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 physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

The small magnets each of magnetic moment $10A-{ m }^{ 2 }$ are placed end on position 0.1m apart from their centres.The force acting between them is :

  1. $0.6\times { 10 }^{ 7 }N$
  2. $0.06\times { 10 }^{ 7 }N$
  3. $0.6N$
  4. $0.06N$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,


$M _1=M _2=10Am^2$


$r=0.1m$

The force acting between the magnet is given by,

$F=\dfrac{\mu _0}{4\pi }.\dfrac{6M _1M _2}{r^4}$

$F=10^{-7}\times \dfrac{6\times 10\times 10}{(0.1)^4}$

$F=0.6N$

The correct option is C.

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A bar magnet having centre O has a length of 4 cm. Point $P _1$ is in the broad side-on and $P _2$ is in the end side-on position with $OP _1=OP _2=10$ metres. The ratio of magnetic intensities H at $P _1$ and $P _2$ is

  1. $H _1:H _2=16:100$
  2. $H _1:H _2=1:2$
  3. $H _1:H _2=2:1$
  4. $H _1:H _2=100:16$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a short magnet, magnetic intensity H at broad side-on (equatorial) is H1 = M / (4 * pi * d^3) and at end side-on (axial) is H2 = 2 * M / (4 * pi * d^3). The ratio H1:H2 = 1:2.

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A magnet of moment  $80A{m^2}$ is placed in a uniform magnetic field of induction $1.8 \times {10^{ - 5}}T$. If each pole of the magnet experiences a force of  $25 \times {10^{ - 3}}N$, the length of the magnet is:

  1. 0.292cm

  2. 5.76cm

  3. 0.362cm

  4. 2.262cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Magnetic moment M = pole strength (m) * length (L). Force on each pole F = m * B. Given M = 80, B = 1.8e-5, F = 25e-3. m = F / B = 25e-3 / 1.8e-5 = 25000 / 18 = 1388.89. L = M / m = 80 / 1388.89 = 0.0576 m = 5.76 cm.

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

Which of the following cannot be the shape of the path of a charged particle moving in a uniform magnetic field?

  1. straight line

  2. parabolic

  3. circular

  4. helical
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

It will be a circular path with radius

$r=\left( \frac { mc }{ qB }  \right) $

and time period of

$T=\left( 2\times pi\times \frac { m }{ qB }  \right) $

This you can get from study of force on moving charge in magnetic field.

This is similar to cyclotron except that, in cyclotron, there is presence of electric field in addition to magnetic field. Hence in cyclotron, the radius of circular path keeps on increasing as speed of charge increases.

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A charge particle moves in a uniform magnetic field. The velocity of the particle at some instant makes right angle with the magnetic field. The path of the particle will be

  1. A straight line

  2. A circle

  3. Any curved path

  4. Zig-zag
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When a charged particle enters a uniform magnetic field with a velocity perpendicular to the field, the magnetic force acts as a centripetal force, causing the particle to move in a circular path.

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

When a charged particle moves perpendicular to a uniform magnetic field, its 

  1. energy and momentum both change.

  2. energy changes but momentum remains unchanged.

  3. momentum changes but energy remains unchanged.

  4. energy and momentum both do not change.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since the force exerted by the magnetic field is perpendicular to the direction of the particle the speed of the particle cannot change but its velocity changes .

so C option is the correct answer

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A proton with kinetic energy K describes a circle of radius r in a uniform magnetic. An $\alpha$- particle with kinetic energy K moving in the same magnetic field will describe a circle of radius? 

  1. $\cfrac {r} {2}$

  2. r
  3. $2r$
  4. $4r$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

WE KNOW THAT 


charge on proton = q
charge on alpha particle = 2q

mass of proton =m 
mass of alpha particle =4m

for proton

 $ r =\dfrac{m\times v}{q\times B}$


for alpha particle

$ R = \dfrac{{4m}\times v}{2q\times B}$

$ R = \dfrac{2m\times v}{q\times B}$

$ R = 2r$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A bar magnet of moment $4Am^{2}$ is placed in a non-uniform magnetic field. If the field strength at poles are 0.2 T and 0.22 T then the maximum couple acting on it is

  1. 0.04Nm

  2. 0.84Nm

  3. 0.4 Nm

  4. 0.44Nm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Average of the two field strengths at the poles is $\frac{0.2 +0.22}{2}=0.21$
Copuling( torque): $MB=4 \times 0.21 : Nm= 0.84:Nm$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A magnet of length $30\ cm$ with pole strength $10\ A-m$ is freely suspended in a uniform horizontal magnetic field of induction $40 \times 10^{-6} T$ . If the magnet is deflected by $60^{o}$ from its equilibrium position, the restoring couple acting on it is :

