu \mathrm{Vm}, \quad \frac{\mathrm{Q}}{3 \epsilon_{0}} \mu \mathrm{Vm}$

The correct answer is:
$\frac{\mathrm{Q}}{6 \epsilon_{0}} \mu \mathrm{Vm}, \quad \frac{\mathrm{Q}}{3 \epsilon_{0}} \mu \mathrm{Vm}$

Solution:
By using Gauss's Law.
It is given as
$\Phi=\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_{0}}$
Now, the flux passing through all the six surfaces would be
$\Phi=6 \phi=\frac{q}{\epsilon_{0}}$
And the flux passing through each surface would be
$\phi=\frac{q}{6 \in_{0}}$

Topic: Physics, Electrostatics, JEE Main
A charge Q is distributed over three concentric spherical shells of radii a, b, c a<b<c such that their surface charge densities are equal to one another.
The total potential at a point at distance r from their common centre, where r<a, would be:

Options:
A Q4πϵ0a+b+c
B Q a2+b2+c24πϵ0a3+b3+c3
C Q12πϵ0ab+bc+caabc
D Qa+b+c4πϵ0a2+b2+c2

The correct answer is:
Qa+b+c4πϵ0a2+b2+c2

Solution:
Surface charge densities of all sphere is given as q1a2=q2b2=q3c2=constant, where q1, q2 and q3 is charge on sphere of radius a, b and c respectively.
q1+q2+q3=Q
q1=a2(Q)a2+b2+c2
q2=b2(Q)a2+b2+c2
q3=c2Qa2+b2+c2

Potential at a point where r<a is given as V=Kq1a+Kq2b+Kq3c, where K is a constant.
V=14πε0Qa+b+ca2+b2+c2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A charge Q is distributed over two concentric conducting thin spherical shells radii r and R R>r. If the surface charge densities on the two shells are equal, the electric potential at the common centre is :

Options:
A 14πϵ0R+r2R2+r2Q
B 14πϵ02R+rR2+r2Q
C 14πϵ0R+2rQ2R2+r2
D 14πϵ0(R+r)R2+r2Q

The correct answer is:
14πϵ0(R+r)R2+r2Q

Solution:
Q1=σ4πr2

Q2=σ4πr2

∴Q=σ4πr2+R2∴ σ=Q4πr2+R2

VC=KQ1r+KQ2R

=K04πr2r+K04πR2R

=KQ4π(r+R)4πr2+R2

=KQ(r+R)r2+R2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A charge $Q$ is enclosed by a Gaussian spherical surface of radius $R$. If the radius is doubled, then the outward electric flux will

Options:
A be reduced to half
B remain the same
C be doubled
D increase four times

The correct answer is:
remain the same

Solution:
Total flux $=\frac{\text { Net enclosed charge }}{\varepsilon_0}$
Hence, we can say the electric flux depends only on net enclosed charge by surface.

Topic: Physics, Electrostatics, NEET, NEET 2011 (Screening)
A charge Q is enclosed by a Gaussian surface of radius $R$. If the radius is doubled then the outward electric flux will

Options:
A be reduced to half
B be doubled
C remain the same
D increase four times

The correct answer is:
remain the same

Solution:
We know from Gauss's law that the electric flux through any closed spherical surface is given by
$\phi=\frac{\mathrm{q}_{\text {enclosed. }}}{\varepsilon_0}$
So electric flux is independent of the radius of the closed surface and only depends on the charge enclosed.
Therefore if the radius of Gaussian spherical surface is doubled then there will be no effect on flux and the outward electric flux will remain the same.

Topic: Physics, Electrostatics, MHT CET, MHT CET 2022 (07 Aug Shift 1)
A charge Q is moving dl→ distance in the magnetic field B→. Find the value of work done by B→.

Options:
A 1
B Infinite
C Zero
D -1

The correct answer is:
Zero

Solution:
Since force on a point charge by magnetic field is always perpendicular to v→F→=qv→×B→

∴ Work by magnetic force on the point charge is zero.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (16 Mar Shift 2)
A charge Q is placed at a distance a2 above the centre of a square surface of side length a. The electric flux through the square surface due to the charge would be?

