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

A charged particle enters a uniform magnetic field with velocity vector at an angle of $45 ^o$ with the magnetic field. The pitch of the helical path followed by the particle is $p.$ The radius of the helix will be

  1. $\dfrac { p } { \sqrt { 2 } \pi }$
  2. $\sqrt { 2 } p$
  3. $\dfrac { p } { 2 \pi }$
  4. $\dfrac { \sqrt { 2 } p } { \pi }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Pitch p = v*cos(theta)T = v*cos(theta)(2*pi*m)/(qB). Radius r = (mv*sin(theta))/(qB). Given theta = 45 degrees, sin(45) = cos(45) = 1/sqrt(2). Thus, p = (v*2*pi*m)/(sqrt(2)*qB) and r = (mv)/(sqrt(2)*qB). Comparing these, r = p/(2*pi).

Multiple choice physics electric fields introduction to electrostatic force electric force charging and discharging

A light beam travelling in the x-direction is described by the electric field ${ E } _{ y }=(300V{ m }^{ -1 })sin\quad \omega (t-x/c).$ An electron is constrained to move along the y-direction with a speed of $2.0\times  {10}^7 { m }^{ -1 }$ Find the maximum electric force and the maximum magnetic force on the electron

  1. $4.8\times {10}^{-17} N, zero$
  2. $4.2\times {10}^{-18} N, 1.8\times 10^{-8}N$
  3. $4.8\times {10}^{-17} N, 3.2\times 10^{-18}N$

  4. Zero, Zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
A light beam travelling 
${ E } _{ y }=\left( 300V/m \right) sinw\left( t-x/c \right) $.
$V=2.0\times { 10 }^{ 7 }{ m }^{ -1 }$
${ B } _{ 0 }=\dfrac { { F } _{ 0 } }{ C } =\dfrac { 300 }{ 2\times { 10 }^{ 7 } } =150\times { 10 }^{ -7 }T$
${ F } _{ m }={ B } _{ 0 }qV$
       $=9.1\times { 10 }^{ -31 }\times 150\times { 10 }^{ -7 }\times 2\times { 10 }^{ 7 }$
       $=2.8\times { 10 }^{ -28 }\times 2$
       $=4.8\times { 10 }^{ -17 }N$
magnetic force $=0$
because they are perpendicular to each other.
Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

If a charge particle projected in a gravity-free room it does not deflect, 

  1. electric field and magnetic field must be zero

  2. both electric field and magnetic field may be present

  3. electric field will be zero and magnetic field may be zero

  4. electric field may be zero and magnetic field may be zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a charged particle moves through a region without deflection, the net force must be zero. The Lorentz force is F = q(E + v x B). If E = 0 and B = 0, the force is zero. While other configurations exist (like E and B being parallel to v), the most fundamental condition for no deflection regardless of velocity is that both fields are zero.

Multiple choice evs - i substances, objects and energy renewable resources alternative fuels and energy sources alternative sources of energy

An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,

  1. the two magnetic forces are equal and opposite, so they produce no net effect

  2. the magnetic forces do no work on each particle.

  3. the magnetic forces do equal and opposite (but non-zero) work on each particle

  4. the magenetic forces are necessarily negligible
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The electric and magnetic forces are defined via the Lorentz force on a charged particle

$F =q(E +v ×B )$
The magnetic force comes from the second term which defines it to be perpendicular to the velocity and therefore displacement $dr$ so it doesn't do any work.
Multiple choice evs - i substances, objects and energy renewable resources alternative fuels and energy sources alternative sources of energy

A proton and a deutron both enter in a region uniform magnetic field B, moving at right angles to the field $\vec { \mathrm { B } }$  . If the radius of circular orbits for both the particles is equal and kinetic energy acquired by deutron is I Me V, then kinetic energy acquired by the proton particle will be . 

