Physics

Magnetism and Magnetic Effects

218 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 motional emf physics

The amplitude of a magnetic field, which is part of a harmonic electromagnetic wave in vacuum, is  $\mathrm { B } _ { 0 } =510\mathrm { nT } .$  What is the amplitude of the electric field of the wave? 

  1. $140 \mathrm { NC } ^ { - 1 }$
  2. $153 \mathrm { NC } ^ { - 1 }$
  3. $163 \mathrm { NC } ^ { - 1 }$

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

In an electromagnetic wave, the relationship between electric field amplitude E_0 and magnetic field amplitude B_0 is E_0 = c * B_0, where c is the speed of light (3 * 10^8 m/s). E_0 = (3 * 10^8) * (510 * 10^-9) = 153 N/C.

Multiple choice motional emf physics

A charged particle enters in a uniform magnetic field with velocity at an angle of ${ 60 }^{ o }$ with the magnetic field. The pitch of helical path is x, the radius of helix is 

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

Pitch x = v * cos(theta) * T = v * cos(theta) * (2 * pi * m) / (q * B). Radius R = (m * v * sin(theta)) / (q * B). Dividing R by x gives R / x = (sin(theta)) / (2 * pi * cos(theta)) = tan(60) / (2 * pi) = sqrt(3) / (2 * pi). Thus R = (sqrt(3) * x) / (2 * pi).

Multiple choice motional emf physics

When a charged particle is projected perpendicular to a magnetic field then the

  1. Velocity of the particle remain constant

  2. Momentum of the particle remain constant

  3. Kinetic energy of the particle remain constant

  4. Path of particle is straight line
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When a charged particle enters a magnetic field perpendicularly, the magnetic force acts perpendicular to the velocity. This force changes the direction of velocity but not its magnitude, meaning kinetic energy remains constant.

Multiple choice physics electromagnetism magnetic field due to a circular current carrying conductor magnetic field produced in a circular loop field due to a current carrying coil oersted experiment

Axis of a solid cylinder of infinite length and radius $R$ lies along $y$-axis. It carries a uniformly distributed corrent I along +$y$ direction. Magnetic field at a point $(R/2,y,R/2)$ is 

  1. $\cfrac{\mu _{0}I}{4\pi R}\left(\hat{i}-\hat{k}\right)$
  2. $\cfrac{\mu _{0}I}{2\pi R}\left(\hat{j}-\hat{k}\right)$
  3. $\cfrac{\mu _{0}I}{4\pi R}\hat{j}$
  4. $\cfrac{\mu _{0}I}{4\pi R}\left(\hat{j}+\hat{k}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Ampere's Law for an infinite cylinder, the magnetic field at a point inside is determined by the current enclosed. The vector direction is perpendicular to the radial vector from the axis.

Multiple choice eddy currents motional emf electromagnetic induction electromagnetic induction and alternating currents physics

Eddy currents are used in

  1. electrolysis

  2. making a galvanometer dead beat

  3. electroplating

  4. to increase the sensitivity of galvanometer
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In general the coil of galvanometer oscillates about it's equilibrium due to rotational inertia which consumes some time. To avoid this coils is bound over  a metallic frame or plate oscillates in a magnetic field the eddy currents generated in the frame or plate oppose the motion and bring the frame to rest as the oscillations die out quickly. This is known as making galvanometer dead beat. 

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

If an electron covers 1/4th of revolution in a circle having radius R and in the presence of perpendicular magnetic field B, the amount of energy acquired by electron will be :

  1. $\dfrac{1}{4}mv^2$
  2. $0$
  3. $\dfrac{1}{8}mv^2$
  4. $(\pi\times R/2) \times (Bev)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Magnetic Force is perpendicular to electron's velocity, so it can only change direction of electron's velocity but cannot change its acceleration

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

An electron beam moving with a speed of $2.5\times 10^{7}ms^{-1}$  enters into the magnetic field directed perpendicular to its direction of motion. The magnetic induction of the field is $4\times 10^{-3}  wb/m^2$. The intensity of the electric field applied so that the electron remains undeflected due to the magnetic field is.

