Tag: magnetism

Questions Related to magnetism

In deflection magnetometer, to find dipole moment $M$ of a magnet, angle of deflection should be

uniform magnetic field lines of earth magnetism physics

  1. $0^0$

  2. $90^0$

  3. $45^0$

  4. any angle

Show answer
Correct Option: C
Explanation:

In deflection magnetometer, $\dfrac{\mu _o}{4\pi}\dfrac{2M}{d^3}= H tan\theta$ 


Thus for $\theta= 0          \implies M=0$ (always)   and  for $\theta= 90          \implies tan 90^o = \infty$

Thus for proper working of instrument, $\theta$ should be $45^o$   as   $tan45^o=1$

In a deflection magnetometer experiment in $tan A$ position, a short bar magnet placed at $18cm$ from the centre of the compass needle produces a deflection of $30^{0}$. If another magnet of same length, but $16$ times pole strength that of first magnet is placed in $tan B$ position at $36cm$, then the deflection is

uniform magnetic field lines of earth magnetism physics

  1. 30$^{0}$

  2. 45$^{0}$

  3. 60$^{0}$

  4. 75$^{0}$

Show answer
Correct Option: A
Explanation:

Given that,
$d _1 = 18 cm$,
$d _2 = 36 cm$, 
$l _1 = l _2$
$m _2 = 16 m _1$,
we have, 
For Tan A position
$\dfrac{2\mu _0 M _1}{4\pi (d _1)^3} = B _H \tan \theta _1$
and for Tan B position
$\dfrac{\mu _0 M _2}{4\pi (d _1)^3} = B _H \tan \theta _2$
$\Rightarrow \dfrac{2M _1}{M _2} =  \dfrac{d _1^3 \tan \theta _1}{d _2^3 \tan \theta _2}$
But, M = ml, hence,
$\dfrac{2m _1}{16m _1} =  \dfrac{d _1^3 tan \theta _1}{d _2^3 tan \theta _2}$
$\dfrac{tan \theta _2}{tan \theta _1} = \dfrac {8 \times d _1^3}{d _2^3}$
$\dfrac{tan \theta _2}{tan \theta _1} = \dfrac {8 (18)^3}{(36)^3}$
$\dfrac{tan \theta _2}{tan \theta _1} = 1$
$tan \theta _2 = tan \theta _1$
$\theta _2 = \theta _1 = 30^\circ$

The ratio of the magnetic moments of two bar magnets is 4 : 5. If the deflection produced by the first one in magnetometer in tan B position is 45$^{o}$, the deflection due to second magnet kept at the same distance is

uniform magnetic field lines of earth magnetism physics

  1. $0^{o} < \theta < 30^{o}$

  2. $30^{o} < \theta < 45^{o}$

  3. $45^{o} < \theta < 60^{o}$

  4. $60^{o} < \theta < 45^{o}$

Show answer
Correct Option: C
Explanation:

$\dfrac{M _1}{M _2}=\dfrac{\tan\theta _1}{\tan\theta _2}$

$\Rightarrow \dfrac{\tan\theta _1}{\tan\theta _2}=\dfrac{4}{5}\Rightarrow \tan\theta _2=\dfrac{5\tan\theta _1}{4}=\dfrac{5\tan 45^o}{4}=1.25$
$\theta _2=\tan^{-1}(1.25)=51.34^0$
$45^o<\theta _2<60^o$

Two bar magnets are placed in a Vibration Magnetometer and allowed to vibrate. They make $20$ oscillations per minute when their similar poles are on the same side and they make $15$ oscillations per minute with their opposite poles lie on the same side. The ratio of their moments is :               

uniform magnetic field lines of earth magnetism physics

  1. $9:5$

  2. $25:7$

  3. $16:9$

  4. $5:4$

Show answer
Correct Option: B
Explanation:

$\dfrac{T _2}{T _1} = \sqrt{\dfrac{(M _1 +M _2)}{(M _1 -M _2)}}$

$ \therefore \frac{f _1}{f _2} = \sqrt{\dfrac{(M _1 +M _2)}{(M _1 -M _2)}}$

$ \therefore \dfrac{20}{15} = \sqrt{\dfrac{(M _1 +M _2)}{(M _1 -M _2)}}$

$ \therefore \left(\dfrac{4}{3}\right)^2 = \dfrac{(M _1 +M _2)}{(M _1 -M _2)}$

$ \therefore \dfrac{16 +9}{16-9} = \dfrac{2M _1}{2M _2}=\dfrac{M _1}{M _2}$

$ \therefore \dfrac{M _1}{M _2} =\dfrac{25}{7}$

With a standard rectangular bar magnet, the time period in a vibration magneto meter is $4\  sec.$ The bar magnet is cut parallel to its length into $4$ equal pieces. The time period in vibration magnetometer when the piece is used $($in sec$) ($bar magnet breadth is small$)$            

uniform magnetic field lines of earth magnetism physics

  1. $16$

  2. $8$

  3. $4$

  4. $2$

Show answer
Correct Option: C
Explanation:

