MCQ Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 2

The restoring couple in the moving coil galvanometer is due to

0%
current in the coil
0%
magnetic field of the magnet.
0%
material of the coil.
0%
twist produced in the suspension wire.

There is

$10$ units of charge at the centre of a circle of radius

$10m$ . The work done in moving one unit of charge around the circle once is

0%
$0$
0%
$10$ units
0%
$100$ units
0%
$1$ unit

Explanation

Work done is zero because the direction of movement of charge and direction of force are perpendicular to each other.
work=

$F.r. \cos(\theta)$
so if the angle is

$90^o$ then work done

$=0$

A circular coil of one turn and area

$A$ carrying a current has a magnetic dipole moment

$M$ . The current through the coil is :

0%
$M A$
0%
$\dfrac{A}{ M}$
0%
$\dfrac{M}{A}$
0%
$\dfrac{M}{A ^{2}}$

Explanation

Magnetic moment of the circular current carrying coil,

$M = NiA$

$i = \dfrac{M}{NA}$

$= \dfrac{M}{1\times A}$

$= \dfrac{M}{A}$

The magnetic force acting on a charged particle of charge

$-2\mu c$ in a magnetic field of 2T acting in y direction, when the particle velocity is

$(2\hat{i}+3\hat{j})\times10^6\;ms^{-1}$ , is :

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8N in z direction
0%
8N in -z direction
0%
4N in z direction
0%
8N in y direction

Explanation

$\vec{F}=q(\vec{v}\times\vec{B})$

$=-2\times10^{-6}[(2\hat{i}+3\hat{j})\times10^6\times2\hat{j}$

$=-(8N)\hat{k}$

A proton is fired with a speed of

$2\times10^6$ m/s at an angle of

$60^o$ to the X- axis. If a uniform magnetic field of

$0.1T$ is applied along the Y- axis, the force acting on the proton is

0%
$1.63 \times 10^{-14} N$
0%
$1.6 \times 10^{-14} N$
0%
$3.23 \times 10^{-14} N$
0%
$3.2 \times 10^{-14} N$

Explanation

$F=q (\vec {V}\times \vec {B})$

$\theta=30^o$ angle between velocity & field

$=1\cdot 6\times 10^{-19}\times 2\times 10^6\times 0\cdot 1\times \sin 30^o$

$=1\cdot 6\times 10^{-14} N$

A flat circular coil carrying current has a magnetic moment

$\mu$ . Then ,
a.

$\mu$ has only magnitude, it does not have direction
b. The direction of

$\mu$ is along the normal to the plane of the coil
c. The direction of

$\mu$ depends on the direction of current flow
d. The direction of

$\mu$ does not change if the current in the coil is reversed

0%
a only is correct
0%
b and c only are correct
0%
d only is correct
0%
b and d only are correct

Explanation

Magnetic moment

$(\mu) = iA$
The magnetic moment can be considered to be a vector quantity with direction perpendicular to the current loop in the right-hand-rule direction where fingers curl in the direction of current flow

The radial magnetic field is used in a suspended coil galvanometer to provide

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a uniform torque on the coil
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maximum torque on the coil in all positions
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a uniform and maximum torque in all positions

of the coil
0%
a non uniform torque on the coil

A moving coil type of galvanometer is based upon the principle that

0%
coil carrying current experiences a torque in magnetic field.
0%
a coil carrying current produces a magnetic field.
0%
a coil carrying current experiences impulse in a magnetic field.
0%
a coil carrying current experiences a force in a magnetic field.

An electron is moving vertically downwards at any place. The direction of magnetic force acting on it due to horizontal component of earth's magnetic field will be

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towards east
0%
towards west
0%
towards north
0%
towards south

When the radius of the circular current carrying coil is doubled and the current in it is halved,then magnetic dipole moment of the coil originally four units becomes

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4 units
0%
8 units
0%
16 units
0%
32 units

The correct statement about magnetic moment is :

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it is a vector quantity
0%
its unit is $Am^2$
0%
its dimensions are $[AL^2]$
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all of the above

Explanation

$(\overrightarrow { m } )$

$(qm)\times (2\overrightarrow { l } )$

Magnetic moment = Pole strength

$\times$ Distance between poles

(directed from south pole to north pole)

Its directed is fixed so it's a vector quantity.

