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Multiple choice uniform magnetic field lines of earth magnetism physics

A combination of two bar magnets, in vibration magnetometer, makes $10$ oscillations per second if their like poles are tied together and $2$ oscillations per second when unlike poles are tied together. If induced magnetism is neglected, then the ratio of their magnetic moments is 

  1. $\displaystyle \frac{3}{2}$
  2. $\displaystyle \frac{13}{12}$
  3. $\displaystyle \frac{8}{9}$
  4. $\displaystyle \frac{12}{11}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
From the given data, we can figure out that $ T _1=\dfrac{1}{2}=0.5 s $ and $ T _2=\dfrac{1}{10}=0.1 s $

$\therefore \displaystyle \dfrac{M _1}{M _2}=\dfrac{T^2 _1+T^2 _2}{T^2 _2-T^2 _1}=\dfrac{(0.5)^2+(0.1)^2}{(0.5)^2-(0.1)^2}=\dfrac{.25+.01}{.25-.01}=\dfrac{13}{12}$
Multiple choice uniform magnetic field lines of earth magnetism physics

When two magnets are  placed $15\ cms$ and $20\ cms$ away from a deflection magnetometer on two arms, no deflection is observed. The ratio of magnetic dipole moments is 

  1. $\displaystyle\dfrac{3}{4}$
  2. $\displaystyle\dfrac{9}{16}$
  3. $\displaystyle\dfrac{27}{64}$
  4. $\displaystyle\dfrac{81}{256}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle\dfrac{M _1}{M _2}= \left( \dfrac{d _1}{d _2}\right)^3=\left(\dfrac{15}{20}\right)^3=\left(\dfrac{3}{4}\right)^3=\dfrac{27}{64}$. (using the standard result)

Multiple choice uniform magnetic field lines of earth magnetism physics

A vibration magnetometer placed in magnetic merlian has a small bar magnet. The magnet executes oscillations with a time period of $2 \,s$ in earth's horizontal magnetic field of $24 \,mu T$. When a horizontal field of $18 \,mu T$ is produced opposite to the earth's field by placing a current carrying wire, the new time period of the magnet will be then

  1. $1 \,s$
  2. $2 \,s$
  3. $3 \,s$
  4. $4 \,s$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$T = 2\pi \sqrt{\dfrac{I}{MB}} T \alpha \dfrac{1}{\sqrt{B}}$

$\dfrac{T _1}{T _2} = \sqrt{\dfrac{B _2}{B _1}}$

$\dfrac{T _1}{2} = \sqrt{\dfrac{24}{24 - 18}} = \sqrt{\dfrac{24}{6}} = 2$

$T _1 = 4$

Multiple choice uniform magnetic field lines of earth magnetism physics

When two short magnets having magnetic moments in the ratio $125 : 216$ are placed on the opposite arms of the Deflection Magnetometer, there is no deflection recorded. The distance between the centres of the magnets is $22  cm$. The distance of the weaker magnet from the center of D.M is :

  1. $11 cm$
  2. $16 cm$
  3. $18 cm$
  4. $10 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the formula,
$\dfrac{M _1}{M _2} = \dfrac{d _1^3}{d _2^3}$
$ \therefore \dfrac{125}{216} =\bigg ( \dfrac{d _1}{d _2}\bigg )^3$
$\Rightarrow  \dfrac{d _1}{d _2} = \dfrac{5}{6}$
$ \Rightarrow \dfrac{d _1}{22-d _1} = \dfrac{5}{6}$
$\Rightarrow \dfrac{22}{d _1} = \dfrac{11}{5}$
$ \Rightarrow d _1 = 10: cm$

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  1. $0^0$
  2. $90^0$
  3. $45^0$

  4. any angle
Reveal answer Fill a bubble to check yourself
C Correct answer
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   $tan45^o=1$

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  1. 30$^{0}$
  2. 45$^{0}$
  3. 60$^{0}$
  4. 75$^{0}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,
$\dfrac{2m _1}{16m _1} =  \dfrac{d _1^3 tan \theta _1}{d _2^3 tan \theta _2}$
$\dfrac{tan \theta _2}{tan \theta _1} = \dfrac {8 \times d _1^3}{d _2^3}$
$\dfrac{tan \theta _2}{tan \theta _1} = \dfrac {8 (18)^3}{(36)^3}$
$\dfrac{tan \theta _2}{tan \theta _1} = 1$
$tan \theta _2 = tan \theta _1$
$\theta _2 = \theta _1 = 30^\circ$

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  1. $0^{o} < \theta < 30^{o}$
  2. $30^{o} < \theta < 45^{o}$
  3. $45^{o} < \theta < 60^{o}$
  4. $60^{o} < \theta < 45^{o}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :               

  1. $9:5$
  2. $25:7$
  3. $16:9$
  4. $5:4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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}$

Multiple choice uniform magnetic field lines of earth magnetism physics

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$)$            

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

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. $50\ A m^{2}$
  2. $100\ Am^{2}$
  3. $200\ Am^{2}$
  4. $1000\ A m^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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{}}$ $36 \times 10^{-4}\times \dfrac{10^8}{32}$

$m=100 \ Am^2$