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.

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Magnetism and Magnetic Effects Questions

Multiple choice physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

If r be the distance of a point on the axis of a bar magnet from its centre, the magnetic field at this point is proportional to :

  1. (1/r)
  2. (1/r$^2$)
  3. (1/r$^3$)
  4. (1/r$^5$)
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a short Bar Magnet, the magnetic induction at a point on the axix at a distance $r$ from centre is given by  the formula

$B = $   $(\dfrac{\mu _0}{4\pi} )\dfrac{2M}{r^3}$

$\Rightarrow$ $B= \dfrac{K}{r^3}$

$\Rightarrow$ $B\propto \dfrac{1}{r^3}$
Therefore, C is correct option.

Multiple choice physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A bar magnet of magnetic moment 'M' has a magnetic length '2d'. Find magnetic induction on its equatorial line at a distance $'\sqrt{13 d}'$.

  1. $\displaystyle \frac{\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$
  2. $\displaystyle \frac{2\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$
  3. $\displaystyle \frac{4\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$
  4. $\displaystyle \frac{8\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$r = \sqrt{13} d ;  2l  = 2d$
$B _eq = \displaystyle \frac{\mu _0}{4 \pi} \times \frac{\mu}{(r^2 + 1^2)^{\frac{3}{2}}}$
$\displaystyle =\frac{\mu _0}{4\pi} \times \frac{M}{\left ((\sqrt{13}d)^2 + (2d)^2 \right )^{\frac{3}{2}}}$
$=\displaystyle \frac{\mu _0}{4\pi} \times \frac{M}{(17 d^2)^{\frac{3}{2}}} = \frac{\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

Multiple choice physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A magnetic induction due to a short bar magnet of magnetic moment 5.4 A m$^2$ at a distance of 30 cm on the equatorial line is :

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

$B _{equi} = \displaystyle \frac{\mu _0}{4 \pi } \frac{\mu}{r^3} = \frac{10^{-7}\times 5.4}{(0.3)^3} = 2 \times 10^{-5}T$

Multiple choice physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A short bar magnet with the north pole facing north forms a neutral point at P in the horizontal plane. If the magnet is rotated by $90^o$ in the horizontal plane, the net magnetic induction at $P$ is ( Horizontal component of earth's magnetic field $= B _H$):

  1. zero
  2. $2 B _H$
  3. $\displaystyle \dfrac{\sqrt{5}}{2} B _H$
  4. $\sqrt{5}B _H$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When the north pole of short bar magnet is facing North pole of the earth, at the neutral point P, which is on equatorial line. 
$B _H = \dfrac {\mu _0 M}{4\pi d^3} = B _1$ ............(1)
When the magnet is rotated by $90^o$, the magnetic induction at P which is on axial line,
$B _H = \dfrac {\mu _0 2M}{4\pi d^3} = B _2$ ............(2)
Therefore, net magnetic induction at P is
$B _{net} = \sqrt {(B _1^2 + B _2^2)}$
$B _{net} = \sqrt {(1^2 + 2^2)}B _H = \sqrt 5 B _H$

Multiple choice physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

The magnetic field strength at a point at a distance d from the centre on the axial line of a very short bar magnet of Magnetic moment $M$ is $B$. The Magnetic induction at a distance $2d$ from the centre on the equatorial line of a Magnetic Moment $8M$ will be

  1. $4B$
  2. $0.25B$
  3. $0.5B$
  4. $2B$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$B=\frac{2\mu _0m}{4 \pi d^{3}}$

$B equatorial =\frac{\mu _0m}{4 \pi d^{3}}$

$d _1=d$
$d _2=2d$
Also,
$m _1=M$
$m _2=8M$

$B _{1}=\frac{2\mu _0m}{4 \pi d^{3}}............1$

$B _{2}=\frac{2\mu _0m}{4 \pi d^{3}} = \frac{\mu _0 8M}{4 \pi (2d)^{3}}.........2$
From 1 & 2 we get that,
$B _{equatorial} =\frac{B}{2}$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

Two bar magnet are kept together and suspended freely in earth's magnetic field. When both like poles are aligned, the time period is 6 sec. When opposite poles are aligned, the time period is 12 sec. The ratio of magnetic moments of the two magnets is :