Options:
A Q6ϵ0
B Q2ϵ0
C Q3ϵ0
D Qϵ0

The correct answer is:
Q6ϵ0

Solution:
Charged particle can be considered at centre of a cube of side a, and given surface represents its one face. So, flux ϕ=Q6ϵ0.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2018 (15 Apr)
A charge $Q$ is placed at a distance $\mathrm{a} / 2$ above the centre of the square surface of edge a as shown in the figure. The electric flux through the square surface is:

Options:
A
$\frac{Q}{3 \varepsilon_0}$

B
$\frac{Q}{6 \varepsilon_0}$

C
$\frac{Q}{2 \varepsilon_0}$

D
$\frac{Q}{\varepsilon_0}$

The correct answer is:
$\frac{Q}{6 \varepsilon_0}$

Solution:
When cube is of side a and point charge $Q$ is at the center of the cube then the total electric flux due to this charge will pass evenly through the six faces of the cube. So, the electric flux through one face will be equal to $1 / 6$ of the total electric flux due to this charge.
Flux through 6 faces $=\frac{Q}{\epsilon_0}$
$\therefore$ Flux through 1 face, $=\frac{Q}{6 \epsilon_0}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero, then the $Q / q$ equals

Options:
A
$-2 \sqrt{2}$

B
$-1$

C
1

D
$-\frac{1}{\sqrt{2}}$

The correct answer is:
$-2 \sqrt{2}$

Solution:
Three forces $F_{41}, F_{42}$ and $f_{43}$ acting on $Q$ are shown
Resultant of $F_{41}+F_{43}$
$$
\begin{aligned}
& =\sqrt{2} F_{\text {each }} \\
& =\sqrt{2} \frac{1}{4 \pi \varepsilon_0} \frac{Q q}{d^2}
\end{aligned}
$$
Resultant on $Q$ becomes zero only when ' $q$ ' charges are of negative nature.
$$
\begin{aligned}
& F_{4,2}=\frac{1}{4 \pi \varepsilon_0} \frac{Q \times Q}{(\sqrt{2} d)^2} \\
& \Rightarrow \sqrt{2} \frac{d Q}{d^2}=\frac{Q \times Q}{2 d^2} \\
& \Rightarrow \sqrt{2} \times q=\frac{Q \times Q}{2} \\
& \therefore q=-\frac{Q}{2 \sqrt{2}} \text { or } \frac{Q}{q}=-2 \sqrt{2}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2009
A charge q is placed at one corner of a cube as shown in figure. The flux of electrostatic field E→ through the shaded area is:

Options:
A q8ε0
B q24ε0
C q48ε0
D q4ε0

The correct answer is:
q24ε0

Solution:
flux through cube =q8ϵ0

flux through surfaces ABEH,ADGH,ABCD will be zero

ϕEFGH=ϕDCFG=ϕEBCF=13q8ϵ0

=q24ϵ0

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (25 Feb Shift 2)
A charge $q$ is placed at the center of one of the surface of a cube. The flux linked with the cube is:

Options:
A $\frac{q}{2 \epsilon_0}$
B $\frac{q}{8 \epsilon_0}$
C Zero
D $\frac{q}{4 \epsilon_0}$

The correct answer is:
$\frac{q}{2 \epsilon_0}$

Solution:
From
$\begin{aligned} & 2 \phi=\frac{\mathrm{q}}{\epsilon_0} \\ & \phi=\frac{\mathrm{q}}{2 \epsilon_0}\end{aligned}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (04 Apr Shift 2)
A charge $q$ is placed at the centre of a cube. Then the flux passing through one face of cube will be

Options:
A $\frac{q}{\varepsilon_0}$
B $\frac{q}{2 \varepsilon_0}$
C $\frac{q}{4 \varepsilon_0}$
D $\frac{q}{6 \varepsilon_0}$

The correct answer is:
$\frac{q}{6 \varepsilon_0}$

Solution:
Net flux through the cube $\phi_{\text {net }}=\frac{Q}{\varepsilon_0}$; so flux through one face $\phi_{\text {faxe }}=\frac{q}{6 \varepsilon_0}$

Topic: Physics, Electrostatics, JEE Main
A charge $+q$ is placed at the origin $O$ of $X-Y$ axes as shown in the figure. The work done in taking a charge $Q$ from $A$ to $B$ along the straight line $A B$ is

Options:
A $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a-b}{a b}\right)$
B $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{b-a}{a b}\right)$
C $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{b}{a^{2}}-\frac{1}{b}\right)$
D $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a}{b^{2}}-\frac{1}{b}\right)$

