Physics

Magnetism and Magnetic Effects

230 Questions

Magnetism and magnetic effects focus on the forces exerted by magnetic fields on moving charges and magnetic materials. Questions cover magnetic dipoles, flux density, the motion of charged particles, and electromagnetic relationships. It is a vital physics topic for government competitive exams.

Magnetic dipolesCharged particle motionMagnetic flux densityBar magnetsEarth magnetism

Magnetism and Magnetic Effects Questions

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

If X amount of work is required to rotate a bar magnet in a magnetic field by $60^0$, from a position parallel to the field, What is the torque required to maintain it in new position.

  1. $\sqrt3 5X$
  2. $\sqrt3 X$
  3. $\sqrt3 2X$
  4. $\sqrt3 3X$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

work done in rotating a magnet from $0$ to $60$ will be,

                        $X =$$M\times B\times (1- cos\theta)$
                            =$M\times B\times (0.5)$
                         $2X=M\times B$              taking $\theta =60$
      torqur required to maintain in that position will be ,
                               $\tau=M\times B\times Sin\theta$
                               $\tau=2X\times \dfrac{\sqrt(3)}{2}$     taking $\theta =60$
                                        =${\sqrt3} X$

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

At a place the horizontal component of earth's field is $0.5 \times 10^{-4}T$. A bar magnet suspended horizontally perpendicular to earth's field experiences a torque of $4.5\times 10^{-4}N-m$ at that place. The magnetic moment of the magnet is: 

  1. $2.25\times 10^{-8}J/T$
  2. $1/9 J/T$
  3. $2.25 J/T$
  4. $9 J/T$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\tau = MB sin\theta$
$4.5 \times 10^{-4}= M\times 0.5 \times 10^{-4}. sin 90^0$
$M= \dfrac{4.5 \times 10^{-4}}{0.5\times 10^{-4}}= 9 J/T$

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

A bar magnet when placed at an angle of $30^o$ to the direction of magnetic field of induction of $ 5\times10^{-5} T$, experiences a moment of a couple $2.5\times10^{-6} N-m$. If the length of the magnet is $5\ cm$ its pole strength is:

  1. $2\times10^{-2}\ Am$
  2. $5\times10^{-2}\ Am$
  3. $2\ Am$
  4. $5\ Am$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Torque experienced by bar magnet is $\tau =MB\sin \theta $ where $M$magnetic moment, $B$ is magnetic field induction and $\theta $ is the angle between $M\,\,and\,\,B$

It is given that

  $ \tau =2.5\times {{10}^{-6}}N $

 $ B=5\times {{10}^{-5}}T $

 $ L=0.05m $

 $ 2.5\times {{10}^{-6}}=M\times 5\times {{10}^{-5}}\,\times \dfrac{1}{2} $

 $ M={{10}^{-1}} $

 $ M=m\times L $

 $ {{10}^{-1}}=m\times 0.05 $

 $ m=2 Am $

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is:

  1. $30^o$
  2. $45^o$
  3. $60^o$
  4. $90^o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that,

$ \tau =MB\sin \theta  $

$ \tau \propto \sin \theta  $

So,

$ \dfrac{{{\tau } _{1}}}{{{\tau } _{2}}}=\dfrac{MB\sin {{\theta } _{1}}}{MB\sin {{\theta } _{2}}} $

$ \dfrac{{{\tau } _{1}}}{{{\tau } _{2}}}=\dfrac{\sin {{\theta } _{1}}}{\sin {{\theta } _{2}}} $

$ \dfrac{\tau }{\frac{\tau }{2}}=\dfrac{\sin {{90}^{0}}}{\sin {{\theta } _{2}}} $

$ \sin {{\theta } _{2}}=\dfrac{1}{2} $

$ {{\theta } _{2}}={{30}^{0}} $

Hence, the angle is ${{30}^{0}}$


Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

A charged particle is moving with uniform velocity $V\hat {j}$ through a uniform magnetic field $B(-\hat {i})$ and a unifom electric field $\vec {E}$. Then $\vec {E}$ is

  1. $-Bv\hat {k}$
  2. $Bv\hat {k}$
  3. $\dfrac {v}{B}\hat {k}$
  4. $\dfrac {-B}{v}\hat {k}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

At some location on earth the horizontal components of earth's magnetic field is $18 \times 10^-6 T.$ At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes $45^0$ angle with horizontal in equilibrium to keep this needle horizontal, the vertical force that should be applied at one of its ends is:

  1. $3.6 \times 10^-5 N$
  2. $6.5 \times 10^-5 N$
  3. $1.3 \times 10^-5 N$
  4. $1.8 \times 10^-5 N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At $45 ^o, B _H = B _V$

$F \dfrac{l}{2} = MB _V = m \times l \times B _V$
$F = \dfrac{2 m l B _V} { l} = 3.6 \times 18 \times 10^{-6} $
$ = 6.5 \times 10^{-5} N$

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

Two bar magnets with magnetic moments2$\mathrm { M }$ and $\mathrm { M }$ are fastened together at right angles to each other at their centres to forma crossed system, which can rotate freelyabout a vertical axis through the centre. The crossed system sets in earth's magnetic fieldmaking an angle $\theta$ with the magnetic merid-ian such that

  1. $\theta = \tan ^ { - 1 } \left( \dfrac { 1 } { \sqrt { 3 } } \right)$
  2. $\theta = \tan ^ { - 1 } ( \sqrt { 3 } )$
  3. $\theta = \tan ^ { - 1 } \left( \dfrac { 1 } { 2 } \right)$
  4. $\theta = \tan ^ { - 1 } \left( \dfrac { 3 } { 4 } \right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The system is in equilibrium when the net torque from the two magnets balances the torque from the Earth's magnetic field. The magnetic moments are 2M and M at 90 degrees. The resultant magnetic moment is sqrt((2M)^2 + M^2) = M sqrt(5), and the angle alpha it makes with the 2M magnet is tan(alpha) = M/2M = 1/2. The equilibrium angle theta with the meridian satisfies tan(theta) = M_perp / M_parallel = 1/2.

