Tag: magnetic effects of current and magnetism

Questions Related to magnetic effects of current and magnetism

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 :  

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

  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

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Correct Option: C

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:

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

  1. $2t$

  2. $T$

  3. $\cfrac { t }{ \sqrt { 2 } } $

  4. $\cfrac { t }{ 2 } $

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Correct Option: C
Explanation:

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

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

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

  1. $ 11.5 N $

  2. $23 N $

  3. $46 N $

  4. $230 N $

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Correct Option: A
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} $

$= 11.5{\rm{N}}$

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.

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

  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. $[uB/ E is an integer]$

Show answer
Correct Option: A
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.

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 

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

  1. zero

  2. $-E _0 x _0$

  3. $-2 \, E _0 x _0$

  4. $-\sqrt{5} E _0 x _0$

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Correct Option: C

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 :

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

  1. $45m$

  2. $4.5m$

  3. $0.45m$

  4. $0.045m$

Show answer
Correct Option: C
Explanation:
$V=\dfrac{E}{B} $ For no deflection to occur
$V=\dfrac{3.2\times 10^5}{2\times 10^{-3}}=1.6\times 10^8m/s$
$R=\dfrac{mv}{qB}$
$R=\dfrac{9.1\times10^{-31}\times1.6\times 10^8}{1.6\times10^{-19}\times 2\times10^{-3}}$
$R=0.45m$


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:

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

  1. $\displaystyle \dfrac{3\mathrm{R}}{2}$

  2. $\sqrt{\dfrac{3}{2}}\mathrm{R}$

  3. $\sqrt{\dfrac{2}{9}}\mathrm{R}$

  4. $\sqrt{\dfrac{4}{3}}\mathrm{R}$

Show answer
Correct Option: C
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 $\sqrt{\dfrac{2}{9}}R$

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

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

  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

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Correct Option: A
Explanation:

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

A magnetic pole of pole strength 9.2 A m. is placed in a field of induction 50x10$^{-6}$ Tesla. The force experienced by the pole is :

electrostatic and magnetic analogy magnetism and matter magnetic effects of current and magnetism physics

  1. 46N

  2. 46$\times$10$^{-4}$N

  3. 4.6$\times$10$^{-4}$N

  4. 460N

Show answer
Correct Option: C
Explanation:

$m=9.2Am$
$B=50\times 10^{-6}T$
$F=B\times m$
$F=9.2\times 50\times 10^{-6}$
    $=4.6\times 10^{-4}N$

Two equal poles repel each other with a force of 10$^{-3}$ N. When placed 2cm apart in air, the pole strength of each is (in amp-m).

electrostatic and magnetic analogy magnetism and matter magnetic effects of current and magnetism physics

  1. 4$\pi $

  2. 2

  3. 4

  4. 2$\pi $

Show answer
Correct Option: B
Explanation:

$F=10^{-3}N$
$d=2cm$
$m _{1}=m _{2}=m$
$F=\dfrac{\mu _{0}m _{1}m _{2}}{4\pi d^{2}}$

$10^{-3} = \dfrac{10^{-7}\times m^{2}}{(2\times 10^{-2})^2}$
$m=2Am$