Tag: uniform magnetic field lines of earth

Questions Related to uniform magnetic field lines of earth

A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 s in earths horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earths field by placing a current carrying wire, the new time period of magnet will be:                  

uniform magnetic field lines of earth magnetism physics

  1. (a) 4 s

  2. (b) 1 s

  3. (c) 2s

  4. (d) 3 s

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Correct Option: A

An axle or truck is $2.5$ m long.If the truck is moving due North at ${ ms }^{ -1 }$ at a place where the vertical component of the earth's magnetic field is 90 $\mu T$, the potential difference between the two ends of the axle is

uniform magnetic field lines of earth magnetism physics

  1. 6.75 mV with West end positive

  2. 6.75 mV with East end positive

  3. 6.75 mV with North end positive

  4. 6.75 mV with South end positive

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Correct Option: C

A deflection magnetometer is adjusted in the usual way. When a magnet is introduced, the deflection observed is, and the period of oscillation of the needle in the magnetometer is $T$. When the magnet is removed the period of oscillation is $T _{o}$. The reaction between $T$ and $T _{o}$ is :

uniform magnetic field lines of earth magnetism physics

  1. $T^{2}={T} _{o}^{2}cos\theta$

  2. $T=T _{o}cos\theta$

  3. $T=\cfrac{T _{o}}{cos\theta}$

  4. $T^{o}=\cfrac{{T} _{o}^{2}}{cos\theta}$

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Correct Option: A

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 

uniform magnetic field lines of earth magnetism physics

  1. $\displaystyle \frac{3}{2}$

  2. $\displaystyle \frac{13}{12}$

  3. $\displaystyle \frac{8}{9}$

  4. $\displaystyle \frac{12}{11}$

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Correct Option: B
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 $

$\therefore \displaystyle \dfrac{M _1}{M _2}=\dfrac{T^2 _1+T^2 _2}{T^2 _2-T^2 _1}=\dfrac{(0.5)^2+(0.1)^2}{(0.5)^2-(0.1)^2}=\dfrac{.25+.01}{.25-.01}=\dfrac{13}{12}$

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 

uniform magnetic field lines of earth magnetism physics

  1. $\displaystyle\dfrac{3}{4}$

  2. $\displaystyle\dfrac{9}{16}$

  3. $\displaystyle\dfrac{27}{64}$

  4. $\displaystyle\dfrac{81}{256}$

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Correct Option: C
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)

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

uniform magnetic field lines of earth magnetism physics

  1. $1 \,s$

  2. $2 \,s$

  3. $3 \,s$

  4. $4 \,s$

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Correct Option: D
Explanation:

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

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

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

$T _1 = 4$

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 :

uniform magnetic field lines of earth magnetism physics

  1. $11  cm$

  2. $16  cm$

  3. $18  cm$

  4. $10  cm$

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Correct Option: D
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$

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

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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$^{0}$

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

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Correct Option: C
Explanation:

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

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