Tag: uniform magnetic field lines of earth

Questions Related to uniform magnetic field lines of earth

The tangent of deflection of angle of the needle of a DMM, taken along the y-axis is plotted against the distance d between the needle and a short magnet. The slope of the curve varies as

uniform magnetic field lines of earth magnetism physics

  1. d

  2. d$^{-1}$

  3. d$^{2}$

  4. d$^{-3}$

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Correct Option: D
Explanation:

The deflection of the magnetic needle in tan A position by a short magnet is given by 

$\dfrac{\mu _o}{4 \pi} \dfrac{2M _A}{d^3} = B _H tan \theta _A  $

$ \theta  = \dfrac{K}{d^3} $

The ratio of magnetic moments of two bar magnets is $5 : 2$. If the deflection produced by the first magnet in the D.M. in $\tan A$ position is $60^{o}$ , the deflection due to the second magnet kept at the same distance in tan A position is :

uniform magnetic field lines of earth magnetism physics

  1. greater than $45^{o}$

  2. less than $45^{o}$

  3. less than $30^{o}$

  4. greater than $90^{o}$

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Correct Option: B
Explanation:

The deflection off the magnetic needle in tan A position by a short magnet is given by 
$\dfrac{\mu _o}{4 \pi} \dfrac{2M _A}{d^3} = B _H \tan \theta _A  $

$\dfrac{ \tan \theta _A}{\tan \theta _B} = \dfrac{ M _A}{M _B} $

$\tan \theta _B  =  \dfrac{ M _B}{M _A} \times \sqrt{3} $

$\tan \theta _B  \approx 0.7 $
$\theta  $ less than $45^o$

A deflection magnetometer is in Tan A position in a region where the Earth's horizontal component of magnetic induction is $60\times 10^{-6}T$. When a magnet is placed at a suitable distance, a deflection of $45^{0}$ is obtained. The induction field strength of the magnet is :

uniform magnetic field lines of earth magnetism physics

  1. $60\times 10^{-5}T $

  2. $6\times 10^{-5}T $

  3. $0.6\times 10^{-5}T $

  4. $6\times 10^{-6}T $

Show answer
Correct Option: B
Explanation:
In $\tan A$ position; magnetometer is set perpendicular to magnetic meridian
$B = B _4\tan\theta$                  [$B _4 = 60\times{10}^{-6}T, \theta=45°]$
$\Rightarrow B= 60\times{10}^{-6}\times \tan45°$
$\Rightarrow B = 6\times{10}^{-5}T.$
Hence, the answer is $6\times{10}^{-5}T.$

Two short magnets are kept on opposite arms of the DMM at 12 cm and 16 cm. If there is no deflection in the needle, the ratio of the magnetic moments is :

uniform magnetic field lines of earth magnetism physics

  1. 3 : 4

  2. 4 : 3

  3. 9 : 14

  4. 27 : 64

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Correct Option: D
Explanation:
The deflection of the magnetic needle in tan A position by a short magnet is given by ,

$\dfrac{\mu _o}{4 \pi} \dfrac{2M _A}{d^3} = B _H tan \theta _A  $

Since the deflection is zero , 
$ \dfrac{M _A}{M _B} = \dfrac{d _A ^3}{d _B ^3}$

$\dfrac{M _A}{M _B}  = 27 : 64 $

A DMM is arranged at the magnetic pole of earth in $\tan A$ position. If a bar magnet is placed at some distance from the needle, deflection is

uniform magnetic field lines of earth magnetism physics

  1. $0^{o}$

  2. $90^o$

  3. $45^{o}$

  4. $180^{o}$

Show answer
Correct Option: B
Explanation:

A DMM is arranged at the magnetic pole of earth in $\tan A$ position. If a bar magnet is placed at some  distance from the needle, deflection is $90^{\circ}$.