  1. $10^{4}N/C$
  2. $10^{5}N/C$
  3. $10^{7}N/C$
  4. $10^{3}N/C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For no deflection
VB = E
$\rightarrow E = 2.5*10^7*4*10^{-3}$
$= 10*10^4$
$= 10^5 {N}/{C}$

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

A particle of mass $1\times 10^{-26} \ kg $ and charge $1.6\times 10^{-19} \ C$ travelling with a velocity $1.28\times 10^6 \ m/s$ along the positive $X$-axis  enters a region in which a uniform electric field $\vec E$ and a uniform magnetic field of induction $\vec B$ are present. if $\vec E=-102.4 \times 10^3 \ \hat k \ NC^{-1}$ and $\vec B=8 \times 10^{-2} \ \hat j \ Wbm^{-2}$, the direction of motion of the particles is 

  1. along the positive X-axis

  2. along the negative X-axis
  3. at $45^{O}$ to the positive X-axis
  4. at $135^{O}$to the postive X-axis
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Write Force
F = q(vXB) + qE.
F =0N.
Hence the particle will continue moving towards positive X-axis.

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

A proton and an $\alpha$ particle enter a magnetic field in a direction perpendicular to it. If the force acting on the proton is twice that acting on the $\alpha$- particle, the ratio of their velocities is

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

We know that Force $(F _m)$ on a particle with charge "q" moving with a velocity of v at an angle $\theta$ to the magnetic field $(B)$ can be expressed as

$F _m=qvB \sin{\theta}$
Given, $\cfrac{F _m (p)}{F _m(\alpha)}$=2
$q _{\alpha}=2 \times q _{p}$
$\Rightarrow \ \cfrac{q _{p}v _pB}{q _{\alpha}v _{\alpha}B}=2$
We get 
$\cfrac{v _{p}}{v _{\alpha}}=4$

Multiple choice properties of magnet magnetic field moving charges and magnetism magnetic effects of current and magnetism physics

If E and B denote electronic and magnetic field respectively, which of the following is dimensionless?

  1. $\sqrt { { \mu } _{ 0 }{ \varepsilon } _{ 0 } } \dfrac { E }{ B } $
  2. $ { { \mu } _{ 0 }{ \varepsilon } _{ 0 } } \dfrac { E }{ B } $
  3. ${ \mu } _{ 0 }{ \varepsilon } _{ 0 }{ \left( \dfrac { B }{ E } \right) }^{ 2 }$
  4. $\dfrac { E }{ { \varepsilon } _{ 0 } } \dfrac { { \mu } _{ 0 } }{ B } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that Velocity of light  $C _0=\dfrac{1}{\sqrt{\mu _0 \epsilon _0}}$

And also velocity of electromagnetic wave $V=\dfrac{E}{B}$
 $[C _0]=\left[\dfrac{1}{\sqrt{\mu _0 \epsilon _0}}\right]=[LT^{-1}]$

 $[V]=\left[\dfrac{E}{B}\right]=[LT^{-1}]$

$\left[\sqrt { { \mu } _{ 0 }{ \varepsilon } _{ 0 } } \dfrac { E }{ B } \right]=[LT^{-1}]^{-1}[LT^{-1}]=[M^0L^0T^0]$    (Dimention less),

Option A

Multiple choice properties of magnet magnetic field moving charges and magnetism magnetic effects of current and magnetism physics

Two particles having the same specific change (q/m) enter a uniform magnetic field with the same speed but at angles of $30^ \circ$ and $60^\circ$ with the field. Let a, b and c be the ratios of their pitches, radii and periods of their helical paths respectively, then

  1. $abc = 1$
  2. $a + b = 2 \sqrt c$
  3. $a^2 = c$
  4. $ab = c$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Pitch p = (2 * pi * m * v * cos(theta)) / (q * B), radius r = (m * v * sin(theta)) / (q * B), period T = (2 * pi * m) / (q * B). Ratios: a = p1/p2 = cos(30)/cos(60) = sqrt(3), b = r1/r2 = sin(30)/sin(60) = 1/sqrt(3), c = T1/T2 = 1. Thus, a * b = sqrt(3) * (1/sqrt(3)) = 1, and c = 1. So a * b = c.