The correct answer is:
$\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a-b}{a b}\right)$

Solution:
We know that

and $\quad V=\frac{Q}{4 \pi \varepsilon_{0} r}$
$\begin{aligned} \text { Hence, } \quad W &=q\left[\frac{Q}{4 \pi \varepsilon_{0} b}-\frac{Q}{4 \pi \varepsilon_{0} a}\right] \\ &=\frac{Q q}{4 \pi \varepsilon_{0}}\left[\frac{a-b}{a b}\right] \end{aligned}$

Topic: Physics, Electrostatics, JEE Main
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 magnetic moment of the loop is:

Options:
A 32 qωr2
B 12 qωr2
C qωr2
D 43 qωr2

The correct answer is:
12 qωr2

Solution:
ML=q2m⇒M=q2m×mr2ω⇒M=qωr22.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2018 (16 Apr Online)
A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is :

Options:
A Q 4 π ∈ 0 L ln 2
B Q ln 2 4 Ï â 0 L
C Q 8 π ∈ 0 L
D 3 Q 4 π ∈ 0 L

The correct answer is:
Q ln 2 4 Ï â 0 L

Solution:
V = ∫ L 2 L kdq x
= ∫ L 2 L 1 4 π ε 0 Q L dx x
= Q 4 π ε 0 L ℓ n 2

Topic: Physics, Electrostatics, JEE Main
A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is :

Options:
A Q 4 π ∈ 0 L ln 2
B Q ln 2 4 Ï â 0 L
C Q 8 π ∈ 0 L
D 3 Q 4 π ∈ 0 L

The correct answer is:
Q ln 2 4 Ï â 0 L

Solution:
V = ∫ L 2 L kdq x
= ∫ L 2 L 1 4 π ε 0 Q L dx x
= Q 4 π ε 0 L ℓ n 2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2013 (07 Apr)
A charge $Q$ is uniformly distributed over the surface of non conducting disc of radius $R$. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity $\omega$. As a result of this rotation a magnetic field of induction $B$ is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure

Options:
A

B

C

D

The correct answer is:

Solution:
Consider ring like element of disc of radius $r$ and thickness $\mathrm{dr}$.
If $\sigma$ is charge per unit area, then charge on the element
$d q=\sigma(2 \pi r d r)$
current ' $i$ ' associated with rotating charge $\mathrm{dq}$ is
$\mathrm{i}=\frac{(\mathrm{dq}) \mathrm{w}}{2 \pi}=\sigma \mathrm{wrdr}$
Magnetic field $\mathrm{dB}$ at center due to element
$\mathrm{dB}=\frac{\mu_0 \mathrm{i}}{2 \mathrm{r}}=\frac{\mu_0 \sigma \omega \mathrm{dr}}{2}$
$\begin{aligned}
& \mathrm{B}_{\text {net }}=\int \mathrm{dB}=\frac{\mu_0 \sigma \omega}{2} \int_0^{\mathrm{R}} \mathrm{dr}=\frac{\mu_0 \sigma \omega \mathrm{R}}{2} \\
& \Rightarrow \quad \mathrm{B}_{\text {net }}=\frac{\mu_0 \mathrm{Q} \omega}{2 \pi \mathrm{R}} \quad\left[\because \mathrm{Q}=\sigma \pi \mathrm{R}^2\right]
\end{aligned}$
So if $Q$ and $w$ are unchanged then
$\mathrm{B}_{\mathrm{net}} \propto \frac{1}{\mathrm{R}}$
Hence variation of $B_{\text {net }}$ with $R$ should be a rectangular hyperbola as represented in (A).

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2012 (Offline)
A charge ' $q$ ' moves in a region where the electric field and the magnetic field both exist, then the force on it is:

Options:
A $q(\vec{v} \times \vec{B})$
B $q \vec{E}+q(\vec{v} \times \vec{B})$
C $q \vec{E}+q(\vec{B} \times \vec{v})$
D $q \vec{E}+q(\vec{B} \times \vec{v})$

The correct answer is:
$q \vec{E}+q(\vec{v} \times \vec{B})$

Solution:
Lorentz force $\vec{F}_v=\vec{F}_e+\vec{F}_m$
$=q \vec{E}+q(\vec{V} \times \vec{B})$

