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Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. 3 : 4

  2. 4 : 3

  3. 9 : 14

  4. 27 : 64
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $
Multiple choice uniform magnetic field lines of earth magnetism physics

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

  1. $0^{o}$
  2. $90^o$
  3. $45^{o}$
  4. $180^{o}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $90^{\circ}$.

Multiple choice uniform magnetic field lines of earth magnetism physics

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 $(B _{H}=3.6\times 10^{-5}T)$

  1. $4.8 \times 10^{-5}T$
  2. $8\times 10^{-5}$
  3. $4 \times 10^{-5}T$
  4. $3 \times 10^{-5}T$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $B=B _H\tan\theta$

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

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  1. $\tan^{-1}( 3/2)$
  2. $\tan^{-1}( 3/4)$
  3. $\tan^{-1}( 2)$
  4. $\tan^{-1}( 1/2)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\theta = \tan^{-1} 2$

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. $\dfrac{8}{27}$
  2. $\dfrac{27}{8}$
  3. $\dfrac{9}{7}$
  4. $\dfrac{4}{9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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} $

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. $3 : 1$
  2. $5 : 18$
  3. $18 : 5$
  4. $25 : 7$
Reveal answer Fill a bubble to check yourself
D Correct answer
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$

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  1. $16 cm$
  2. $24 cm$
  3. $12 cm$
  4. $18 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.  
$\dfrac {M}{8M} = \dfrac {(8)^3}{(d _2)^3}$
$(d _2)^3 = 512 \times 8 = 4096$
$\therefore d _2 = 16 cm$

Multiple choice uniform magnetic field lines of earth magnetism physics

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)

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  1. $30^{o}$
  2. $60^{o}$
  3. $45^{o}$
  4. $0^{o}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\theta _A= 30^{\circ}$

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. ${T} _{2}$ is infinite
  2. ${T} _{2}={T} _{1}$
  3. ${T} _{2}>{T} _{1}$
  4. ${T} _{2}<{T} _{1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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