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 uniform magnetic field lines of earth magnetism physics

A short magnet with its N-pole pointing towards north produces a null point at a distance 15 cm from its mid-point. If this magnet is used in tan A position of deflection magnetometer at a distance 15 cm from the magnetic needle, what will be the deflection

  1. $tan^{-1}(\frac{1}{2})$
  2. $tan^{-1}(\frac{3}{2})$
  3. $tan^{-1}(\frac{3}{4})$
  4. $tan^{-1} (2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice uniform magnetic field lines of earth magnetism physics

A compass needle placed at a distance r from a short magnet in tan A position shows a deflection of $60^0$. If the distance is increased to $r(3)^{1/3}$, then the deflection of the compass needle is:

  1. $30^0$
  2. $60^0 \times (3)^{1/3}$
  3. $60^0 \times (3)^{2/3}$
  4. $60^0 \times (3)^{3/3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice uniform magnetic field lines of earth magnetism physics

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

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

  4. 25
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice uniform magnetic field lines of earth magnetism physics

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice uniform magnetic field lines of earth magnetism physics

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 

  1. $\displaystyle\dfrac{3}{4}$
  2. $\displaystyle\dfrac{9}{16}$
  3. $\displaystyle\dfrac{27}{64}$
  4. $\displaystyle\dfrac{81}{256}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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)

Multiple choice uniform magnetic field lines of earth magnetism physics

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 :

  1. $11 cm$
  2. $16 cm$
  3. $18 cm$
  4. $10 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
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$

Multiple choice uniform magnetic field lines of earth magnetism physics

In deflection magnetometer, to find dipole moment $M$ of a magnet, angle of deflection should be

  1. $0^0$
  2. $90^0$
  3. $45^0$

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

In deflection magnetometer, $\dfrac{\mu _o}{4\pi}\dfrac{2M}{d^3}= H tan\theta$ 


Thus for $\theta= 0          \implies M=0$ (always)   and  for $\theta= 90          \implies tan 90^o = \infty$

Thus for proper working of instrument, $\theta$ should be $45^o$   as   $tan45^o=1$

Multiple choice uniform magnetic field lines of earth magnetism physics

In a deflection magnetometer experiment in $tan A$ position, a short bar magnet placed at $18cm$ from the centre of the compass needle produces a deflection of $30^{0}$. If another magnet of same length, but $16$ times pole strength that of first magnet is placed in $tan B$ position at $36cm$, then the deflection is

  1. 30$^{0}$
  2. 45$^{0}$
  3. 60$^{0}$
  4. 75$^{0}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,
$d _1 = 18 cm$,
$d _2 = 36 cm$, 
$l _1 = l _2$
$m _2 = 16 m _1$,
we have, 
For Tan A position
$\dfrac{2\mu _0 M _1}{4\pi (d _1)^3} = B _H \tan \theta _1$
and for Tan B position
$\dfrac{\mu _0 M _2}{4\pi (d _1)^3} = B _H \tan \theta _2$
$\Rightarrow \dfrac{2M _1}{M _2} =  \dfrac{d _1^3 \tan \theta _1}{d _2^3 \tan \theta _2}$
But, M = ml, hence,
$\dfrac{2m _1}{16m _1} =  \dfrac{d _1^3 tan \theta _1}{d _2^3 tan \theta _2}$
$\dfrac{tan \theta _2}{tan \theta _1} = \dfrac {8 \times d _1^3}{d _2^3}$
$\dfrac{tan \theta _2}{tan \theta _1} = \dfrac {8 (18)^3}{(36)^3}$
$\dfrac{tan \theta _2}{tan \theta _1} = 1$
$tan \theta _2 = tan \theta _1$
$\theta _2 = \theta _1 = 30^\circ$

Multiple choice uniform magnetic field lines of earth magnetism physics

The ratio of the magnetic moments of two bar magnets is 4 : 5. If the deflection produced by the first one in magnetometer in tan B position is 45$^{o}$, the deflection due to second magnet kept at the same distance is

  1. $0^{o} < \theta < 30^{o}$
  2. $30^{o} < \theta < 45^{o}$
  3. $45^{o} < \theta < 60^{o}$
  4. $60^{o} < \theta < 45^{o}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\dfrac{M _1}{M _2}=\dfrac{\tan\theta _1}{\tan\theta _2}$

$\Rightarrow \dfrac{\tan\theta _1}{\tan\theta _2}=\dfrac{4}{5}\Rightarrow \tan\theta _2=\dfrac{5\tan\theta _1}{4}=\dfrac{5\tan 45^o}{4}=1.25$
$\theta _2=\tan^{-1}(1.25)=51.34^0$
$45^o<\theta _2<60^o$