Topic: Physics, Magnetic Effects of Current, JEE Main
A charged ball B hangs from a silk thread $S$ which makes an angle $\theta$ with a large charged conducting sheet $\mathrm{P}$, as show in the figure. The surface charge density $\sigma$ of the sheet is proportional to

Options:
A
$\cos \theta$

B
$\cot \theta$

C
$\sin \theta$

D
$\tan \theta$

The correct answer is:
$\tan \theta$

Solution:
$\tan \theta=\frac{q \sigma}{\left(2 \varepsilon_{\circ}\right) m g}$

Topic: Physics, Electrostatics, JEE Main
A charged ball B hangs from a silk thread $S$ which makes an angle $\theta$ with a large charged conducting sheet $\mathrm{P}$, as show in the figure. The surface charge density $\sigma$ of the sheet is proportional to

Options:
A
$\cos \theta$

B
$\cot \theta$

C
$\sin \theta$

D
$\tan \theta$

The correct answer is:
$\tan \theta$

Solution:
$\tan \theta=\frac{q \sigma}{\left(2 \varepsilon_{\circ}\right) m g}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2005
A charged dust particle of radius 5Ã10-7 m is located in a horizontal electric field having an intensity of 6.28Ã105 V m-1. The surrounding medium in air with coefficient of viscosity Î·=1.6Ã10-15N s m-2. If this particle moves with a uniform horizontal speed of 0.01 m s-1, the number of electrons on it will be

Options:
A 20
B 15
C 25
D 30

The correct answer is:
15

Solution:
n e E=6Ï Î· r v or n=6Ï Î· r ve E
=6Ã3.14Ã1.6Ã10-5Ã5Ã10-7Ã0.011.6Ã10-19Ã6.28Ã105=15

Topic: Physics, Electromagnetic Waves, JEE Main
A charged oil drop is suspended in a uniform field of $3 \times 10^4 \mathrm{~V} / \mathrm{m}$ so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge $=9.9 \times 10^{-15} \mathrm{~kg}$ and $\mathrm{g}=$ $10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A
$3.3 \times 10^{-18} \mathrm{C}$

B
$3.2 \times 10^{-18} \mathrm{C}$

C
$1.6 \times 10^{-18} \mathrm{C}$

D
$4.8 \times 10^{-18} \mathrm{C}$

The correct answer is:
$3.3 \times 10^{-18} \mathrm{C}$

Solution:
Since ball is hanging in equilibrium, force by gravity is balanced by electric force.
$$
\begin{aligned}
& \mathrm{qE}=\mathrm{mg} \\
& \Rightarrow \mathrm{q}=\frac{\mathrm{m} \times \mathrm{g}}{\mathrm{E}} \\
& \Rightarrow \frac{9.9 \times 10^{-15} \times 10}{3 \times 10^4} \\
& \therefore \mathrm{q}=3.3 \times 10^{-18} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2004
A charged particle carrying charge 1 μC is moving with velocity (2i^+3j^+4k^) m s-1. If an external magnetic field of (5i^+3j^-6k^)×10-3T exists in the region where the particle is moving then the force on the particle is F→×10-9 N . the vector F→ is :

Options:
A -0.30i^+0.32j^-0.09k^
B -30i^+32j^-9k^
C -300i^+320j^-90k^
D -3.0i^+32j^-0.9k^

The correct answer is:
-30i^+32j^-9k^

Solution:
Using the equation of force as,

F→=qV→×B→

F→=10-6×10-3i^j^k^23453-6

=-30i^+32j^-9k^×10-9 N

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (03 Sep Shift 1)
A charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v$. The associated magnetic moment $\mu$ is given by:

Options:
A $q v R^2$
B $q v R^2 / 2$
C $q v R$
D $q v R / 2$.

The correct answer is:
$q v R / 2$.

Solution:
Magnetic moment $\mu=I A$
Since $T=\frac{2 \pi R}{v}$
also $I=\frac{q}{T}=\frac{q v}{2 \pi R}$
$\begin{aligned}
\therefore \quad \mu & =\left(\frac{q v}{2 \pi R}\right)\left(\pi R^2\right) \\
& =\frac{q v R}{2}
\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main
A charged particle enters a magnetic field $H$ with its initial velocity making an angle of $45^{\circ}$ with $H$. The path of the particle will be

Options:
A A straight line
B A circle
C An ellipse
D A helix

The correct answer is:
A helix

Solution:
At point A, charge is entering in a magnetic field in which direction of field is shown in the figure. The velocity of particle is \(u\) making an angle of \(45^{\circ}\) with field. We resolve it in two directions, one along the field and other perpendicular to it. Since \(\mathrm{u} \sin 45^{\circ}\) is perpendicular to \(\mathrm{H}\), it will create a rotatory effect on the charge. So charge particle will start rotating with axis along the direction of \(\mathrm{H}\). At the same time it will move forward with velocity \(\mathrm{u}\) \(\cos 45^{\circ}\). Under both these motions, it will have helical path as shown in the figure.

