Physics

Magnetism and Magnetic Effects

230 Questions

Magnetism and magnetic effects focus on the forces exerted by magnetic fields on moving charges and magnetic materials. Questions cover magnetic dipoles, flux density, the motion of charged particles, and electromagnetic relationships. It is a vital physics topic for government competitive exams.

Magnetic dipolesCharged particle motionMagnetic flux densityBar magnetsEarth magnetism

Magnetism and Magnetic Effects Questions

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}} $

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

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

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 physics current electricity meter bridge meter bridge and problems on it galvanometer

An electron is released from the origin at a place where a uniform electric field $\overrightarrow{E}$ and uniform magnetic field $\overrightarrow{B}$ exit along the negative y-axis and the negative z-axis respectively.

  1. At time $t$ the y-component of velocity of the electron becomes $u _y = \dfrac{E}{B}\sin \omega t$ where $\omega = \dfrac{eB}{m}$.
  2. At $t = \pi m / eB$ the electron will have only x-component of velocity.
  3. At $t = \dfrac{2\pi m}{3eB}$, the y-component of velocity becomes zero.
  4. The displacement along y-axis is $\dfrac{2Em}{eB^2}$ when the velocity of electron becomes perpendicular to the y-axis
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

An electromagnetic wave is propagating along x-axis. At x = 1 m and t = 10 s, its electric vector |$\overset{-}{E}|  = 6 V/m$ then the magnitude of its magnetic vector is:

  1. $2 \, \times \, 10^{-8} \, T$
  2. $3 \, \times \, 10^{-7} \, T$
  3. $6 \, \times \, 10^{-8} \, T$
  4. $5 \, \times \, 10^{-7} \, T$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electric and magnetic compounds of an electromagnetic field are related by 

$E = CB$
$B = \dfrac{E}{C}$
$B = \dfrac{6}{3 \times 10^8}$  (when $E$ is given)
$B = 2 \times 10^{-8} T$ 

Multiple choice the nature of electromagnetic waves space exploration and forms of light observing space: telescopes electromagnetic waves physics

A plane electromagnetic wave travels in free space along X-direction. If the value of $\overrightarrow { B } $ (in tesla) at a particular point in space and time is $1.2 \times {10}^{-8} \hat {k}$, the value of $\overrightarrow { E } $ (in V ${m}^{-1}$) at that point is,

  1. $1.2\ \hat {j}$
  2. $3.6\ \hat {k}$
  3. $1.2\ \hat {k}$
  4. $3.6\ \hat {j}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given: The magnetic field of the plane electromagnetic wave is $1.2×10^{-8} \hat k\ T$
The direction of propagation of the electromagnetic wave is along X-direction.

The magnitude of  $\vec E$ is given by:
$E\, = \, B\cdot c\\ \ \ \ \ = (1.2 \, \times \, 10^{-8} T)(3 \, \times \, 10^{-8} m \,  s^{-1})\\ \ \ \ \ = 3.6 \,  V/m$

Since the magnetic field is along $Z-$ direction and the wave propagates along $X -$ direction. Therefore $\vec E$  should be in a direction perpendicular to both $X$ and $Z$ axes.

Using vector algebra  should be along X-direction.

Since $(+\hat j) \times (\hat k )= \hat i$

$\vec E$ is along the $Y-$direction.
Thus,  $\vec E= 3.6\hat j\ Vm^{-1}$