Multiple choice properties of magnet magnetic field moving charges and magnetism magnetic effects of current and magnetism physics

In a given region a charge particle is moving under the effect of electric and magnetic field with uniform velocity $\vec{v}=(\hat{i}+\hat{j}-\hat{k})$ m/s and magnetic field is given as $\vec{B}=(2\hat{i}+\hat{j}-2k)T$. The electric field is given as?

  1. $({i}+{j}-{k})$ V/m
  2. $({i}-{j}+{k})$ V/m
  3. $({i}+k)$ V/m
  4. $(-{i}-{k})$ V/m
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know Force $F=qE$------(1)

and also $F=qV\times B$-----(2)
From 1 and 2 we get 
$V\times B=E$
$V\times B$=\begin{matrix} \hat { i }  & \hat { j }  & \hat { i }  \ 1 & 1 & -1 \ 2 & 1 & -2 \end{matrix}
$V\times B=(-2+1)\hat { i } -(-2+2)\hat { j } +(1-2)\hat { k } $
$V\times B==-\hat { i } -\hat { k } $

Multiple choice properties of magnet magnetic field moving charges and magnetism magnetic effects of current and magnetism physics

A long, straight, $non-$ conducting string, painted with a charge density of $40\mu\ c/m$, is pulled along its length at a speed of $300\ m/s$. The magnetic field at a normal distance of $5\ mm$ from the moving string is $4.8\times {10}^{-1}\ T$

  1. $4.8\times {10}^{-1}\ T$

  2. Zero
  3. $\infty$

  4. Cannot be found
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A moving charged string is equivalent to a current I = lambda * v. Here, lambda = 40 * 10^-6 C/m, v = 300 m/s, so I = 40 * 10^-6 * 300 = 12 * 10^-3 A. B = mu0 * I / (2 * pi * r) = (2 * 10^-7 * 12 * 10^-3) / (5 * 10^-3) = 4.8 * 10^-7 T. The value in the question is 4.8 * 10^-1 T, which is incorrect.

Multiple choice properties of magnet magnetic field moving charges and magnetism magnetic effects of current and magnetism physics

In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to other. A charged particle is released from rest in this region. The path of the particle will be a:

  1. circle

  2. helix

  3. straight line

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

When steady and uniform electric and magnetic fields are parallel to each other and a charged particle is released from rest, the magnetic force is initially zero because the velocity is zero. Only the electric field exerts a force on the particle, accelerating it in a straight line along the direction of the electric field. As it gains velocity, the magnetic force acts perpendicular to the velocity, but since velocity remains parallel to the magnetic field, the cross product v x B remains zero, keeping the path a straight line.

Multiple choice properties of magnet magnetic field moving charges and magnetism magnetic effects of current and magnetism physics

If a charged particle goes unaccelerated in a region containing electric and magnetic fields:

  1. ${\vec E}$ must be parallel to ${\vec B}$
  2. ${\vec V}$ must be perpendicular to Electric field
  3. ${\vec V}$ must be parallel to ${\vec B}$
  4. $E$ must be equal to $vB$.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

velocity must be perpendicular to B and d)E must be equal to vB. Explanation: Lorenz force is computed as F = q (E + v × B)

So, if particle is accelerated we must have qE = qvB and v × B = −vB First is possible when d is true and second is possible when d is true