  1. $5/3$
  2. $2/1$
  3. $3/2$
  4. $3/1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$T _1= 6 = 2\pi \sqrt{\dfrac{1}{(M _1+M _2)B}}$
$T _2= 12 = 2\pi \sqrt{\dfrac{1}{(M _1+M _2)B}}$
$\therefore \dfrac{6}{12}=\dfrac{1}{2}= \sqrt{\dfrac{M- M _2}{M _1+M _2}}$ or $ M _1+M _2= 4( M _1-M _2) $
$ 3M _1= 5M _2$
$\dfrac{M _1}{M _2}=\dfrac{5}{3}$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A bar magnet of length $16 cm$ has a pole strength of 500 milli amp.m. The angle at which it should be placed to the direction of external magnetic field of induction $2.5$ gauss so that it may experience a torque of $\sqrt { 3 } \times{ 10 }^{ -5 }$ N.m. is

  1. $\pi $
  2. $\dfrac { \pi }{ 2 } $
  3. $\dfrac { \pi }{ 3 } $
  4. $\dfrac { \pi }{ 6 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A bar magnet of magnetic moment $M$ and moment of inertia I is freely suspended such that the magnetic axial line is in the direction of magnetic meridian. If the magnet is displaced by a very small angle $(\theta )$ , the angular acceleration is (Magnetic induction of earths horizontal field $ = $ $B _H$)

  1. $\dfrac{MB _{H}\theta }{I}$
  2. $\dfrac{IB _{H}\theta }{M}$
  3. $\dfrac{M\theta }{IB _{H}}$
  4. $\dfrac{I\theta }{MB _{H}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Torque acting on the magnet is  $T=MB _H sin \theta $
Since $ \theta$ is very small , $T = MB _H \theta $
$I {\alpha} =  MB _H \theta $
$ \alpha = \dfrac{MB _H \theta}{I} $

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

When a bar magnet is suspended in a uniform magnetic field, the torque acting on it will be :

a) maximum e) $\theta=45^o$ with the field
b) half the maximum field f) $\theta=60^o$ with the field
c) $\sqrt{3}/2$ times the maximum field g) $\theta=30^o$ with the field
d) $1/\sqrt{2}$ times the maximum field h) $\theta=90^o$ with the field

  1. a-g, b-h, c-d, d-e

  2. a-e, b-f, c-g, d-h

  3. a-f, b-e, c-g, d-h

  4. a-h, b-g, c-f, d-e
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Torque acting on the bar magnet suspended in a uniform field B is given by,

$T=MBsin\theta$  where $M$  is the moment of magnet.
Torque is maximum when $sin\theta=1$ i.e. $\theta=90^o$
$\therefore$ maximum torque $=MB$
Torque is half when $sin\theta=1/2$ i.e. $\theta=30^o$
Torque is $\sqrt{3}/2$ the maximum when $sin\theta=\sqrt{3}/2$ i.e. $\theta=60^o$
Torque is $1/\sqrt{2}$ the maximum when $sin\theta=1/\sqrt{2}$ i.e. $\theta=45^o$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A bar magnet of length $16 cm$ has a pole strength of $500\times 10^{-3}Am$ The angle at which it should be placed to the direction of external magnetic field of induction $2.5 G$ so that it may experience a torque of $\sqrt{3}\times 10^{-5}$ Nm is :

  1. $\pi $
  2. $\dfrac{\pi }{2}$
  3. $\dfrac{\pi }{3}$
  4. $\dfrac{\pi }{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$l=16m$
$P=500\times 10^{-3}Am$
$z=\vec{m}\times \vec{B}$

$B=2.5\;G=2.5\times 10^{-4}T$
$m=lP  $
torque due to magnetic field $z =\sqrt{3}\times 10^{-5}N.m$

$z=mBm \times sin Q$
$\sqrt{3}\times 10^{-5}=lPBsinQ$
$\sqrt{3}\times 16\times 500\times 2.5\times sinQ.$
$SinQ=\dfrac{\sqrt{3}}{2}$
$Q=\pi /3$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

When a bar magnet is placed perpendicular to a uniform a magnetic field, it is acted upon by a couple of magnitude $1.732\times 10^{-5}Nm$. The angle through which the magnet should be turned so that the couple acting on it becomes $1.5\times 10^{-5}Nm$ is

