Tag: moving charges and magnetism

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Questions Related to moving charges and magnetism

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

A positive charge is released from the origin at a place where uniform electric field $E$ and a uniform magnetic field be exist along the positive $y-$axis and positive $z-$axis respectively, then :  

  1. initial the charge particle tends to move along positive $z-$axis
  2. initial the charge particle tends to move along negative $Y-$direction
  3. initial the charge particle tends to move along positive $y-$direction
  4. the charge particle moves in $y-z$ plane
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Lorentz force is F = q(E + v x B). Initially, the velocity v is zero, so the magnetic force is zero. The electric field E is along the positive y-axis, so the initial force F = qE is also along the positive y-axis, causing the particle to accelerate in that direction.

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

A charged particle placed in an electric field falls from rest through a distance $d$ in time $t$. If the charge on the particle is doubled, the time of fall through the same distance will be:

  1. $2t$
  2. $T$
  3. $\cfrac { t }{ \sqrt { 2 } } $
  4. $\cfrac { t }{ 2 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial velocity $u=0$
Force on charge$=F=qE$
$F=ma$
$qE=ma$
$\cfrac { qE }{ m } =a$
$s=ut+\cfrac { 1 }{ 2 } a{ t }^{ 2 }$
$\quad d=\cfrac { 1 }{ 2 } \cfrac { qE }{ m } { t } _{ 1 }^{ 2 }....(1)$
$d=\cfrac { 1 }{ 2 } \cfrac { qE }{ m } { t } _{ 2 }^{ 2 }....(2)$
$\cfrac { 1 }{ 2 } \cfrac { qE }{ m } { t } _{ 1 }^{ 2 }=\cfrac { 1 }{ 2 } \cfrac { qE }{ m } { t } _{ 2 }^{ 2 }\Rightarrow \cfrac { { t } _{ 1 } }{ \sqrt { 2 }  } ={ t } _{ 2 }$

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

Three equal charges, each having a magnitude of $ 4 \mu C$ , are placed at the three corners of a right-angled triangle of sides $6 cm, 8 cm$ and $10 cm.$ The force on the charge at the right-angle corner will be

  1. $ 11.5 N $
  2. $23 N $
  3. $46 N $
  4. $230 N $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Two forces are,

${F _1} = \dfrac{{k{Q _1}{Q _2}}}{{{R _1}^2}}$

$ = \dfrac{{9 \times {{10}^9} \times {{\left( {4 \times {{10}^{ - 6}}} \right)}^2}}}{{{{\left( {6 \times {{10}^{ - 2}}} \right)}^2}}}$

$ = 10{\rm{N}}$

${F _2} = \dfrac{{k{Q _1}{Q _2}}}{{{R _2}^2}}$

$= \dfrac{{9 \times {{10}^9} \times {{\left( {4 \times {{10}^{ - 6}}} \right)}^2}}}{{{{\left( {8 \times {{10}^{ - 2}}} \right)}^2}}}$

$= \dfrac{{360}}{{64}} = 5.62{\rm{N}}$

Resultant force at the right angle vertex is,

$F = \sqrt {\left( {{F _1}^2 + {F _2}^2} \right)} $

$= \sqrt {{{10}^2} + {{5.62}^2}} $

$ = \sqrt {131.56} $

$= 11.5{\rm{N}}$

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

A particle of specific charge (qm) is projected from the origin of coordinate with initial velocity $\left[ u\hat { i } -v\hat { j }  \right] $ Uniform electric magnetic fields exist in the region along the +y direction, of magnitude E and B. The particle will definitely return to the origin once if.

  1. $\dfrac{ BE}{2E } $ is an integer
  2. $\left( { u }^{ 2 }+{ v }^{ 2 }\quad ^{ 1/2 } \right) \left[ B/E \right] $
  3. $[VB/ E]$ in an integer
  4. $[uB/ E is an integer]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Charge $={ q } _{ m }$
Velocity $=u\hat { i } -u\hat { j } $
along the $+y$ direction $E$ and $B$.
The particle will definitely return to the origin once if.
In $Y$ direction the electric field $E$ exist. So, the force due to electric on the charge is given by 
$F=qE$
Now, the acceleration of the charge is given by $a=\dfrac { qE }{ m } $
now, the acceleration on is opposite to the direction of velocity. So, it will return to origin if displacement in $y-$ direction is zero.
$d={ V } _{ y }\times t+\dfrac { 1 }{ 2 } { at }^{ 2 }$
$0=-V\times t+\dfrac { 1 }{ 2 } \dfrac { qE }{ m } { T }^{ 2 }$
$T=\dfrac { 2mV }{ qE } $
now due to magnitude field it will move in circular path, time period of its circular motion is given by
$T=\dfrac { 2\pi m }{ qB } $
now if complete $N$ number of rounds in above time of return then
$N\times \dfrac { 2\pi m }{ qB } =\dfrac { 2mV }{ qE } $
$N=\dfrac { Bv }{ \pi E } $
So, here above value must be an integer so that it complete integral number of rounds 
integer $=\dfrac { Bv }{ \pi E } =\dfrac { Bv }{ 2E } $    ($\because$   $n=2$) in an integer.
Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

If uniform electric field $\vec{E} = E _0 \hat{i} + 2E _0 \hat{j}$ where $E _0$ is a constant, exists in a region of space and at (0, 0) the electric potential V is zero, then the potential at $(x _0, 0)$ will be 

  1. zero
  2. $-E _0 x _0$
  3. $-2 \, E _0 x _0$
  4. $-\sqrt{5} E _0 x _0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electric field is given by E = E0 i + 2E0 j. The potential difference between two points is V(x,y) - V(0,0) = - integral (E dot dl). Integrating along the x-axis from 0 to x0 gives V(x0,0) - 0 = - E0 * x0, resulting in -E0 x0.

