Tag: uniform magnetic field lines of earth

Questions Related to uniform magnetic field lines of earth

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. $\text{none of these}$

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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. $1 : 6$

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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. $2 : 1$

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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. $60^{o}$

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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. $\dfrac{2}{1}$

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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} $

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:

uniform magnetic field lines of earth magnetism physics

  1. 400 Am$^{2}$

  2. 800 Am$^{2}$

  3. 1200 Am$^{2}$

  4. 1600 Am$^{2}$

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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 :

uniform magnetic field lines of earth magnetism physics

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

  2. 1.51Am$^{2}$

  3. 3.01Am$^{2}$

  4. 1.31Am$^{2}$

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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

uniform magnetic field lines of earth magnetism physics

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

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

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

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

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Correct Option: D