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Multiple choice uniform magnetic field lines of earth magnetism physics

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$)

  1. $2.77\ Am^2$
  2. $8\ Am^2$
  3. $0.2\ Am^2$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In Tan A position, B = (mu_0 / 4pi) * (2M / d^3). Also B = B_H * tan(theta). Given B_H = 0.4 * 10^-4 T, d = 0.2 m, theta = 60 degrees. tan(60) = sqrt(3). M = (B_H * tan(60) * d^3) / (2 * 10^-7) = (0.4 * 10^-4 * 1.732 * 0.008) / 2 * 10^-7 = 2.77 Am^2.

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  1. $2:1$
  2. $1:2$
  3. $1:1$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$B = B _Htan\theta$

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

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

$\dfrac{tan\theta _1}{tan\theta _2} = B _1:B _2 = 2:1$ $(l<<d)$
Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. $2\ s$
  2. ${2}/{3}\ s$
  3. $2\sqrt{3}\ s$
  4. ${2}/{\sqrt{3}}\ s$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)  } $
$\Rightarrow { T }^{ \prime  }=\dfrac { 1 }{ 3 } \times T$
${ T }^{ \prime  }=\dfrac { 2 }{ 3 } s$

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $B _H$

Multiple choice uniform magnetic field lines of earth magnetism physics

Two short magnets have equal pole strengths but one is twice as long as the other. The shorter magnet is placed $20\ cm$ in $\tan A$ position from the compass needle. The longer magnet must be placed on the other side of the magnetometer for no deflection at a distance equal to

  1. $20\ cm$
  2. $20\times (2)^{1/3} cm$
  3. $20\times (2)^{2/3} cm$
  4. $20\times (2)^{3/3} cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For no deflection in $\tan A$ position
$\dfrac {\mu _{0}}{4\pi} \dfrac {2M _{1}}{d _{1}^{3}} \dfrac {2M _{2}}{d _{2}^{3}}$
$\therefore \dfrac {M _{1}}{M _{2}} = \left (\dfrac {d _{1}}{d _{2}}\right )^{3}$
or $\dfrac {1}{2} = \left (\dfrac {20}{d _{2}}\right )^{3}$
or $d _{2} = 20\times (2)^{1/3}cm$

Multiple choice uniform magnetic field lines of earth magnetism physics

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 sec$. The magnet is cut along its length into $3$ equal parts and the $3$ parts are then placed on each other with their like poles together. The time period of this combination will be

  1. $2 sec$
  2. $\dfrac{2}{9}sec$
  3. $2\sqrt{3}sec$
  4. $\dfrac{2}{\sqrt{3}}sec$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\dfrac{T _1}{T _2} =\sqrt{\frac{I _1}{I _2}} \sqrt{\dfrac{M _2}{M _1}}   ........(1)$
When the magnet is cut into three pieces, pole strength of the cut pieces is same as the original pole strength.
$I _1 =\dfrac{ml^2}{12}$
$I _2 = \dfrac{\dfrac{m}{3}(\dfrac{l}{3})^2}{12}$
$ \therefore = \dfrac{I _1}{I _2} = \dfrac{27}{1}$
$M _1 = pole \times l$ where pole is the pole strength.
$M _2= pole \times \dfrac{l}{3}$
$\dfrac{M _2}{M _1} = 3$
Eqn$(1)$  becomes
$\dfrac{T _1}{T _2}= \sqrt{\dfrac{27}{1}}\sqrt{3}=9$
Given $T _1=2$
$ \therefore T _2 = \dfrac{2}{9}$

Multiple choice uniform magnetic field lines of earth magnetism physics

With a standard rectangular bar magnet 'the time period of a vibration magnetometer is $4 s$. The bar magnet is cut parallel to its length into four equal pieces. The time period of vibration magnetometer when one piece is used (in second) (bar magnet breadth is, small) is

  1. $16$
  2. $8$
  3. $4$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Time period of magnet, $T=2\pi \sqrt { \dfrac { I }{ MB }  } $
When magnet is cut parallel to its length into four equal pieces.
Then new
magnetic moment, ${ M }^{ \prime  }=\dfrac { M }{ 4 } $
New moment of inertia, ${ I }^{ \prime  }=\dfrac { I }{ 4 } $
$\therefore $ New time period, ${ T }^{ \prime  }=2\pi \sqrt { \dfrac { { I }^{ \prime  } }{ { M }^{ \prime  }{ B }^{ \prime  } }  } $
$\Rightarrow \quad T={ T }^{ \prime  }=4s$

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The particle of a medium vibrates about their mean position whenever a wave travels through that medium. The phase difference between the vibrations of two such particles

  1. varies with time only

  2. varies with distance separating them only

  3. varies with time as well as distance

  4. is always zero
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The phase difference between the vibrations of two particles of the medium is given by :

            $\Delta \phi=\dfrac{2\pi}{\lambda}\Delta x$ 
it is clear that phase difference varies as the path difference between the particles varies, which is the distance, separating the particles.