Unit of pole strength

$=A-m$

Unit of distance too poles =

$m$

Unit of magnetic moment = (Unit of pole strength)

$\times$ Unit of distance b\w poles.

$=Am \times m$

units of

$\overrightarrow { m }$

$=A{ m }^{ 2 }$ .

Dimension of

$\overrightarrow { m }$ = Dimension pole strength

$\times$ Distance b\w poles.

$\Rightarrow$

$(\overrightarrow { m } )=(A.L) \times (L)$

$=(A{ L }^{ 2 })$ .

Hence, the answer are: it is a vector quantity, its unit is

$=A{ m }^{ 2 }$ , its dimensions are

$(A{ L }^{ 2 }).$

The correct expression for Lorentz force is

0%
$q[\vec E+(\vec B\times \vec V)]$
0%
$q[\vec E+(\vec V\times \vec B)]$
0%
$q(\vec V\times \vec B)$
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$q\vec E$

Explanation

In electromagnetic field both electric and magnetic forces are experienced by the charge,
F = electric force + magnetic force

$F = qE + q(\vec v \times B)$

$F = q[E + (\vec v \times B)]$

A current carrying conductor

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Experiences a force when it is in magnetic field
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Does not experience a force when it is in magnetic field
0%
Experiences the force only when the field is electromagnetic in nature
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none of the above

In a moving coil galvanometer, the magnetic pole pieces are made cylindrical and a soft iron core is placed at the centre of the coil,the purpose for doing so is:

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to make the magnetic field strong
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to make the magnetic field strong and radial
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to make the magnetic field uniform
0%
to make the magnetic field strong and uniform

A magnetic field cannot exert any force on a

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Moving magnet
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Stationary magnet
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Moving charge
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Stationary charge

A magnetic field can exert force on a

0%
stationary magnet
0%
moving charge
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moving magnet
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all of these

Explanation

Hint:

The magnetic field is a vector quantity that describes the magnetic influence on moving charges, electric currents, and magnetic materials.

Explanation:

$\bullet$ Magnetic field exerts a force on a magnet irrespective of its motion, which is given by

$F=mB$

Where $m$ is the magnetic dipole moment of a magnet, $B$ is Magnetic Field.

$\bullet$ Magnetic field exerts a force on moving charges given by,

$\overrightarrow{F}=q\left( \overrightarrow{v}\times \overrightarrow{B} \right)$

Where $q$ is charge, is velocity of charge, $\overrightarrow{B}$ is magnetic Field

Thus, Magnetic Field exerts a force on stationary magnets, moving charges, moving magnets.

The correct expression for Ampere's law is :

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$\int B.dl=\sum i$
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$\int B.dl=\dfrac {1}{\sum i}$
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$\int B.dl=\mu_0\sum i$
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$\int B.dl=\dfrac {\sum i}{\mu_0}$

Explanation

Amper's law is

$\oint \vec{B}.\vec{dl} = \mu_0 \sum i$ where

$i$ is the sum of enclosed currents by the loop.

An electron and a proton travel with equal speeds and in the same direction, at

$90^o$ to a uniform magnetic field. They experience forces which are initially

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in opposite direction and differ by a factor of about 1840
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in the same direction and differ by a factor of about 1840
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equal in magnitude but in opposite directions
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identical

In a moving cell galvanometer, we use a radial magnetic field so that the galvanometer scale is

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logarithmic
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exponential
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linear
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none of the above

If a long hollow copper pipe carries a direct current, the magnetic field associated with the current will be :

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only inside the pipe
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only outside the pipe
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both inside and outside the pipe
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neither inside not outside the pipe

Find the magnetic field intensity due to a thin wire carrying current

$I$ in the figure :

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$\dfrac {\mu_0i}{2\pi R}(\pi -a+tan a)$
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$\dfrac {\mu_0i}{2\pi R}(\pi -a)$
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$\dfrac {\mu_0i}{2\pi R}(\pi +a)$
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$\dfrac {\mu_0}{2\pi R}(\pi +a-tan a)$

Explanation

Magnetic field intensity due to arc part of wire

$B_{arc}=\dfrac {\mu_0I}{4\pi R}(2\pi-2a)$ ,
Magnetic field intensity due to linear part of wire

