Physics

Magnetism and Electromagnetism

1,019 Questions

This hub provides practice questions on magnetism and electromagnetism. It covers magnetic flux density, electromagnets, magnetic lines of force, and electromagnetic induction. These physics concepts frequently appear in technical and non-technical competitive exams.

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Magnetism and Electromagnetism Questions

Multiple choice uniform magnetic field lines of earth magnetism physics

Among the following statements: 

A) In $\tan A$ position of deflection magnetometer $B$ and $B _{H}$ are perpendicular 

B) In $\tan B$ position of deflection magnetometer $B$ and $B _{H}$ parallel

  1. A true & B false

  2. A false & B true

  3. A and B are true

  4. A and B are false
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In $\tan A$ position , the arms are in east-west direction and the compass needle is in north-south direction and the bar magnet is kept parallel to the arms, perpendicular to the compass needle.

In $\tan B$ position , the arms are in north-south direction and the compass needle is in same direction , the magnet is kept in east - west direction , perpendicular to the compass needle.

Multiple choice uniform magnetic field lines of earth magnetism physics

The restoring couple for a magnet oscillating in the vibration magnetometer is provided by

  1. horizontal component of earths magnetic field

  2. gravity

  3. torsion in the suspended thread

  4. magnetic field of magnet
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the bar magnet in vibration magnetometer is displaced from the earth's magnetic field , the torque acts on it to allign it with the earth's magnetic field . So the earth's magnetic field is  the restoring force.

Multiple choice uniform magnetic field lines of earth magnetism physics
Assertion (A) : In deflection magnetometer a short magnetic needle is arranged in the compass box
Reason (R) : The magnetic needle is found in the uniform magnetic field produced by earth and bar magnet

  1. Both A and R are true and R is the correct explanation of A

  2. Both A and R are true and R is not correct explanation of A

  3. A is true, But R is false

  4. A is false, But R is true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In Deflection magnetometer ,the bar magnet is kept at a distance from the magnetic needle and the deflection thus observed is due to the earth's and bar magnet's magnetic field..

Multiple choice uniform magnetic field lines of earth magnetism physics

Deflection magnetometer is held in $\tan B$ position. A magnet placed on one of its arms produces no deflection. This implies that the axis of the magnet is

  1. in the east - west direction

  2. in the north-south direction

  3. perpendicular to the wooden bench

  4. North - East
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $\tan B$ position , the arms are in the north south direction and the aluminium pointer are in east- west direction. So the compass needle is in north- south direction. For the magnetic field lines of the magnet to be ineffective then they must be in the north south direction so that they will not create a torque.

If the magnet is in east- west direction , a couple acts on the magnet and thus causes deflection.

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  1. greater for long magnet

  2. greater for small magnet

  3. equal for both

  4. zero
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  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.
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
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.

Multiple choice uniform magnetic field lines of earth magnetism physics

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice uniform magnetic field lines of earth magnetism physics

Vibration magnetometer works on the principle of

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. increase by 19%

  2. decrease by 19%

  3. increase by 11%

  4. decrease by 11%
Reveal answer Fill a bubble to check yourself
C Correct answer
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 \%$
Multiple choice uniform magnetic field lines of earth magnetism physics

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

  1. $2 s$
  2. $\frac{2}{3} s$
  3. $\sqrt 3 s$
  4. $\frac{2}{\sqrt3} s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice uniform magnetic field lines of earth magnetism physics

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:                  

  1. (a) 4 s

  2. (b) 1 s

  3. (c) 2s

  4. (d) 3 s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

T is proportional to 1/sqrt(B_net). Initially B_net = 24. New B_net = 24 - 18 = 6. T_new = T_old * sqrt(B_old / B_new) = 2 * sqrt(24 / 6) = 2 * sqrt(4) = 4 s.

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. $T^{2}={T} _{o}^{2}cos\theta$
  2. $T=T _{o}cos\theta$
  3. $T=\cfrac{T _{o}}{cos\theta}$
  4. $T^{o}=\cfrac{{T} _{o}^{2}}{cos\theta}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the magnet is present, the restoring force is due to both the Earth's field and the magnet's field. The effective field becomes B_H / cos(theta). Since T is proportional to 1/sqrt(B), T^2 = T_0^2 * cos(theta).

Multiple choice uniform magnetic field lines of earth magnetism physics

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 

  1. $\displaystyle \frac{3}{2}$
  2. $\displaystyle \frac{13}{12}$
  3. $\displaystyle \frac{8}{9}$
  4. $\displaystyle \frac{12}{11}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

$\therefore \displaystyle \dfrac{M _1}{M _2}=\dfrac{T^2 _1+T^2 _2}{T^2 _2-T^2 _1}=\dfrac{(0.5)^2+(0.1)^2}{(0.5)^2-(0.1)^2}=\dfrac{.25+.01}{.25-.01}=\dfrac{13}{12}$
Multiple choice uniform magnetic field lines of earth magnetism physics

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

  1. $1 \,s$
  2. $2 \,s$
  3. $3 \,s$
  4. $4 \,s$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

$T _1 = 4$

Multiple choice uniform magnetic field lines of earth magnetism physics

The length of a bar magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2   s$. The magnet is cut along 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 :

  1. $2\ s$
  2. ${2}/{3}\ s$
  3. $2\sqrt{3}\ s$
  4. ${2}/{\sqrt{3}}\ s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The time period of oscillations of magnet
$T = 2 \pi  \sqrt { \left( \dfrac { I }{ MH }  \right)  } $            .....(i)
where $I =$ moment of inertia of magnet
             $=\dfrac { m{ L }^{ 2 } }{ 12 } $  ($m$, being the mass of magnet)
$M =$ pole strength $\times L$
and $H =$ horizontal component of earth's magnetic field.
When the three equal parts of magnet are place on one another with their like poles together, then
${ I }^{ \prime  }=\dfrac { 1 }{ 12 } \left( \dfrac { m }{ 3 }  \right) \times { \left( \dfrac { L }{ 3 }  \right)  }^{ 2 }\times 3$
$=\dfrac { 1 }{ 12 } \dfrac { m{ L }^{ 2 } }{ 9 } =\dfrac { I }{ 9 } $
and ${ M }^{ \prime  }=pole\quad strength\times \dfrac { L }{ 3 } \times 3=M$
Hence,  ${ T }^{ \prime  }=2\pi \sqrt { \left( \dfrac { { I }/{ 9 } }{ MH }  \right)  } $
$\Rightarrow { T }^{ \prime  }=\dfrac { 1 }{ 3 } \times T$
${ T }^{ \prime  }=\dfrac { 2 }{ 3 } s$