Tag: magnetism

Questions Related to magnetism

The magnets of same magnetic moment $M$ and different lengths are placed in $tan\; A$ position. The fields at equal distances from them are :

uniform magnetic field lines of earth magnetism physics

  1. greater for long magnet

  2. greater for small magnet

  3. equal for both

  4. zero

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Correct Option: A
Explanation:

In tan A position , the magnetic field due to the bar magnet is given by
$\dfrac{\mu _o}{4 \pi} \dfrac{2Md}{(d^2 - l^2)^{3/2}} = B$ for a bar magnet
where $ l $ is the length of the magnet and $ d $ is the distance of the point from the centre of the magnet.
For the same distance $ d $ and the same magnetic moment, strength of the magnetic field
$ B \propto \dfrac{1}{(d^2-l^2)^\dfrac{3}{2}} $
So larger the length of the magnet, greater is the value of $ \dfrac{1}{(d^2-l^2)^\dfrac{3}{2}} $ and consequentially, the strength of the magnetic field.
So , magnetic field at a point is larger for long bar magnet

A D.M.M is in tan A position in a region where $B _H$  is 50$\mu $T. When a magnet is placed at a suitable distance the deflection obtained is 45$^{0}$. The resultant magnetic field at the centre of the compass is

uniform magnetic field lines of earth magnetism physics

  1. 50$\mu $T

  2. $50\sqrt{2}\mu \top $

  3. 25 $\mu $ T

  4. 100$\mu $ T

Show answer
Correct Option: B
Explanation:

The magnetic field due to the magnet is B $ = B _H  tan \theta $
$ B = B _H tan 45 = B _H$
Total magnetis field at the centre is 
$ B _T= \sqrt{B^2 + B _H ^2}= \sqrt{2}{B _H} $
$B _T = 50 \sqrt{2}\mu $T

To measure the magnetic moment of a bar magnet, one may use.

uniform magnetic field lines of earth magnetism physics

  1. a tangent galvanometer.

  2. a deflection galvanometer if the earth's horizontal field is known.

  3. an oscillation magnetometer if the earth's horizontal field is known.

  4. both deflection and oscillation magnetometer if the earth's horizontal field is not known.

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Correct Option: B,C,D
Explanation:

To measure the magnetic moment of a bar magnet,
a deflection galvanometer is used  if the earth's horizontal field is known.
An oscillation magnetometer  can be used if the earth's horizontal field is known.
Both deflection and oscillation magnetometer can be used  if the earth's horizontal field is not known since there are two variables.

When two magnets are placed $20\ \text{cms}$ and $15\ \text{cms}$ away on the two arms of a deflection magnetometer, it shows no deflection. The ration of magnetic moments is :

uniform magnetic field lines of earth magnetism physics

  1. $\displaystyle\dfrac{M _1}{M _2}=\dfrac{64}{27}$

  2. $\displaystyle\dfrac{M _1}{M _2}=\dfrac{4}{3}$

  3. $\displaystyle\dfrac{M _1}{M _2}=\dfrac{16}{9}$

  4. $\text{none of these}$

Show answer
Correct Option: A
Explanation:

Magnetic moment ratio,
$\dfrac{M _1}{M _2}=\dfrac{d _1^3}{d _2^3}$

$\Rightarrow \dfrac{M _1}{M _2}=\left(\dfrac{20}{15}\right)^3$
$\Rightarrow \dfrac{M _1}{M _2}=\left(\dfrac{4}{3}\right)^3$
$\Rightarrow \dfrac{M _1}{M _2}=\dfrac{64}{27}$

The factor on which the period of oscillation of a bar magnet in uniform magnetic field depends is

uniform magnetic field lines of earth magnetism physics

  1. nature of suspension fibre

  2. length of the suspension fibre

  3. vertical component of earths magnetic induction

  4. moment of inertia of the magnet

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Correct Option: D
Explanation:

The time period of oscillation is given by
$T =2 \pi \sqrt{ \dfrac{I}{mB _H} }$
Therefore $T $ is proportional to $ \sqrt{I}$

When a D.M. is set in $\tan A$ position, the deflection is $30^{o}$ for a magnet A placed at a distance of $40\ cm$ from the midpoint of the D.M. When the D.M. is kept in $\tan B$ position another magnet B produces a deflection of $60^{o}$, when placed at the same distance. The ratio of the magnetic moments of A and B is :

uniform magnetic field lines of earth magnetism physics

  1. $1 : 2$

  2. $1 : 3$

  3. $2 : 3$

  4. $1 : 6$

Show answer
Correct Option: D
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  $

The deflection off 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  $

$\dfrac{ \tan \theta _A}{\tan \theta _B} = \dfrac{2 M _A}{M _B} $

$ \dfrac{M _A}{M _B} =\dfrac{1}{2} \times   \dfrac{\tan 30}{\tan 60}  = \dfrac{1}{6} $

Two magnets when placed in $\tan A$ position at the same distance cause deflections of $30^{o}$ and $60^{o}$. The ratio of their magnetic moments is :

uniform magnetic field lines of earth magnetism physics

  1. $3 : 1$

  2. $1 : 3$

  3. $1 : 2$

  4. $2 : 1$

Show answer
Correct Option: B
Explanation:

In $\tan A$ position , the deflection of the needle is given by

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

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

$\dfrac{M _A}{M _B} = \dfrac{1}{3} $

Vibration magnetometer works on the principle of

uniform magnetic field lines of earth magnetism physics

  1. torque acting on the bar magnet and rotational inertia

  2. force acting on the bar magnet and rotational inertia

  3. both the force and torque acting on the bar magnet

  4. neither force nor torque

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Correct Option: A
Explanation:

When the bar magnet in the deflection magnetometer is displaced , a torque acts on it due to the horizontal earth's magnetic filed. So the magnet vibrates and alligns parallel to the earth's magnetic field.

A short magnet when placed at a distance of $15 cm$ in $\tan A$ position produces a deflection of $60^{o}$. If the magnet is cut into $3$ equal parts and one of them is kept at the same distance in $\tan A$ position, the deflection is :

uniform magnetic field lines of earth magnetism physics

  1. $20^{o}$

  2. $30^{o}$

  3. $45^{o}$

  4. $60^{o}$

Show answer
Correct Option: B
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}{d^3} = B _H \tan \theta _A  $

$ M _B = \dfrac{M}{3}$

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

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

$\theta _B = 30 ^o $

Two bar magnets of same size with magnetic moments M$ _{1}$ and M$ _{2}$ (M$ _{1}$ > M$ _{2}$ ) are simultaneously used at the tan A position in a DMM. When the magnets are placed with unlike poles in contact the deflection is 30$^{0}$ and when like poles are in contact the deflection is 60$^{0}$ . Then $\dfrac{M _{1}}{M _{2}} :$

uniform magnetic field lines of earth magnetism physics

  1. $\dfrac{3}{1}$

  2. $\dfrac{3}{4}$

  3. $\dfrac{6}{1}$

  4. $\dfrac{2}{1}$

Show answer
Correct Option: D
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}{d^3} = B _H tan \theta  $

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

$\dfrac{ tan \theta _A}{tan \theta _B} = \dfrac{M _1 - M _2}{M _1 + M _2} $

$  \dfrac{M _1 - M _2}{M _1 + M _2} = \dfrac{1}{3}$

$\dfrac{M _1}{M _2} = \dfrac{2}{1} $