Topic: Physics, Magnetic Effects of Current, JEE Main
A charged particle enters a uniform magnetic field with a velocity vector at an angle of 45° with the magnetic field. The pitch of the helical path followed by the particle is p. The radius of the helix will be -

Options:
A p2π
B 2p
C p2π
D 2 pπ

The correct answer is:
p2π

Solution:
Pitch of helical path is,
p=v cosâ¡Î¸ Ã 2Ï mq B=2 Ï mv cosâ¡45Â°q B
=2 Ï mv2 q B
Also radius of helix is,
r=mv sinâ¡Î¸qB=mv 2 q B
â´ rp=mv2 q B Ã 2 q B2Ï mv=12Ï
or r=p2Ï

Topic: Physics, Magnetic Effects of Current, JEE Main
A charged particle going around in a circle can be considered to be a current loop. A particle of a mass m carrying charge q is moving in a plane with speed v under the influence of magnetic field B→. The magnetic moment of this moving particle is :

Options:
A mv2B→2B2
B -mv2B→2πB2
C −mv2B→B2
D −mv2B→2B2

The correct answer is:
−mv2B→2B2

Solution:
i=qT=qv2πr

M=i×πr2

=2v2πr×πr2

=12qv×mvqB

Direction of M→ is opposite of B→ therefore

M→=-mv2B→2B2

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (06 Sep Shift 2)
A charged particle having charge q and mass m moves rectilinearly under the action of an electric field E(x)=(4 â3x) N/C where x is the distance in metre from the point where the particle was initially at rest. The distance travelled by the particle till it comes instantaneously to rest again for the first time, is

Options:
A 23m
B 43m
C 2Â m
D 83 m

The correct answer is:
83 m

Solution:
F=qE(x)
F=q(4 – 3x)
a=qm4-3x
dvdx dxdt=qm4-3x

∫00vdv=qm∫0x4dx-3xdx

(The initial and final value of velocity is zero)

v22=qm4x-3x22
v2=2qm4x-3x22
As the particle comes to rest
4x-3x22=0
32x=4
x=83 m

Topic: Physics, Electrostatics, JEE Main
A charged particle (mass m and charge q) moves along X axis with velocity V0. When it passes through the origin it enters a region having uniform electric field E→=-Ej^ which extends upto x=d. Equation of path of electron in the region x>d is:

Options:
A y=qEdmv02x-d
B y=qEdmv02d2-x
C y=qEdmv02
D y=qEd2mv02x

The correct answer is:
y=qEdmv02d2-x

Solution:
x>d path is straight line

-y-12at2x-d=atV0

-y-12at2at=x-dV0

-yat-12dV0=xV0-dV0

-myV0qEd=xV0-d2V0

y=-qEdmV0xV0-d2V0

y=qEdmV02d2-x.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (02 Sep Shift 1)
A charged particle moves through a magnetic field in a direction perpendicular to it. Then the:

Options:
A speed of the particle remains unchanged
B direction of the particle remains unchanged
C acceleration remains unchanged
D velocity remains unchanged

The correct answer is:
speed of the particle remains unchanged

Solution:
A charged particle moves through a magnetic field in a direction perpendiculars to it. Then the speed of the particle remains unchanged.