  1. $10.39\times 10^{-5}\ Nm$
  2. $\sqrt{3} \times 10^{-5}Nm$
  3. $6\times 10^{-5} Nm$
  4. $\sqrt{5}\times 10^{-5}Nm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$l=30\ cm$
$P=10\ Am$
$B=40\times 10^{-6}$
$\theta =60^o$
$m=pl$
$\vec{\tau}=\vec{m}\times \vec{B}$
$=mB\sin\theta $
$=10^{-6}\times \dfrac{\sqrt{3}}{2}$
$=1.039\times 10^{-4}Nm$
$=10.39\times 10^{-5}Nm$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

Two unlike magnetic poles are distance "d" apart, and mutually attract with a force "F". If one of the pole strength is doubled and to maintain the same force between them, the new separation between the poles must be

  1. 2 d
  2. $ \sqrt {2} $ d
  3. $ d / \sqrt {2} $

  4. d / 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The force between magnetic poles is F = k(m1*m2)/d^2. If m1 becomes 2*m1, to keep F constant, the new distance d' must satisfy (2*m1*m2)/(d')^2 = (m1*m2)/d^2. This simplifies to 2/d'^2 = 1/d^2, so d' = d/sqrt(2).

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $60^o$, and one of the fields has a magnitude of $1.2\times 10^{-2} T$. If the dipole comes to stable equilibrium at an angle of $15^o$ with this field, what is the magnitude of the other field?

  1. $3\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$
  2. $\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$
  3. $6\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$
  4. $2\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}{\text{T}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Here, $\theta=60^{0}, B1=1.2\times 10^{−2}tesla,$
$\theta _1=15^{0}, \theta _{2}=60^{0}−15^{0}=45^{0}.$
In equilibrium, torque due to two fields must balance i.e.
$\tau _{1}=\tau _{2}$
$MB _{1}sin\theta _1=MB2sin\theta _2$

$\implies 1.2\times 10^{-2}\times sin15^{\circ}=B _2sin(60-15)^{\circ}=B _2sin45^{\circ}$
$\implies B _2=4.4\times 10^{-3}T$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A short bar magnet experiences a torque of magnitude $0.64\ J$. When it is placed in a uniform magnetic field of $0.32\ T$, making an angle of $30^{\circ}$ with the direction of the field. The magnetic moment of the magnet is

  1. $2\ Am^{2}$
  2. $4\ Am^{2}$
  3. $6\ Am^{2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Torque, $\tau = 0.64\ J, B = 0.32\ T, \theta = 30^{\circ}$
Torque, $\tau = MB\sin \theta$
$0.64 = M\times 0.32\sin 30^{\circ}$
$0.64 = M\times 0.32\times \dfrac {1}{2}$
$M = \dfrac {2\times 0.64}{0.32} = 4\ Am^{2}$.

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A short bar magnet placed with its axis at $30^o$ with a uniform external magnetic field of $0.35$ T experiences a torque of magnitude equal to $4.5\times 10^{-2}$N m. The magnitude of magnetic moment of the given magnet is?

  1. $26$J $T^{-1}$
  2. $2.6$J $T^{-1}$
  3. $0.26$J $T^{-1}$
  4. $0.026$J $T^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given:
The angle made by the magnetic field is, $θ=30^o$
The magnetic field in the region is, $B=0.35\ T$
The torque acting on the bar magnet is $\tau=4.5\times 10^{-2}\ Nm$

The torque acting on the magnet is given by:

$\tau=MB\ sin\ θ$

$ 4.5\times 10^{-2}= M\times 0.35 \times (sin 30^o)$

$⟹M=0.26\ JT^{-1} $
Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

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 $30^o$, will be:

  1. $30$ N m
  2. $30\sqrt{3}$ N m
  3. $60$ N m
  4. $60\sqrt{3}$ N m
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
$\tau =MB \sin \theta$
Here, $M=200A m^2, B=0.30N A^{-1}m^{-1}$ and $\theta =30^o$
$\therefore \tau =200\times 0.30\times \sin 30^o$
$\tau =30$N m

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A magnet of magnetic moment $10 \hat{i} A-m^2 $ is placed along the x-axis in a magnetic field $ \overline{B} = ( 2 \hat {i} + 3 \hat{j} )  T $ . The torque acting on bar magnet is :

  1. $ 20 \hat{i} + 30 \hat{k} N-m $
  2. $20 \hat{k} N-m $
  3. $ 30 \hat{k} N-m $
  4. $ 20 \hat{i} + 30 \hat{j} N-m $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The torque acting on bar magnet is given as,

$\tau  = M \times B$

$\tau  = \left( {10\hat i} \right) \times \left( {2\hat i + 3\hat j} \right)$

$\tau  = 30\hat k\;{\rm{N}} \cdot {\rm{m}}$

Thus, the torque acting on bar magnet is $30\hat k\;{\rm{N}} \cdot {\rm{m}}$.