Tag: motion of charged particle in magnetic field and electric field

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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 and a proton are injected into a uniform magnetic field perpendicular to it with the same momentum. If the two particles are injected into a uniform transverse electric field with same kinetic energy, then

  1. electron trajectory is more curved

  2. proton trajectory is more curved

  3. both trajectories are equally curved

  4. both trajectories are straight lines
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let, Kinetic energies of electron and proton be K.
Transverse electric force on particle provides the centripetal force to support the circular motion in this case.
If r is radius of curvature, $ F _e = F _c $


i.e. qE = m $\dfrac{v^2}{r} $.

Solving the above equation we get r = $ \dfrac {2K}{qE} $ where K = Kinetic Energy = $\dfrac {mv^2}{2}$


Here, q,E and K are constant. So, radius of curvature is same for both of them.

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

An infinitely long current carrying wire carries current $i$. A charge of mass $m$ and charge $q$ is projected with speed $v$ parallel to the direction of current at a distance $r$ from it. Then, the radius of curvature at the point of projection is

  1. $\cfrac { 2rmv }{ q{ \mu } _{ 0 }i } $

  2. $\cfrac { 2\pi rmv }{ q{ \mu } _{ 0 }i } $

  3. $r$

  4. cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Magnetic field due an infinitely long straight wire at a distance $r$    $B = \dfrac{\mu _o i}{2\pi r}$
Magnetic force acting on the charge  $F=qvB$ $=\cfrac { { qv\mu  } _{ 0 }i }{ 2\pi r } $
This force makes the charge to trace a curved path of radius of curvature $R$.
Now  $F _{centripetal} = F$

$\therefore$ $\cfrac { m{ v }^{ 2 } }{ R } =\cfrac { { qv\mu  } _{ 0 }i }{ 2\pi r }$
$ \Rightarrow R=\cfrac { 2\pi rmv }{ { q\mu  } _{ 0 }i } $

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

A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. This particle will

  1. get deflected in vertically upward direction

  2. move in circular path with an increased speed

  3. move in a circular path with decreased speed

  4. move in a circular path with uniform speed
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When a positively charged enters in a region of uniform magnetic field directed vertically upwards, it experiences a centripetal force which moves the particle in circular path with a uniform speed (in clockwise direction).

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

If a charged particle goes unaccelerated in a region containing electric and magnetic fields, then $($more than one may be correct$)$

  1. $\overrightarrow{E}$ must be perpendicular to $\overrightarrow{B}$

  2. $\overrightarrow{v}$ must be perpendicular to $\overrightarrow{B}$

  3. $\overrightarrow{v}$ must be perpendicular to $\overrightarrow{E}$

  4. $\overrightarrow{E}$ must be parallel to $\overrightarrow{B}$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

Charge particle in a region goes unaccelerated only if magnetic force and electric force action on it cancle each other.
therefore, $q\overrightarrow E = q(\overrightarrow v \times \overrightarrow B)$
therefore, $\overrightarrow{E}$ is definitely perpendicular to $\overrightarrow{B}$
there is no necessity of $\overrightarrow{v}$ being perpendicular to $\overrightarrow{B}$ but $\overrightarrow{v}$ should not be parallel to $\overrightarrow{B}$ otherwise magnetic force will be zero leading to imbalancing the effect of electric force and particle will accelerate therefore, option(A) is true but option (B) is not true from this point of view which may be a mistake.
But since magnetic force will always perpendicuar to the direction of motion of charge particle therefore, Electric force should also be perpendicular and opposite to the magnetic force to balance the magnetic force therefore, $\overrightarrow{v} $ should be perpendicular to $\overrightarrow{E}$
and since $\overrightarrow{E}$ and $\overrightarrow{B}$ are perpendicular to each other
therefore, $\overrightarrow{v},\overrightarrow{E}, \ and\  \overrightarrow{B}$ all are mutually perpendicular to each other.

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

A charge particle goes undeflected in a region containing electric and magnetic fields. It is possible that $($more than one may be correct$)$

  1. $\overrightarrow{E} || \overrightarrow{B},\ \overrightarrow{v} || \overrightarrow{E}$

  2. $\overrightarrow{E}$ is not parallel to $\overrightarrow{B}$

  3. $\overrightarrow{v} || \overrightarrow{B}$ but $\overrightarrow{E}$ is not parallel to $\overrightarrow{B}$

  4. $\overrightarrow{E} || \overrightarrow{B}$ but $\overrightarrow{v}$ is not parallel to $\overrightarrow{E}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\vec{F}=q(\vec{v} \times \vec{B})+q \vec{E}$
In the option (A), the force due to magnetic field is zero and the force is along electric field.

