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 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 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

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

Multiple choice physics properties of a magnetic field let's make a magnet making magnets making and handling magnets

A domain in a ferromagnetic substance is in form of a cube of side length $1 \,\mu m$. It is contains $8 \times 10^{10}$ atoms and each atomic dipole has a moment of $9 \times 10^{-24} A \,m^2$, then the magnetization of the domain is then

  1. $7.2 \times 10^5 \,A \,m^{-1}$
  2. $7.2 \times 10^3 \,A \,m^{-1}$
  3. $7.2 \times 10^9 \,A \,m^{-1}$
  4. $7.2 \times 10^{12} \,A \,m^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Magnitisation $= \dfrac{M}{v}$

$= \dfrac{8 \times 10^{10} \times 9 \times 10^{-24}}{(10 - 6)^3}$

$= 7.2 \times 10^5 \,Am^{-1}$

Multiple choice magnetic compass magnetic poles and magnetic compass properties of magnet effect of electric current physics

The tangent law is applicable only when

  1. There are atleast two magnetic field

  2. There are two uniform magnetic fields mutually perpendicular to each other

  3. One strong magnetic field and the other weak magnetic field

  4. In the present magnetic fields one should be horizontal component of the earth's magnetic field
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The tangent law states that if a magnetic needle is placed in two mutually perpendicular uniform magnetic fields B and H, it will come to rest at an angle theta such that B = H tan(theta).

Multiple choice physics magnetic effect of electric current fleming's left hand rule magnetic force magnetic force on a moving charge and current carrying wire

An electron having a charge e moves with a velocity v in X-direction. A magnetic field acts on it in Y-direction. The force on the electron acts in

  1. positive direction of Y-axis

  2. negative direction of Y-axis

  3. positive direction of Z-axis

  4. negative direction of Z-axis
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
As electron moving in positive x-direction, so the current  is moving in negative x-direction ( the direction of your middle finger) and the magnetic field acts on positive Y-direction ( the direction of your index finger) then thumb will be in negative Z-direction which is the direction of force.