A DMM set in tan A position. A small magnet is placed at a certain distance and the deflection observed in the needle is $53^{o}$ . The magnetic field at the site of the needle is $(B _{H}=3.6\times 10^{-5}T)$

uniform magnetic field lines of earth magnetism physics

  1. $4.8 \times 10^{-5}T$ 

  2. $8\times 10^{-5}$

  3. $4 \times 10^{-5}T$

  4. $3 \times 10^{-5}T$ 

Show answer
Correct Option: A
Explanation:

According to tangent law, when two uniform magnetic fields act at right angles to each other on a magnetic needle, it comes to rest in the direction of $B=B _H\tan\theta$

$=3.6\times 10^{-5}\times \tan53^{\circ}T$
$=4.8\times 10^{-5}T$

A short bar magnet with its $N -$ pole pointing north produces a null point at a distance $15 cm$ from its midpoint. If this magnet is used in $\tan A$ position of deflection magnetometer at a distance $15 cm$ from the magnetic needle, the deflection is

uniform magnetic field lines of earth magnetism physics

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

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

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

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

Show answer
Correct Option: C
Explanation:

For null deflection in $\tan A$ position,
$\dfrac{\tan \theta _1}{\tan \theta _2} = 2$
Here, the deflection obtain at same distance for the same magnet used.
$\therefore \tan\theta = 2$
or $\theta = \tan^{-1} 2$

The ratio of the magnetic moment of two short magnets when they give zero deflection in $\tan B$ position when placed at $12 cm$ and $18 cm$ from centre of a deflection magnetometer is :

uniform magnetic field lines of earth magnetism physics

  1. $\dfrac{8}{27}$

  2. $\dfrac{27}{8}$

  3. $\dfrac{9}{7}$

  4. $\dfrac{4}{9}$

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Correct Option: A
Explanation:

The deflection of the magnetic needle in $\tan B$ position by a short magnet is given by ,


$\dfrac{\mu _o}{4 \pi} \dfrac{M _B}{d^3} = B _H \tan \theta _B  $

Since the deflection is zero , 

$ \dfrac{M _A}{M _B} = \dfrac{d _A ^3}{d _B ^3}$

$\dfrac{M _A}{M _B}  = \dfrac{8}{27} $

Two bar magnets are placed together in a vibration magnetometer vibrates with a time period is $3s$ . If one magnet is reversed, the combination takes $4s$ for one vibration. The ratio of their magnetic moments is :

uniform magnetic field lines of earth magnetism physics

  1. $3 : 1$

  2. $5 : 18$

  3. $18 : 5$

  4. $25 : 7$

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Correct Option: D
Explanation:

The time period of oscillation of a bar magnet freely suspended in air is given by
$T =2 \pi \sqrt{ \dfrac{I}{mB} } $

$T _1 =2 \pi \sqrt{ \dfrac{I}{(m _1 + m _2)B} }  $

$T _2 =2 \pi \sqrt{ \dfrac{I}{(m _1 - m _2)B} }  $

$\dfrac{T _1 ^2}{T _2 ^2} = \dfrac{m _1 - m _2}{m _1 + m _2} $

$ \dfrac{9}{16} =\dfrac{m _1 - m _2}{m _1 + m _2} $

$ m _1 : m _2 = 25 : 7$

Two small magnets of moments $M$ and $8M$ produce no deflection in $\tan A$ position when $M$ is at a distance $8 cm$. The distance of the magnet of moment $8M$ is

uniform magnetic field lines of earth magnetism physics

  1. $16 cm$

  2. $24 cm$

  3. $12 cm$

  4. $18 cm$

Show answer
Correct Option: A
Explanation:

For null deflection in $\tan A$ position,
$\dfrac {M _1}{M _2} = \dfrac {(d _1)^3}{(d _2)^3}$
where, $M _1$, $M _2$ are magnetic moments of the magnets, $d _1$, $d _2$ are distance of magnet from magnetometer.  
$\dfrac {M}{8M} = \dfrac {(8)^3}{(d _2)^3}$
$(d _2)^3 = 512 \times 8 = 4096$
$\therefore d _2 = 16 cm$