Therefore the particle is undeflected.
In all other cases, force is not along the direction of the particle.

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

A $10eV$ electron is circulating in a plane at right angles to a uniform field at a magnetic induction $10^{-4} Wb/m^2 (= 1.0 gauss)$. The orbital radius of the electron is

  1. 12 cm

  2. 16 cm

  3. 11 cm

  4. 18 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given : $K=10 eV$   $B =10^{-4}$  $Wb/m^2$                 

Using : $mv=Bqr$   where  $mv = \sqrt{2Km}$

$\implies$$r =\sqrt{ \dfrac{2Km}{B^2q^2}}$

$\therefore$    $r =\sqrt{ \dfrac{2\times 9.1\times 10^{-31}\times 10 e}{(10^{-8})e^2}}  = 1.06\times 10^{-2}\ m$ $ \approx 11\ cm$  

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 carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circle of same radius in the same field?

  1. 2 MeV

  2. 1 MeV

  3. 0.5 MeV

  4. 4 MeV
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a charged particle's motion  in  a magnetic field

$\displaystyle  F _C = F _m \Rightarrow \frac{mv^2}{R} = qVB$
$\Rightarrow \displaystyle  R _p = \frac{m _pv _p}{q _pB} = \frac{P _p}{q _pB} = \frac{\sqrt{mK _p}}{qB}$
$\displaystyle  R _{\alpha} = \frac{\sqrt{2 (4m) K _{\alpha}}}{2qB}$
$ \displaystyle  \frac{R _p}{R _{\alpha}} = \sqrt{\frac{K _p}{K _{\alpha}}}$

but $R _p = R _{\alpha} $ (given)

Thus $K _p = K _{\alpha} = 1 Me V$

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, deutron and an $\alpha$-particle enter a magnetic field perpendicular to field with same velocity. What is the ratio of the radii of circular paths?

  1. 1 : 2 : 2

  2. 2 : 1 : 1

  3. 1 : 1 : 2

  4. 1 : 2 : 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\text{Force on a charged particle due circular motion: }\ F=\dfrac{mv^2}{r}$


$\text{Force on a charged particle due to magnetic field: }\ F _B=qvB$

$F=F _B\Rightarrow \dfrac{mv^2}{r}=qvB\Rightarrow r=\dfrac{mv}{qB}$

$\text{Here}\ v\ \text{and}\ \text{are constant.}$

$\Rightarrow r \propto \dfrac{m}{q}$

$\text{For proton:}\ r _p=1\times k$

$\text{For deutron:}\ r _d=2k$

$\text{For an-}\alpha\ \text{particle:}\ r _{\alpha}=2k$

$\text{Therefore, }\ r _p:r _d:r _{\alpha}=1:2:2$

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 moves in a circular arc of radius 10 m at a constant speed of $2 \times 10^7 ms^{-1}$ with its plane of motion normal to magnetic flux density of $10^{-5}$T. What will be the value of specific charge of the electron?

  1. $2 \times 10^4 C kg^{-1}$

  2. $2 \times 10^5 C kg^{-1}$

  3. $5 \times 10^6 C kg^{-1}$

  4. $2 \times 10^{11} C kg^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\text{Force due to circular motion:} F=\dfrac{mv^2}{r}$


$\text{Force due to magnetic field:}\ F _B=qvB$

$\Rightarrow F=F _B$

$\Rightarrow \dfrac{mv^2}{r}=qvB$

$\Rightarrow \dfrac{q}{m}=\dfrac{v}{rB}$

$\text{Let}\ r=10\ \text{m},\ v=2\times 10^7\ \text{ms}^{-1},\ B=10^{-5}\ \text{T}$

$\Rightarrow \dfrac{q}{m}=\dfrac{2\times 10^7}{10\times 10^{-5}}$

$\Rightarrow \dfrac{q}{m}=2\times 10^{11}\ \text{C kg}^{-1}$

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

A cathode ray beam is bent in a circle of radius 2 cm by a magnetic induction $4.5 \times 10^{-3} weber/m^2 $. The velocity of electron is

  1. $3.43 \times 10^7 m/s$

  2. $5.37 \times 10^7 m/s$

  3. $1.23 \times 10^7 m/s$

  4. $1.58 \times 10^7 m/s$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle v = \dfrac{Bqr}{m} $

$= \dfrac{4.5 \times 10^{-3} \times 1.6 \times 10^{-19} \times 2 \times 10^{-2}}{9.1 \times 10^{-32}} $

$= 1.58 \times 10^7 m/s$