Tag: uniform magnetic field lines of earth

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Multiple choice uniform magnetic field lines of earth magnetism physics

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:                  

  1. (a) 4 s

  2. (b) 1 s

  3. (c) 2s

  4. (d) 3 s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

T is proportional to 1/sqrt(B_net). Initially B_net = 24. New B_net = 24 - 18 = 6. T_new = T_old * sqrt(B_old / B_new) = 2 * sqrt(24 / 6) = 2 * sqrt(4) = 4 s.

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

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

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

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

  4. $T^{o}=\cfrac{{T} _{o}^{2}}{cos\theta}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the magnet is present, the restoring force is due to both the Earth's field and the magnet's field. The effective field becomes B_H / cos(theta). Since T is proportional to 1/sqrt(B), T^2 = T_0^2 * cos(theta).

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

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

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