Tag: uniform magnetic field lines of earth

Questions Related to uniform magnetic field lines of earth

A DMM is placed with its arms in $N-S$ direction.The distance at which a short bar magnet having $\dfrac {M}{B _{H}}=80Am^{2}/T$ should be placed, so that the needle can stay in any position is (nearly)

uniform magnetic field lines of earth magnetism physics

  1. $2.5 cm$ from the needle, $N-$pole pointing GS

  2. $2 cm$ from the needle, $N -$ pole pointing GN

  3. $4 cm$ from the needle, $N -$ pole pointing GN

  4. $2 cm$ from the needle, $N -$ pole pointing GS

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Correct Option: D
Explanation:

Here, DMM is placed in $\tan B$ position, we have
$\dfrac {\mu _0 M}{4\pi d^3} = B _H \tan \theta$
where, variables have their usual meanings.
$d^3 \tan \theta = \dfrac {\mu _0 M}{4\pi B _H}$

$d^3 \tan \theta = \dfrac {4\pi \times 10^{-7} \times 80}{4\pi}$
$d^3 \tan \theta  = 8 \times 10^{-6}$
$d \tan \theta  = 2 \times 10^{-2} = 2cm$
and the needle is in position with $N - pole$ pointing Gaussian South.

A short magnet produces a deflection of $30^{o}$ when  placed at some distance in $\tan A$ position of the magnetometer. If another magnet of same length and double the pole strength is kept at the same distance in $\tan B$ position, the deflection produced is

uniform magnetic field lines of earth magnetism physics

  1. $30^{o}$

  2. $60^{o}$

  3. $45^{o}$

  4. $0^{o}$

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Correct Option: A
Explanation:

Given,

$\dfrac{\mu _0}{4\pi}\dfrac{2Md}{d^3} = B _H\tan \theta _A$
$\dfrac{\mu _0}{4\pi}\dfrac{  2 \times Md}{d^3} = B _H\tan \theta _B$

From the above two equations, $\tan \theta _A = \tan \theta _B$
$\Rightarrow \theta _B = 30^{\circ}$ since given $\theta _A= 30^{\circ}$

Two magnets of a magnetic moments $M$ and $2M$ are placed in a vibration magnetometer, with the identical poles in same direction. The time period of vibration is ${T} _{1}$. If the magnets are placed with opposite pole together and vibrate with time period ${T} _{2}$ then :

uniform magnetic field lines of earth magnetism physics

  1. ${T} _{2}$ is infinite

  2. ${T} _{2}={T} _{1}$

  3. ${T} _{2}>{T} _{1}$

  4. ${T} _{2}<{T} _{1}$

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Correct Option: C
Explanation:

$\dfrac{T _2}{T _1} = \sqrt{\dfrac{M _1 +M _2}{M _1-M _2}}$
$\dfrac{T _2}{T _1} = \sqrt{\dfrac{2M +M}{2M-M}}=\sqrt{\dfrac{3M}{M}}=\sqrt{3}$

$\Rightarrow  T _2 \gt T _1$
$T _1$ is the time period when like poles touch each other
$T _2$ is the time period when unlike poles touch each other

A magnetic needle of pole strength $20\sqrt{3}$ Am is pivoted at its centre.Its N -pole is pulled eastward by a string.The horizontal force required to produce a deflection of $30^o$ from magnetic meridian (taken $B _H=10^{-4}T$) is :

uniform magnetic field lines of earth magnetism physics

  1. $4\times 10^{-3}N$

  2. $2\times 10^{-3}N$

  3. $\dfrac{2}{\sqrt{3}}\times 10^{-3}N$

  4. $4\sqrt{3}\times 10^{-3}N$

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Correct Option: A
Explanation:

Given,


$B _H=10^{-4}T$


$m=20\sqrt{3}Am$

$\theta=30^0$ deflection from the magnetic meridian

The horizontal required  force ,

$F=mB$. . . . . .. . . . . .(1)

we know that, the magnetic meridian $B _H$ 

$B _H= Bcos\theta$

$B=\dfrac{B _H}{cos\theta}=\dfrac{10^{-4}}{cos30^0}$

$B=1.1547\times 10^{-4}T$

From equation (1),

$F=20\sqrt{3}\times 1.1547\times 10^{-4}N$

$F=4\times 10^{-3}N$

The correct option is A.

A short magnet with its N-pole pointing towards north produces a null point at a distance 15 cm from its mid-point. If this magnet is used in tan A position of deflection magnetometer at a distance 15 cm from the magnetic needle, what will be the deflection

uniform magnetic field lines of earth magnetism physics

  1. $tan^{-1}(\frac{1}{2})$

  2. $tan^{-1}(\frac{3}{2})$

  3. $tan^{-1}(\frac{3}{4})$

  4. $tan^{-1} (2)$

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Correct Option: A

A compass needle placed at a distance r from a short magnet in tan A position shows a deflection of $60^0$. If the distance is increased to $r(3)^{1/3}$, then the deflection of the compass needle is:

uniform magnetic field lines of earth magnetism physics

  1. $30^0$

  2. $60^0 \times (3)^{1/3}$

  3. $60^0 \times (3)^{2/3}$

  4. $60^0 \times (3)^{3/3}$

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Correct Option: B

The time period of a thin magnet is 4 s. If it is divided into two equal halves, then the time period of each part will be:

uniform magnetic field lines of earth magnetism physics

  1. 4s

  2. 1s

  3. 2s

  4. 8s

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Correct Option: C
Explanation:

In case of vibration magnetometer when a magnet is cut n equal parts by cutting normal to its length. Then the time period of each part of magnet will be 
$T'=\frac{T}{n}$       ...(i)   (here, $T=4s, n = 2$)
Now, Putting the given values in Eq. (i), we get
$T'=\frac{4}{2}=2s$

A bar magnet used in a vibration magnetometer is heated so as to reduce its magnetic moment by 19%. The periodic time of the magnetometer will :

uniform magnetic field lines of earth magnetism physics

  1. increase by 19%

  2. decrease by 19%

  3. increase by 11%

  4. decrease by 11%

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Correct Option: C
Explanation:
The time period of oscillation is given by
$T = 2 \pi \sqrt {\dfrac{I}{MB}} $
$ m _2 = m _1 - 0.19 m _1 $
$ m _2 = 0.81 m _1 $

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

$ \dfrac{T _2}{T _1} = \dfrac{1}{0.9} $

$ \dfrac{ \Delta T}{T _1} \times 100 \approx 11 \%$

The length of a magnet is very large as compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2$ s. The magnetic is cut perpendicular to 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. $\frac{2}{3} s$

  3. $\sqrt 3 s$

  4. $\frac{2}{\sqrt3} s$

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Correct Option: B

A magnet makes 12 oscillation per minute at a place where horizontal component of earth's field is $6.4 \times 10^{-3}$ T. It is found to require 8 seconds per oscillation at another place X. The vertical component of earths field at X where resultant field makes angle $60^0$ with horizontal is $ ---- \times 10^{-4}$ T

uniform magnetic field lines of earth magnetism physics

  1. $\frac{25}{\sqrt{3}}$

  2. $\sqrt{3}$

  3. $25\sqrt{3}$

  4. 25

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Correct Option: A