Multiple choice uniform magnetic field lines of earth magnetism physics

Two bar magnets are placed in a Vibration Magnetometer and allowed to vibrate. They make $20$ oscillations per minute when their similar poles are on the same side and they make $15$ oscillations per minute with their opposite poles lie on the same side. The ratio of their moments is :               

  1. $9:5$
  2. $25:7$
  3. $16:9$
  4. $5:4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\dfrac{T _2}{T _1} = \sqrt{\dfrac{(M _1 +M _2)}{(M _1 -M _2)}}$

$ \therefore \frac{f _1}{f _2} = \sqrt{\dfrac{(M _1 +M _2)}{(M _1 -M _2)}}$

$ \therefore \dfrac{20}{15} = \sqrt{\dfrac{(M _1 +M _2)}{(M _1 -M _2)}}$

$ \therefore \left(\dfrac{4}{3}\right)^2 = \dfrac{(M _1 +M _2)}{(M _1 -M _2)}$

$ \therefore \dfrac{16 +9}{16-9} = \dfrac{2M _1}{2M _2}=\dfrac{M _1}{M _2}$

$ \therefore \dfrac{M _1}{M _2} =\dfrac{25}{7}$

Multiple choice uniform magnetic field lines of earth magnetism physics

In an experiment with vibration magneto-meter  the value of $4\pi ^{2}\dfrac{I}{T^{2}}$ for a short bar magnet is observed as $36 \times 10^{-4}$. In the experiment with deflection magnetometer with the same magnet  the value of $\dfrac{4\pi d^{3}}{2\mu _{0}}$ is observed as $\dfrac{10^8}{36}$. The magnetic moment of the magnet used is :

  1. $50\ A m^{2}$
  2. $100\ Am^{2}$
  3. $200\ Am^{2}$
  4. $1000\ A m^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From first concept $mB _H=4 \pi^2 \dfrac{I}{T^2}$...(i)

and second concept $\dfrac{m}{B _H} = \dfrac{4 \pi d^3}{2 \mu _\circ{}}$....(ii)

Multiplying (i) and (ii)

$m^2 = $$4 \pi^2 \dfrac{I}{T^2}$$\times \dfrac{4 \pi d^3}{2 \mu _\circ{}}$ $36 \times 10^{-4}\times \dfrac{10^8}{32}$

$m=100 \ Am^2$

Multiple choice uniform magnetic field lines of earth magnetism physics

A small magnet of dipole moment $M$ is kept on the arm of a deflection magnetometer set in $\tan A$ position at a distance of $0.2\ m$. If the deflection is $60^o$, the value of $P$ is : ($ B _H=0.4\times 10^{-4}\ T$)

  1. $2.77\ Am^2$
  2. $8\ Am^2$
  3. $0.2\ Am^2$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In Tan A position, B = (mu_0 / 4pi) * (2M / d^3). Also B = B_H * tan(theta). Given B_H = 0.4 * 10^-4 T, d = 0.2 m, theta = 60 degrees. tan(60) = sqrt(3). M = (B_H * tan(60) * d^3) / (2 * 10^-7) = (0.4 * 10^-4 * 1.732 * 0.008) / 2 * 10^-7 = 2.77 Am^2.

Multiple choice uniform magnetic field lines of earth magnetism physics

In end on and broadside on position of a deflection magnetometer, if ${\theta} _{1}$ and ${\theta} _{2}$ are the deflections produced by short magnets at equal distances, then $\tan { { \theta  } _{ 1 } } /\tan{{ \theta  } _{ 2 }}$ is

  1. $2:1$
  2. $1:2$
  3. $1:1$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$B = B _Htan\theta$

End On, Tan A: $\dfrac{\mu _02Md}{4\pi (d^2-l^2)^2}$

Tan B: $\dfrac{\mu _0M}{4\pi (d^2+l^2)^{3/2}}$

$\dfrac{tan\theta _1}{tan\theta _2} = B _1:B _2 = 2:1$ $(l<<d)$
Multiple choice uniform magnetic field lines of earth magnetism physics

Two short magnets have equal pole strengths but one is twice as long as the other. The shorter magnet is placed $20\ cm$ in $\tan A$ position from the compass needle. The longer magnet must be placed on the other side of the magnetometer for no deflection at a distance equal to

  1. $20\ cm$
  2. $20\times (2)^{1/3} cm$
  3. $20\times (2)^{2/3} cm$
  4. $20\times (2)^{3/3} cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For no deflection in $\tan A$ position
$\dfrac {\mu _{0}}{4\pi} \dfrac {2M _{1}}{d _{1}^{3}} \dfrac {2M _{2}}{d _{2}^{3}}$
$\therefore \dfrac {M _{1}}{M _{2}} = \left (\dfrac {d _{1}}{d _{2}}\right )^{3}$
or $\dfrac {1}{2} = \left (\dfrac {20}{d _{2}}\right )^{3}$
or $d _{2} = 20\times (2)^{1/3}cm$