Tag: magnetic effects of current and magnetism

Questions Related to magnetic effects of current and magnetism

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

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

  1. electron trajectory is more curved

  2. proton trajectory is more curved

  3. both trajectories are equally curved

  4. both trajectories are straight lines

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Correct Option: C
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.

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

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

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

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

  3. $r$

  4. cannot be determined

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Correct Option: B
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 } $

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

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

  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

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Correct Option: D
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).

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

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

  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 $\overrightarrow{B}$

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Correct Option: A,B,C
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.

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

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

  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 $\overrightarrow{E}$

Show answer
Correct Option: A
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.

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

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

  1. 12 cm

  2. 16 cm

  3. 11 cm

  4. 18 cm

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Correct Option: C
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$  

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?

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

  1. 2 MeV

  2. 1 MeV

  3. 0.5 MeV

  4. 4 MeV

Show answer
Correct Option: B
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 $K _p = K _{\alpha} = 1 Me V$

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?

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

  1. 1 : 2 : 2

  2. 2 : 1 : 1

  3. 1 : 1 : 2

  4. 1 : 2 : 1

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Correct Option: A
Explanation:

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


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

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

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

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

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

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

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

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

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?

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

  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. $2 \times 10^{11} C kg^{-1}$

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Correct Option: D
Explanation:

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


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

$\Rightarrow F=F _B$

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

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

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

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

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

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

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

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

Show answer
Correct Option: D
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}} $

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