  1. $60^{o}$
  2. $45^{o}$
  3. $30^{o}$
  4. $75^{o}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\theta _{1}=\pi /2$
$z _{1}=1.732\times 10^{-5}$
$z _{2}=1.5\times 10^{-5}$
$z=mB\sin\theta $
$z _{max}=mB$
$=1.732\times 10^{-5}$
$z _{2}=mB\sin\theta ^{1}$
$1.5\times 10^{-5}=1.732\times 10^{-5}\sin\theta ^{1}$
$\sin\theta ^{1}=\dfrac{\sqrt{3}}{2}$
$\theta ^{1}=60$
Angle turned$=\pi /2-\theta ^{1}$
$=30^o$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A bar magnet of length 0.2 m and pole strength 5 A.m. should be kept in a uniform magnetic field of induction 15 tesla at angle ..... radians to the field so that the torque experienced by it will be 7.5N-m

  1. $\dfrac{\pi }{2}$
  2. $\dfrac{\pi }{3}$
  3. $\dfrac{\pi }{6}$
  4. $\dfrac{\pi }{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$l=0.2$
$P=50m$
$B=15T$
$z=7.5Nm$
$\vec{z}=\vec{m}\times \vec{B}   m=lP$
$z=mBsin\theta $
$7.5=0.2\times 5\times 15\times sin\theta $
$sin\theta =\dfrac{1}{2}$
$\theta =\pi /6$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A magnet of magnetic moment $50\hat{i} A-m^2$ is placed along x- axis in a magnetic field$ \overrightarrow {B} =( 0.5 \hat{i} + 3.0 \hat{j})$ tesla. The torque acting on the magnet is 

  1. $175 \hat{k} N-m$
  2. $75 \hat{k} N-m$
  3. $150 \hat{k} N-m$
  4. $25 \sqrt{37}\hat{k} N-m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

Magnetic field $\vec{B}\,=\,0.5\hat{i}\,+\,3.0\hat{j}\,\,T$

Magnetic moment $\mu =50\hat{i}\,\,A{{m}^{2}}$

Torque on magnet  $\tau $

$\tau =\mu \times \vec{B}\,\,=\,\,50\hat{i}\,\times \,(0.5\hat{i}\,+\,3.0\hat{j}\,)$

$\tau =\,150\hat{k}\,\,N-m$

Hence, Torque experiences by the magnet present in the magnetic field, $\tau =\,150\hat{k}\,\,N-m$ 

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A bar magnet of dipole moment M is initially parallel to a magnetic field of induction B. The angle through which it should be rotated so that the torque acting on it is half the maximum torque is _____.

  1. $90^0$
  2. $60^0$
  3. $45^0$
  4. $30^0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\tau=MXB=MBsinQ$

At maximum torque the Q is $90^o$ 
$\tau=MBsin(90^o)=MB$----------------1

At the half of maximum torque,
$\dfrac{\tau}{2}=MXB=MBsinT$-----------------2

1 to 2
${\tau/2}{\tau}=\dfrac{MBsinT}{MB}$
$sinT=\dfrac{1}{2}$
$T=30^o$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A magnetic dipole of dipole moment $10(\hat{i}+\hat{j}+\hat{k})$ is placed in a magnetic field $0.6\hat{i}+0.4\hat{j}+0.5\hat{k}$, force acting on the dipole is :-

  1. $\hat{i}-\hat{j}-2\hat{k}$
  2. $\hat{i}+\hat{j}+2\hat{k}$
  3. $Zero$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

Dipole moment $m=10\left( \hat{i}+\hat{j}+\hat{k} \right)$

Magnetic field $B=0.6\hat{i}+0.4\hat{j}+0.5\hat{k}$

We know that,

The force is

  $ \tau =m\times B $

 $ \tau =\left( 10\hat{i}+10\hat{j}+10\hat{k} \right)\times \left( 0.6\hat{i}+0.4\hat{j}+0.5\hat{k} \right) $

 $ \tau =\hat{i}-\hat{j}-2\hat{k} $

Hence, the force acting on the dipole is $\hat{i}-\hat{j}-2\hat{k}$