Tag: magnetism

Quizzes (14) Questions (113)

Questions Related to magnetism

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

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

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

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

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

  4. $4\sqrt{3}\times 10^{-3}N$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice uniform magnetic field lines of earth magnetism physics

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

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

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

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

  4. $tan^{-1} (2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice uniform magnetic field lines of earth magnetism physics

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:

  1. $30^0$

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

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

  4. $60^0 \times (3)^{3/3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice uniform magnetic field lines of earth magnetism physics

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:

  1. 4s

  2. 1s

  3. 2s

  4. 8s
Reveal answer Fill a bubble to check yourself
C Correct answer
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
$T'=\frac{4}{2}=2s$