Tag: magnetic effects of current and magnetism

Questions Related to magnetic effects of current and magnetism

The migration of fine particles of a solid suspended in a liquid to the anode or cathode when an electric field is applied to the suspension is called.

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

  1. Anaphoresis

  2. Electrophoresis

  3. Cataphoresis

  4. Electroviscosity

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Correct Option: B
Explanation:

Electrophoresis is the migration of fine particles of a solid suspended in a liquid to the anode or cathode when an electric field is applied to the suspension.

At a distance $\lambda$ from a uniformly charged long wire, a charged particle is thrown radially outward with a velocity $u$ in the direction perpendicular to the wire. When the particle reaches a distance $2\lambda$ from the wire its speed is found to be $\sqrt{2\ u}$. The magnitude of the velocity, when it is a distance $4\lambda$ away form the wire, is (ignore gravity)

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

  1. $\sqrt{3\ u}$

  2. $2u$

  3. $2\sqrt{3\ u}$

  4. $4u$

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Correct Option: C

A mass particle (mass $=m$ and charge $=q$) is placed between two point charges of charge $q$, separation between these two charges is $2L$. The frequency of oscillation of mass particle, if it is displaced for a small distance along the line joining the charges is ?

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

  1. $\dfrac{q}{2\pi}\sqrt{\dfrac{1}{m\pi\epsilon _{0} L^{3}}}$

  2. $\dfrac{q}{2\pi}\sqrt{\dfrac{4}{m\pi\epsilon _{0}L^{4}}}$

  3. $\dfrac{q}{2\pi}\sqrt{\dfrac{1}{4m\pi\epsilon _{0}L^{3}}}$

  4. $\dfrac{q}{2\pi}\sqrt{\dfrac{1}{5m\pi\epsilon _{0}L^{3}}}$

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Correct Option: A

A particle having charge $q$ and mass $m$ is projected with velocity $v= 2\hat i -3\hat j$ in a uniform electric field $E=E _0\hat j$. Change in momentum $\left| \Delta p \right| $  during any time interval is given by

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

  1. $\sqrt{13}$ m

  2. $qE _0t$

  3. $\cfrac{qE _0t}{m}$

  4. Zero

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Correct Option: B
Explanation:
$v=2\hat { i } -3\hat { j } $
charge $=q$
mass $=m$
field $=E={ E } _{ 0 }\hat { j } $ 
change in moment $(AP)$ during any time interval $=$
$\Delta p=$ momentum 
       $=p\times t$
$p=q{ E } _{ 0 }$
   $=q{ E } _{ 0 }\times t$

A uniform electric field 'E' is directed towards positive X-axis. If at X=0, the electric potential is zero, then the potential at $X=+X _0,$ would be 

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

  1. $\dfrac{E}{X _0}$

  2. $\dfrac{-E}{X _0}$

  3. $-EX _0$

  4. $EX _0$

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Correct Option: A
Explanation:

We know that electric potential $V=\dfrac { dE }{ dx } $

Since $X=x+{ x } _{ 0 }$  so,
$V=\dfrac { E }{ { x } _{ 0 } } $

Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii $R _{1}$ and $R _{2}$ respectively. The ratio of masses of X and Y is-

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

  1. $(R _{1}/R _{2})^{1/2}$

  2. $(R _{2}/R _{1})$

  3. $(R _{1}/R _{2})^{2}$

  4. $(R _{1}/R _{2})$

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Correct Option: C

The path of cathode rays in an electric field can be approximated to a circle of radius r. In order to double the radius of the circular path, we must 

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

  1. reduce the electric field to half

  2. double the electric field

  3. increase electric field four times

  4. reduce electric field to one fourth

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Correct Option: B
Explanation:

Given, path of cathode rays is nearly circular. So, electric force must be acting as centripetal force for the cathode rays and it is given by
$ F _e = F _c $
qE = $ m \omega ^2 r$
we can conclude that $E \alpha r$ ,provided all the other quantities are constant.
Thus, to double the radius we should double electric field strength.

Cathode rays are made to pass between the poles of a magnet. The effect  of magnetic field is 

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

  1. to increase their velocity

  2. to deflect them towards north pole

  3. to deflect them towards south pole

  4. to deflect them out of the plane of magnetic field

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Correct Option: D
Explanation:

When a magnetic field is  applied, the cathode ray is  deflected from its normal  straight path into a curved path which can be identified by right hand thumb rule.

An electron and a proton are injected into a uniform magnetic field perpendicular to it in the same direction. If electron and proton have same kinetic energy then the radius of curvature is 

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

  1. more for proton

  2. more for electron

  3. same for both

  4. none of thesr

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Correct Option: A
Explanation:

let, Kinetic energies of electron and proton be K.
magnetic force on particle provides the centripetal force to support the circular motion.
If r is radius of curvature, $ F _b = F _c $
i.e. qvB = m $\frac{v^2}{r} $.
Solving the above equation we get r = $ \frac {p}{qB} $ where p = momentum = mv
If, q and B are constant we can say that $ r \alpha p $
But, K.E. = $ \frac {mv^2}{2}$.
Derive P in terms of K.E. which turns out to be
p = ${(2m(K.E.))}^{\frac{1}{2}}$
Given that, K.E. is same for either particles. So, $p  \ \alpha \  m^{\frac{1}{2}}$
therefore, $ r \  \alpha \  p \  \alpha \  m^{\frac{1}{2}}$ and radius is more for higher mass particle which is a proton in this case.

An electron and a proton are injected into a uniform magnetic field perpendicular to it with the same momentum. If both particles are fired with same momentum into a transverse electric field, 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 less curved

  2. proton trajectory is less curved

  3. both trajctories are equally curved

  4. both trajectories are straight lines

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Correct Option: B
Explanation:

let, momentum of electron and proton be p.
Transverse electric force on particle provides the centripetal force to support the circular motion.
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 {p^2}{mqE} $ 

where, $p = momentum = mv$

Given $q$ ,$p$ and $E$ are constant. So, we can say that $ r \propto \dfrac{1}{m} $

Radius of curvature is higher for a lighter mass particle which is an electron and vice versa for proton.