$B_{line}=\dfrac {\mu_0I(sin \alpha+sin \alpha)}{4\pi R cos\alpha}$

$B_{net}=B_{arc}+B_{line}$

$B_{net}=\dfrac {\mu_0I}{4\pi R}(2\pi-2a),+\dfrac {\mu_0I(sin \alpha+sin \alpha)}{4\pi R cos\alpha}$

$B_{net}=\dfrac {\mu_0I}{2\pi R}(\pi-a+tan a)$

Two parallel wires carrying current in the same direction attract each other because of

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potential difference between them
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mutual inductance between them
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electric forces between them
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magnetic forces between them

Explanation

Force on a current carrying conductor is

$F=\int i (\vec{dl} \times \vec{B})$
Magnetic field due to one wire at the location of the second wire ( other one ) is

$B_1= \frac{\mu_0 i_1}{2\pi d}$
Forceon the 2nd wire due to

$B_1$ is

$F_{21}= i_2l_2B_1= i_2l_2 \dfrac{\mu_0 i_1}{2\pi d} =\dfrac{\mu_0 i_1i_2l_2}{2\pi d}$
Force per unit length for wire

$l_2$ is

$F_{ per\: unit\: length}=\dfrac{F_{21}}{l_2}$
The force is attractive.

A current-carrying loop lying in a magnetic field behaves like a

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magnetic dipole
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magnetic pole
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magnetic material
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non-magnetic material

Suppose that a proton traveling in vaccum with velocity

$u_1$ at right angles to a uniform magnetic field experiences twice the force that an

$\alpha$ particle experiences when it is traveling along the same path with velocity

$u_2$ .The ratio

$\left (\dfrac {u_1}{u_2}\right )$ is

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$\dfrac {1}{2}$
0%
$1$
0%
$2$
0%
$4$

Explanation

$F_p= q_pu_1B$

$F_{\alpha} =q_{\alpha}u_2B$
Given,

$F_p = 2F_{\alpha}$

$\therefore q_pu_1B = 2q_{\alpha}u_2B$

$\therefore \dfrac{u_1}{u_2} = 2\dfrac{q_{\alpha}}{q_p}=2 \times 2 =4$ since

$q_{\alpha}=2q_p$

A current of

$1\ A$ is flowing in a coil of

$10$ turns having radius

$10\ cms$ . Its magnetic moment will be

0%
$3140\ Am^2$
0%
$100\ Am^2$
0%
$\mu_0 \ Am^2$
0%
$0.314\ Am^2$

Explanation

$A = \pi r^2 = 3.14 \times \left(10 \times 10^{-2}\right)^2= 3.14 \times 10^{-2}$

$N= 10$

$i= 1\: A$

$\therefore \mu = NiA = 10 \times 1 \times 3.14 \times 10^{-2}=0.314\: Am^2$

A bar magnet for dipole moment

$10^4\ JT^{-1}$ is free to rotate around a horizontal plane. A horizontal magnetic field

$4\times 10^{-5}\ T$ exists on the space. Find the work done in rotating the magnet slowly from a direction parallel to the field to a direction

$60^0$ from the field.

0%
$0.1\ J$
0%
$0.2\ J$
0%
$0.4\ J$
0%
$0.5\ J$

Explanation

$W=MB(\cos 0-\cos 60)$

$W=\dfrac{1}{2}MB$

$W=\dfrac{1}{2}\times4\times10^{-5}\times10^4$

$W=0.2\ J$

A proton enters a perpendicular magnetic field of 20 Tesla. If the velocity of proton is

$4 \times 10^7 m/s$ then the force acting on it will be :

0%
1.28 Newton
0%
$1.28 \times 10^{-11}$ Newton
0%
12.8 Newton
0%
$12.8 \times 10^{-11}$ Newton

Explanation

$F = qVB$

so we have

$B = 20 Tesla$ and

$V = 4 \times 10^{7}$

So putting this value

$Force = (1.6 \times 10^{-19})(20)(4 \times 10^{7})$

$Force = 12.8 \times 10^{-11}$ N

Current

$I$ is flowing in a coil of area

$A$ and number of turns is

$N$ , then magnetic moment of the coil is

$M$ equal to

0%
$NIA$
0%
$NI/A$
0%
$NI/\sqrt A$
0%
${N}^{2}AI$

Explanation

Magnetic moment

$M =NIA$

If a bar magnet in magnetic moment

$m$ is deflected from an angle

$\theta$ in a uniform magnetic field of induction

$B$ , the work done in reversing the direction is

0%
$MB\sin\theta$
0%
$MB$
0%
$MB(1-\cos\theta)$
0%
$MB\cos\theta$

Explanation

Now lets assume that the bar magnetic is initially placed parallel to the field the initial potential energy will be