  1. 1 MeV

  2. 2 MeV

  3. 0.5 MeV

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

$R = \dfrac{{mv}}{{qB}} = \dfrac{{\sqrt {2mk} }}{{qB}}{K _p} = \dfrac{{{q^2}{B^2}R _p^2}}{{2mp}}$

$k = \dfrac{{{q^2}{B^2}{R^2}}}{{2m}}$
$\therefore \dfrac{{{K _\infty }}}{{{K _p}}} = {\left( {\dfrac{{{q _\infty }}}{{q{I _0}}}} \right)^2}\left( {\dfrac{{{m _p}}}{{{m _\infty }}}} \right){\left( {\dfrac{{{R _p}}}{{{R _\infty }}}} \right)^2}$
$ \Rightarrow {K _\infty } = {K _p}{\left( {\dfrac{{{q _\infty }}}{{{q _p}}}} \right)^2}\left( {\dfrac{{mp}}{{{m _\infty }}}} \right){\left( {\dfrac{{{R _\infty }}}{{{R _p}}}} \right)^2}$
${K _p} = 1MeV{\left( 2 \right)^2}{\left( {\dfrac{1}{4}} \right)^2}{\left( 1 \right)^2} = 1MeV$
Hence,
option $(A)$ is correct answer.

Multiple choice evs - i substances, objects and energy renewable resources alternative fuels and energy sources alternative sources of energy

An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because

  1. The two magnetic forces are equal and opposite, so they produce no net effect

  2. The magnetic forces do no work on each particle

  3. The magnetic forces do equal and opposite (but non-zero) work on each particle

  4. The magnetic forces are necessarily negligible
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As the magnetic forces due to motion of electron and proton act in a direction perpendicular to the direction of motion, no work is done by these forces. That is why one ignores the magnetic force of one particle on another.

Multiple choice uniform magnetic field lines of earth magnetism physics

In null method of comparison of magnetic moments, the net magnetic field at the centre of the DMM, when null deflection is obtained is

  1. $0$
  2. $B _{H}$
  3. between $0$ and $B _{H}$
  4. above $B _{H}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In null method , two magnets are kept on either side of the arms so that the net deflection is zero . So the magnetic field due to the bar magnets gets cancelled and only earth's horizontal magnetic field is left

Multiple choice uniform magnetic field lines of earth magnetism physics

A D.M.M is in tan A position in a region where $B _H$  is 50$\mu $T. When a magnet is placed at a suitable distance the deflection obtained is 45$^{0}$. The resultant magnetic field at the centre of the compass is

  1. 50$\mu $T
  2. $50\sqrt{2}\mu \top $
  3. 25 $\mu $ T
  4. 100$\mu $ T
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The magnetic field due to the magnet is B $ = B _H  tan \theta $
$ B = B _H tan 45 = B _H$
Total magnetis field at the centre is 
$ B _T= \sqrt{B^2 + B _H ^2}= \sqrt{2}{B _H} $
$B _T = 50 \sqrt{2}\mu $T

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  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. $\text{none of these}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Magnetic moment ratio,
$\dfrac{M _1}{M _2}=\dfrac{d _1^3}{d _2^3}$

$\Rightarrow \dfrac{M _1}{M _2}=\left(\dfrac{20}{15}\right)^3$
$\Rightarrow \dfrac{M _1}{M _2}=\left(\dfrac{4}{3}\right)^3$
$\Rightarrow \dfrac{M _1}{M _2}=\dfrac{64}{27}$

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. $1 : 2$
  2. $1 : 3$
  3. $2 : 3$
  4. $1 : 6$
Reveal answer Fill a bubble to check yourself
D Correct answer
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} $

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. $3 : 1$
  2. $1 : 3$
  3. $1 : 2$
  4. $2 : 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
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} $

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. $20^{o}$
  2. $30^{o}$
  3. $45^{o}$
  4. $60^{o}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

Multiple choice uniform magnetic field lines of earth magnetism physics

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 $\dfrac{M _{1}}{M _{2}} :$

  1. $\dfrac{3}{1}$
  2. $\dfrac{3}{4}$
  3. $\dfrac{6}{1}$
  4. $\dfrac{2}{1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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} $
Multiple choice uniform magnetic field lines of earth magnetism physics

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:

  1. 400 Am$^{2}$
  2. 800 Am$^{2}$
  3. 1200 Am$^{2}$
  4. 1600 Am$^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $
Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. 1.5 $\times $ 10$^{-2}$ Am$^{2}$
  2. 1.51Am$^{2}$
  3. 3.01Am$^{2}$
  4. 1.31Am$^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $