Tag: magnetism

Questions Related to magnetism

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. $18 cm$

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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.  
$\dfrac {M}{8M} = \dfrac {(8)^3}{(d _2)^3}$
$(d _2)^3 = 512 \times 8 = 4096$
$\therefore d _2 = 16 cm$

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

Show answer
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
$T'=\frac{4}{2}=2s$