Tag: magnetism

Questions Related to magnetism

Two bar magnets A and B are placed on the two arms of a deflection magnetometer. When their distances from the centre of the needle are 20 cm and 40 cm respectively, the needle lies in the magnetic meridian. If the moment of the magnet A is 100 Am$^{2}$, then the moment of the magnet B is:

  1. 400 Am$^{2}$

  2. 800 Am$^{2}$

  3. 1200 Am$^{2}$

  4. 1600 Am$^{2}$


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  $

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

$\dfrac{M _A}{M _B}  = \dfrac{1}{8} $

$M _B = 800 A m^2 $

When a short bar magnet is kept at a distance of 20 cm from the centre of D.M., in Tan A position, the deflection is 45$^{0}$ . If $H=30$ A/m, the moment of the magnet is :

  1. 1.5 $\times $ 10$^{-2}$ Am$^{2}$

  2. 1.51Am$^{2}$

  3. 3.01Am$^{2}$

  4. 1.31Am$^{2}$


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  $

$\theta = 45 ^o $
$B = \mu _o \times H  $
$4 \pi \times 10^{-7} \times 30 = 10 ^{-7} \times \dfrac{2M}{d^3} $

$M = 1.51 Am^2 $

A short bar magnet is kept at a distance of 30 cm from the centre of the compass box on D.M, which is in Tan A position. The deflection is 45$^{0}$. If the horizontal component of earth's field strength is 30 A/m, the magnetic moment of the magnet is

  1. $0.128\pi Am^{2}$

  2. $1.28\pi Am^{2}$

  3. $128\pi Am^{2}$

  4. $12.8\pi Am^{2}$


Correct Option: D

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

  1. d

  2. d$^{-1}$

  3. d$^{2}$

  4. d$^{-3}$


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 :

  1. greater than $45^{o}$

  2. less than $45^{o}$

  3. less than $30^{o}$

  4. greater than $90^{o}$


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 :

  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 $


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 :

  1. 3 : 4

  2. 4 : 3

  3. 9 : 14

  4. 27 : 64


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

  1. $0^{o}$

  2. $90^o$

  3. $45^{o}$

  4. $180^{o}$


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

  1. $4.8 \times 10^{-5}T$ 

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

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

  4. $3 \times 10^{-5}T$ 


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

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

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

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

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


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 $\theta = \tan^{-1} 2$