Tag: magnetism

Questions Related to magnetism

A bar magnet used in a vibration magnetometer is heated so as to reduce its magnetic moment by 19%. The periodic time of the magnetometer will :

uniform magnetic field lines of earth magnetism physics

  1. increase by 19%

  2. decrease by 19%

  3. increase by 11%

  4. decrease by 11%

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Correct Option: C
Explanation:
The time period of oscillation is given by
$T = 2 \pi \sqrt {\dfrac{I}{MB}} $
$ m _2 = m _1 - 0.19 m _1 $
$ m _2 = 0.81 m _1 $

$ \dfrac{T _2}{T _1} = \sqrt{ \dfrac{m _1}{m _2} } $

$ \dfrac{T _2}{T _1} = \dfrac{1}{0.9} $

$ \dfrac{ \Delta T}{T _1} \times 100 \approx 11 \%$

The length of a magnet is very large as compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2$ s. The magnetic is cut perpendicular to its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be

uniform magnetic field lines of earth magnetism physics

  1. $2 s$

  2. $\frac{2}{3} s$

  3. $\sqrt 3 s$

  4. $\frac{2}{\sqrt3} s$

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Correct Option: B

A magnet makes 12 oscillation per minute at a place where horizontal component of earth's field is $6.4 \times 10^{-3}$ T. It is found to require 8 seconds per oscillation at another place X. The vertical component of earths field at X where resultant field makes angle $60^0$ with horizontal is $ ---- \times 10^{-4}$ T

uniform magnetic field lines of earth magnetism physics

  1. $\frac{25}{\sqrt{3}}$

  2. $\sqrt{3}$

  3. $25\sqrt{3}$

  4. 25

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Correct Option: A

A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 s in earths horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earths field by placing a current carrying wire, the new time period of magnet will be:                  

uniform magnetic field lines of earth magnetism physics

  1. (a) 4 s

  2. (b) 1 s

  3. (c) 2s

  4. (d) 3 s

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Correct Option: A

An axle or truck is $2.5$ m long.If the truck is moving due North at ${ ms }^{ -1 }$ at a place where the vertical component of the earth's magnetic field is 90 $\mu T$, the potential difference between the two ends of the axle is

uniform magnetic field lines of earth magnetism physics

  1. 6.75 mV with West end positive

  2. 6.75 mV with East end positive

  3. 6.75 mV with North end positive

  4. 6.75 mV with South end positive

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Correct Option: C

A deflection magnetometer is adjusted in the usual way. When a magnet is introduced, the deflection observed is, and the period of oscillation of the needle in the magnetometer is $T$. When the magnet is removed the period of oscillation is $T _{o}$. The reaction between $T$ and $T _{o}$ is :

uniform magnetic field lines of earth magnetism physics

  1. $T^{2}={T} _{o}^{2}cos\theta$

  2. $T=T _{o}cos\theta$

  3. $T=\cfrac{T _{o}}{cos\theta}$

  4. $T^{o}=\cfrac{{T} _{o}^{2}}{cos\theta}$

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Correct Option: A

A combination of two bar magnets, in vibration magnetometer, makes $10$ oscillations per second if their like poles are tied together and $2$ oscillations per second when unlike poles are tied together. If induced magnetism is neglected, then the ratio of their magnetic moments is 

uniform magnetic field lines of earth magnetism physics

  1. $\displaystyle \frac{3}{2}$

  2. $\displaystyle \frac{13}{12}$

  3. $\displaystyle \frac{8}{9}$

  4. $\displaystyle \frac{12}{11}$

Show answer
Correct Option: B
Explanation:
From the given data, we can figure out that $ T _1=\dfrac{1}{2}=0.5 s $ and $ T _2=\dfrac{1}{10}=0.1 s $

$\therefore \displaystyle \dfrac{M _1}{M _2}=\dfrac{T^2 _1+T^2 _2}{T^2 _2-T^2 _1}=\dfrac{(0.5)^2+(0.1)^2}{(0.5)^2-(0.1)^2}=\dfrac{.25+.01}{.25-.01}=\dfrac{13}{12}$

When two magnets are  placed $15\ cms$ and $20\ cms$ away from a deflection magnetometer on two arms, no deflection is observed. The ratio of magnetic dipole moments is 

uniform magnetic field lines of earth magnetism physics

  1. $\displaystyle\dfrac{3}{4}$

  2. $\displaystyle\dfrac{9}{16}$

  3. $\displaystyle\dfrac{27}{64}$

  4. $\displaystyle\dfrac{81}{256}$

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Correct Option: C
Explanation:

$\displaystyle\dfrac{M _1}{M _2}= \left( \dfrac{d _1}{d _2}\right)^3=\left(\dfrac{15}{20}\right)^3=\left(\dfrac{3}{4}\right)^3=\dfrac{27}{64}$. (using the standard result)

A vibration magnetometer placed in magnetic merlian has a small bar magnet. The magnet executes oscillations with a time period of $2 \,s$ in earth's horizontal magnetic field of $24 \,mu T$. When a horizontal field of $18 \,mu T$ is produced opposite to the earth's field by placing a current carrying wire, the new time period of the magnet will be then

uniform magnetic field lines of earth magnetism physics

  1. $1 \,s$

  2. $2 \,s$

  3. $3 \,s$

  4. $4 \,s$

Show answer
Correct Option: D
Explanation:

$T = 2\pi \sqrt{\dfrac{I}{MB}} T \alpha \dfrac{1}{\sqrt{B}}$

$\dfrac{T _1}{T _2} = \sqrt{\dfrac{B _2}{B _1}}$

$\dfrac{T _1}{2} = \sqrt{\dfrac{24}{24 - 18}} = \sqrt{\dfrac{24}{6}} = 2$

$T _1 = 4$

When two short magnets having magnetic moments in the ratio $125 : 216$ are placed on the opposite arms of the Deflection Magnetometer, there is no deflection recorded. The distance between the centres of the magnets is $22  cm$. The distance of the weaker magnet from the center of D.M is :

uniform magnetic field lines of earth magnetism physics

  1. $11  cm$

  2. $16  cm$

  3. $18  cm$

  4. $10  cm$

Show answer
Correct Option: D
Explanation:

Using the formula,
$\dfrac{M _1}{M _2} = \dfrac{d _1^3}{d _2^3}$
$ \therefore \dfrac{125}{216} =\bigg ( \dfrac{d _1}{d _2}\bigg )^3$
$\Rightarrow  \dfrac{d _1}{d _2} = \dfrac{5}{6}$
$ \Rightarrow \dfrac{d _1}{22-d _1} = \dfrac{5}{6}$
$\Rightarrow \dfrac{22}{d _1} = \dfrac{11}{5}$
$ \Rightarrow d _1 = 10: cm$