Topic: Physics, Magnetic Effects of Current, JEE Main
A charged particle moves through a magnetic field in a direction perpendicular to it. Then the:

Options:
A speed of the particle remains unchanged
B direction of the particle remains unchanged
C acceleration remains unchanged
D velocity remains unchanged

The correct answer is:
speed of the particle remains unchanged

Solution:
A charged particle moves through a magnetic field in a direction perpendiculars to it. Then the speed of the particle remains unchanged.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2003
A charged particle moves through a magnetic field perpendicular to its direction. Then

Options:
A
the momentum changes but the kinetic energy is constant

B
both momentum and kinetic energy of the particle are not constant

C
both, momentum and kinetic energy of the particle are constant

D
kinetic energy changes but the momentum is constant

The correct answer is:
the momentum changes but the kinetic energy is constant

Solution:
No Solution Available

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2007
A charged particle moves through a magnetic field perpendicular to its direction. Then

Options:
A both momentum and kinetic energy of the particle are changed
B its momentum changes but its kinetic energy remains constant
C its kinetic energy changes but momentum remains constant
D both momentum and kinetic energy of the particle remains constant

The correct answer is:
its momentum changes but its kinetic energy remains constant

Solution:
Magnetic force is perpendicular to the direction of motion of the particle.
Therefore no external work is done and the kinetic energy remains constant. The particle undergoes uniform circular motion, however due to change in direction of motion velocity is not constant. So the momentum is not constant, during uniform circular motion. Therefore, option (B) is correct.

Topic: Physics, Magnetic Effects of Current, MHT CET, MHT CET 2022 (08 Aug Shift 2)
A charged particle moves through a magnetic field perpendicular to its direction. Then

Options:
A kinetic energy changes but the momentum is constant
B the momentum changes but the kinetic energy is constant
C both momentum and kinetic energy of the particle are not constant
D both momentum and kinetic energy of the particle are constant

The correct answer is:
the momentum changes but the kinetic energy is constant

Solution:
When a charged particle moves through a magnetic field perpendicular to its direction, it attains a circular path. The magnetic field exerts force in a direction perpendicular to its direction of velocity so its changes at each point but the magnitude is same. Thus work done by magnetic force is zero so kinetic energy of charge particle remains same. As external force is acting on the system momentum will not conserve.

Topic: Physics, Electromagnetic Induction, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 1)
A charged particle moves through a magnetic field perpendicular to its direction. Then

Options:
A the momentum changes but the kinetic energy is constant
B both momentum and kinetic energy of the particle are not constant
C both, momentum and kinetic energy of the particle are constant
D kinetic energy changes but the momentum is constant

The correct answer is:
the momentum changes but the kinetic energy is constant

Solution:
The charged particle moves with constant speed on a circular path, i.e. it changes its direction continuously. Hence momentum changes, but as the speed is constant hence the kinetic energy will remain constant.

Topic: Physics, Magnetic Effects of Current, NEET
A charged particle moves with constant velocity in a region where no effect of gravity is felt but an electrostatic field $\overrightarrow{\mathrm{E}}$
together with a magnetic field $\vec{B}$ may be present. Then which of the following cases are possible?

Options:
A $\vec{E} \neq 0, \vec{B} \neq 0$
B $\vec{E} \neq 0, \vec{B}=0$
C $\vec{E}=0, \vec{B}=0$
D $\overrightarrow{\mathrm{E}}=0, \overrightarrow{\mathrm{B}} \neq 0$

The correct answers are:
$\vec{E} \neq 0, \vec{B} \neq 0$, $\vec{E}=0, \vec{B}=0$, $\overrightarrow{\mathrm{E}}=0, \overrightarrow{\mathrm{B}} \neq 0$

Solution:
If a charged particle is moving in a gravity-free space without changing its velocity, then three cases possible
(i) Particle can move with constant velocity in any direction, if \(\vec{E}=0, \vec{B}=0\).
(ii) If \(\vec{E}, \vec{B}\) are having values such that \(q E=q v B\) and hence both forces (Force due to magnetic field and force due to electric field) acts in such a way that they cancel each other, in this case also particle can move with uniform velocity.
(iii) If \(\vec{E}=0, \vec{B} \neq 0\) and \(\vec{B}\) is such that it acts in the direction of velocity then magnetic force would be zero and hence velocity can remain constant.

Topic: Physics, Electromagnetic Waves, WBJEE, WBJEE 2020
A charged particle moving in a uniform magnetic field penetrates the layer of lead and thereby losses one-half of its kinetic energy. The radius of curvature of its path

Options:
A The radius reduces to r2
B The radius reduces to r2
C The radius remains the same
D The radius becomes r2

The correct answer is:
The radius reduces to r2

Solution:
r= 2mEqB and r⁡1=2mE⁡/2qB
So, r⁡ 1 = r⁡ 2

Topic: Physics, Magnetic Effects of Current, NEET