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

A bar magnet of magnetic moment 1.5 J/T is along the direction of the uniform magnetic field of 0.22T. The work done in turning the magnet opposite to the field direction and the torque required to keep in that position are 

  1. $0.33J$ and $0.33 N-m$
  2. $0.66J$ and $0.66 N-m$
  3. $0.33J$ and $0 N-m$
  4. $0.66J$ and $0 N-m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Work done to rotate a magnet by 180 degrees is W = MB(cos(0) - cos(180)) = MB(1 - (-1)) = 2MB. W = 2 * 1.5 * 0.22 = 0.66 J. In the opposite position (180 degrees), the torque tau = MB sin(180) = 0.

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

A very long magnet of pole strength 16 A-m is placed vertically with its one pole on the table. At what distance from the pole, there will be a neutral point on the table. $(B _H =4 \times 10^{-5} \ Wbm^{-2})$

  1. 0.4 m

  2. 0.2 m

  3. 0.5 m

  4. 0.8 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

A charge $q$ is spread uniformly over an insulated loop of radius $r$. If it is rotated with an angular velocity $\omega$ with respect to normal axis then the magnetic moment of the loop is

  1. $\dfrac {1}{2}q \omega r^{2}$
  2. $\dfrac {4}{3}q \omega r^{2}$
  3. $\dfrac {3}{2}q \omega r^{2}$
  4. $q \omega r^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let us take an element at an angle $\theta$ subtending an angle $d \theta$

The charge $dq$ the element has can be written as
$dq = \dfrac{q}{2 \pi} d \theta$
We know that $i = \dfrac{dq}{dt}= \dfrac{q}{2 \pi} \times \dfrac{d \theta}{dt}$
The time $dt$ can be written as
$dt= \dfrac{d\theta}{w} $
Hence $i = \dfrac{q w}{2 \pi}$
magnetic moment = $i A$
Hence Manetic moment $M= \dfrac{qw}{2 \pi} \times \pi r^2=\dfrac{qwr^2}{2}$

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics
The potential energy of a bar magnet of magnetic moment $M$ placed in a magnetic field of induction $B$. The position of stable equilibrium of the magnet is at the angular position given by $\theta$ equal to:
[where $\theta$ is the angle between B and M]
  1. ${0}^{o}$
  2. ${90}^{o}$
  3. ${45}^{o}$
  4. ${180}^{o}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Potential energy U = -M * B = -MB cos(theta). Stable equilibrium occurs at the minimum potential energy, which is when cos(theta) = 1, meaning theta = 0 degrees.

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

A dipole of magnetic moment $\vec{m}=30\hat{j}$A $m^2$ is placed along the y-axis in a uniform magnetic field $\bar{B}=(2\hat{i}+5\hat{j})$T. The torque acting on it is?

  1. $-40\hat{k}$N m
  2. $-50\hat{k}$N m
  3. $-60\hat{k}$N m
  4. $-70\hat{k}$N m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that,

Magnetic moment of a dipole, $\vec m=30\hat j\ Am^2$
Magnetic field, $\vec B= (2\hat i+ 5\hat j)\ T$

Torque acting on the dipole , $\vec{\tau}= \vec m \times \vec B$

$\implies  \vec{\tau}= 30\hat j \times( 2\hat i +5\hat j)= -60 \hat k \ Nm $

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

The torque and magnetic potential energy of a magnetic dipole in most stable position in a uniform magnetic field $(\bar{B})$ having magnetic moment $(\bar{m})$ will be

  1. $-mB, zero$
  2. $mB, zero$
  3. $zero, mB$
  4. $zero, -mB$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Torque, $\bar{T} = \bar{m} \times \bar{B} = mB sin \theta$
and magnetic potential energy $U _m = \bar{m}. \bar{B} = mB cos \theta$ at $\theta = 0^0$ the dipole will be in most stable position
$T = mB sin\theta = mB sin 0^0 = 0$
and $U _m = -mB cos\theta = -mB cos 0^0 = -mB$

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

The magnetic moment of a short bar magnet placed with its magnetic axis at $30^0$ to an external field of $900$G and experiences a torque of $0.02$N m is going to be

  1. $0.35 A m^2$
  2. $0.44 A m^2$
  3. $2.45 A m^2$
  4. $1.5 A m^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, $B = 900 Gauss = 900 \times 10^{-4}T. = 9 \times 10^{-2}T$
$T = 0.02 N m and \theta = 30^0$
$\therefore T = mB sin \theta$
$\Rightarrow 0.02 = m \times 9 \times 10^{-2} \times sin 30^0$
$m = \dfrac{0.02 \times 2 }{9 \times 10^{-2}}= 0.44 A m^2$.

Multiple choice force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

A bar magnet has a magnetic moment of $200 A m^2$. The magnet is suspended in a magnetic field of $0.30 N A^{-1} m^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^0$, will be then

  1. $30$Nm
  2. $30 \sqrt3$Nm
  3. $60$Nm
  4. $60 \sqrt3$Nm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Torque experienced by a magnet suspended in a uniform magnetic field B is given by
$T = MB sin \theta$
Here, $M = 200 A m^2, B = 0.30 N A^{-1}m^{-1}  and \  \theta = 30^0$
$\therefore T = 200 \times 0.30 \times sin 30^0$
$T = 30 N m$