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

An electron $($mass $=9.1\times 10^{-31}$; charge $=-1.6\times 10^{-19}\mathrm{C})$ experiences no deflection if subjected to an electric field of $3.2\times 10^{5}\mathrm{V}/\mathrm{m}$ and a magnetic field of $2.0\times 10^{-3}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$. Both the fields are normal to the path of electron and to each other. Ifthe electric field is removed, then the electron will revolve in an orbit of radius :

  1. $45m$
  2. $4.5m$
  3. $0.45m$
  4. $0.045m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$V=\dfrac{E}{B} $ For no deflection to occur
$V=\dfrac{3.2\times 10^5}{2\times 10^{-3}}=1.6\times 10^8m/s$
$R=\dfrac{mv}{qB}$
$R=\dfrac{9.1\times10^{-31}\times1.6\times 10^8}{1.6\times10^{-19}\times 2\times10^{-3}}$
$R=0.45m$
Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics


An electron having kinetic energy $\mathrm{T}$ is moving in a circular orbit of radius $\mathrm{R}$ perpendicular to a uniform magnetic induction $\vec{\mathrm{B}}$. If kinetic energy is doubled and magnetic induction tripled, the radius will become:

  1. $\displaystyle \dfrac{3\mathrm{R}}{2}$
  2. $\sqrt{\dfrac{3}{2}}\mathrm{R}$
  3. $\sqrt{\dfrac{2}{9}}\mathrm{R}$
  4. $\sqrt{\dfrac{4}{3}}\mathrm{R}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

$KE = T Jule$
$Radius = R$
$Magnetic \ Induction = \overrightarrow B$

Then , the radius
$R=\dfrac{\sqrt{2mT}}{qB}$
when, 
$KE = 2T$
$B =3\overrightarrow B$
$Radius = r'$

When,
 $r'=\dfrac{\sqrt{2m\left ( 2T \right )}}{q\left ( 3B \right )}$

$r' =\sqrt{2}\dfrac{\sqrt{2mT}}{3qB}$

$r'= \dfrac{\sqrt{2}}{3}\dfrac{\sqrt{2mT}}{qB}$

$r'=\dfrac{\sqrt{2}}{3}R$

Therefore, the radius will be $\sqrt{\dfrac{2}{9}}R$

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

A proton and a deutron initially at rest are accelerated with the same uniform electric field of time t.

  1. Both particles will have same momentum

  2. Both particles with have same K.E.

  3. Both particles with have same speed

  4. Both particles will cover same distance
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$qE=ma$
$a=\dfrac{qE}{m}$
$v=\dfrac{qEt}{m}$
$mv\rightarrow same$

Multiple choice physics moving charges and magnetism field due to a current carrying conductor magnetic field due to a straight current carrying conductor magnetic field lines due to current

SI unit of permittivity of free space is:

  1. Farad

  2. Weber
  3. ${ C }^{ 2 }{ N }^{ -1 }{ m }^{ -2 }$
  4. ${ C }^{ 2 }{ N }^{ -1 }{ m }^{ -1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Coulombic force between two charges is given by

F=1/4π€ (q'.q″/r²)

From here, the SI UNIT of permittivity comes out is

= coulomb²/Newton-meter²

C²/N-m²

Multiple choice physics moving charges and magnetism field due to a current carrying conductor magnetic field due to a straight current carrying conductor magnetic field lines due to current

If an electron is moving with velocity $\bar{v}$ produces a magnetic field $\bar{B}$, then

  1. the direction of field $\bar{B}$ will be same as the direction of velocity $\bar{v}$
  2. the direction of field $\bar{B}$ will be opposite as the direction of velocity $\bar{v}$
  3. the direction of field $\bar{B}$ will be perpendicular as the direction of velocity $\bar{v}$
  4. the direction of field $\bar{B}$ does not depend upon the direction of velocity $\bar{v}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Biot-Savart's law, the magnetic field
$\displaystyle \overset{\rightarrow}{B} = \frac{\mu _o}{4 \pi} . \frac{q (\overset{\rightarrow}{v} \times \overset{\rightarrow}{r} ) }{r^3}$
The direction of $\overset{\rightarrow}{B}$ will be along $\overset{\rightarrow}{v} \times \overset{\rightarrow}{r}$ i.e. perpendicular to the plane containing $\overset{\rightarrow}{v}$ and $\overset{\rightarrow}{r}$.