Time period of vibration: $T \propto \sqrt{\dfrac{I}{M}}$ where $I$ is the moment of inertia and $M$ is the magnetic moment
$ \therefore$ $\dfrac{T _1}{T _2}= \sqrt{\dfrac{I _1M _2}{I _2M _1}}$
When the magnet is cut into 4 pieces parallel to its length, magnetic moment remains same since the breadth is very small..
$\therefore M _1= M _2$
Moment of inertial also  does not change.
$I _2=I _1$
$ \therefore \dfrac{T _1}{T _2}=\sqrt{\dfrac{I _1M _2}{I _2\times M _1}}=1$
$ \therefore T _2 = T _1=4$

In an experiment with vibration magneto-meter  the value of $4\pi ^{2}\dfrac{I}{T^{2}}$ for a short bar magnet is observed as $36 \times 10^{-4}$. In the experiment with deflection magnetometer with the same magnet  the value of $\dfrac{4\pi d^{3}}{2\mu _{0}}$ is observed as $\dfrac{10^8}{36}$. The magnetic moment of the magnet used is :

uniform magnetic field lines of earth magnetism physics

  1. $50\ A m^{2}$

  2. $100\ Am^{2}$

  3. $200\ Am^{2}$

  4. $1000\ A m^{2}$

Show answer
Correct Option: B
Explanation:

From first concept $mB _H=4 \pi^2 \dfrac{I}{T^2}$...(i)

and second concept $\dfrac{m}{B _H} = \dfrac{4 \pi d^3}{2 \mu _\circ{}}$....(ii)

Multiplying (i) and (ii)

$m^2 = $$4 \pi^2 \dfrac{I}{T^2}$$\times \dfrac{4 \pi d^3}{2 \mu _\circ{}}$ $36 \times 10^{-4}\times \dfrac{10^8}{32}$

$m=100 \ Am^2$

A small magnet of dipole moment $M$ is kept on the arm of a deflection magnetometer set in $\tan A$ position at a distance of $0.2\ m$. If the deflection is $60^o$, the value of $P$ is : ($ B _H=0.4\times 10^{-4}\ T$)

uniform magnetic field lines of earth magnetism physics

  1. $2.77\ Am^2$

  2. $8\ Am^2$

  3. $0.2\ Am^2$

  4. $none\ of\ these$

Show answer
Correct Option: A

In end on and broadside on position of a deflection magnetometer, if ${\theta} _{1}$ and ${\theta} _{2}$ are the deflections produced by short magnets at equal distances, then $\tan { { \theta  } _{ 1 } } /\tan{{ \theta  } _{ 2 }}$ is

uniform magnetic field lines of earth magnetism physics

  1. $2:1$

  2. $1:2$

  3. $1:1$

  4. None of these

Show answer
Correct Option: A
Explanation:
$B = B _Htan\theta$

End On, Tan A: $\dfrac{\mu _02Md}{4\pi (d^2-l^2)^2}$

Tan B: $\dfrac{\mu _0M}{4\pi (d^2+l^2)^{3/2}}$

$\dfrac{tan\theta _1}{tan\theta _2} = B _1:B _2 = 2:1$ $(l<<d)$

The length of a bar magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2   s$. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be :

uniform magnetic field lines of earth magnetism physics

  1. $2\ s$

  2. ${2}/{3}\ s$

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

  4. ${2}/{\sqrt{3}}\ s$

Show answer
Correct Option: B
Explanation:

The time period of oscillations of magnet
$T = 2 \pi  \sqrt { \left( \dfrac { I }{ MH }  \right)  } $            .....(i)
where $I =$ moment of inertia of magnet
             $=\dfrac { m{ L }^{ 2 } }{ 12 } $  ($m$, being the mass of magnet)
$M =$ pole strength $\times L$
and $H =$ horizontal component of earth's magnetic field.
When the three equal parts of magnet are place on one another with their like poles together, then
${ I }^{ \prime  }=\dfrac { 1 }{ 12 } \left( \dfrac { m }{ 3 }  \right) \times { \left( \dfrac { L }{ 3 }  \right)  }^{ 2 }\times 3$
$=\dfrac { 1 }{ 12 } \dfrac { m{ L }^{ 2 } }{ 9 } =\dfrac { I }{ 9 } $
and ${ M }^{ \prime  }=pole\quad strength\times \dfrac { L }{ 3 } \times 3=M$
Hence,  ${ T }^{ \prime  }=2\pi \sqrt { \left( \dfrac { { I }/{ 9 } }{ MH }  \right)  } $
$\Rightarrow { T }^{ \prime  }=\dfrac { 1 }{ 3 } \times T$
${ T }^{ \prime  }=\dfrac { 2 }{ 3 } s$

To measure the magnetic moment of a bar magnet, one may use

uniform magnetic field lines of earth magnetism physics

  1. a deflection galvanometer if the earth's horizontal field is known

  2. an oscillation magnetometer if the earth's horizontal field is known

  3. both deflection and oscillation magnetometer if the earth's horizontal field is not known

  4. all of the above

Show answer
Correct Option: D
Explanation:

If horizontal component of earth's magnetic field is known, one can use either deflection galvanometer or oscillation magnetometer.
If horizontal component of  earth's magnetic field is  not known, we will need two measurements for two variables $M$ and $B _H$