$U_i=-\overrightarrow M.\overrightarrow B=-MB\cos 0=-MB$
Now when it is deflected by an angle

$\theta$ then

$U_f=-\overrightarrow M.\overrightarrow B=-MB\cos \theta$
Now we know that the work done =

$-\Delta U$

$\Delta U=U_i-U_f$

$\Delta U=-MB(1-\cos \theta)$

therefore, work done

$=MB(1-\cos \theta)$ .

A coil

$(8cm\times4cm)$ carries a current

$2A$ and has

$200$ turns. Find the magnetic dipole moment.

0%
$12.8Am^2$
0%
$1.28Am^2$
0%
$6.4Am^2$
0%
$0.64Am^2$

Explanation

Magnetic dipole moment of a loop is given by

$\overrightarrow { M } =N\overrightarrow { I } .\overrightarrow { S }$

$\overrightarrow {M} = 200\times 2\times S\times \cos { 0 }$

$S=0.0032 sqr m$

$\overrightarrow {M}=1.28 \hat { n } Am^2$ .

Find the ratio of magnetic dipole moment to angular momentum in a hydrogen like atom :

0%
$\displaystyle \frac {e}{m}$
0%
$\displaystyle \frac {e}{2m}$
0%
$\displaystyle \frac {e}{3m}$
0%
$\displaystyle \frac {2e}{m}$
0%
$\displaystyle \frac {3e}{m}$

Explanation

For Hydrogen atom,

$\displaystyle \dfrac {M}{L}$

$= \dfrac {iA}{mvr}$

$=\dfrac {\dfrac{ev}{2\pi r}\times \pi r^{2}}{mvr}$

$=\dfrac{e}{2m}$

A charge q is moving with a velocity v parallel to a magnetic field B. Force on the charge due to magnetic field is

0%
q v B
0%
q B/v
0%
zero
0%
B v/q

Explanation

$F_n = q (\vec v \times \vec B)$

$= qvB sin \theta$

$= 0$ (because

$\theta = 0^o$ )

The magnetic dipole moment of current loop is independent of

0%
number of turns
0%
area of loop
0%
current in the loop
0%
magnetic field in which it is laying

Explanation

$\text{Magnetic moment of the current loop is given by:}$

$M=niA$

$\text{Here,}\ n=\text{Number of turns},\ i=\text{Current},\ A=\text{Area}$

$\text{Therefore,}\ M \text{ is independent of B.}$

What name is given to a cylindrical coil of diameter less than its length ?

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Lactometer
0%
Barometer
0%
Bolenoid
0%
Solenoid

A charged particle moves through a magnetic field in a direction perpendicular to it. Then the

0%
velocity remains unchanged
0%
speed of the particle remains unchanged
0%
direction of the particle remains unchanged
0%
acceleration remains unchanged

An electron moving to the east in a horizontal plane is deflected towards south by a magnetic field. The direction of this magnetic field is

0%
Towards north
0%
Towards west
0%
Downwards
0%
Upwards

The magnetic field produced by a current-carrying wire at a given point depends on

0%
the current passing through it.
0%
the voltage across it
0%
the power through it
0%
all

When a current in a circular loop is equivalently replaced by a magnetic dipole

0%
the pole strength $m$ of each pole is fixed
0%
the distance $d$ between the poles is fixed
0%
the product $md$ is fixed
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None of these

In a current carrying conductor, if the direction of magnetic field and that of the current are mutually perpendicular to each other, then the force acting on the conductor will be

0%
perpendicular to both
0%
perpendicular to direction of current
0%
perpendicular to direction of field.
0%
none

1 gauss is equal to

0%
$10^4T$
0%
$10^{-4}T$
0%
$10^3 T$
0%
none of these

Explanation

The Gauss is defined by the Ampère's force law when it is applied to CGS units. A wire 1 centimetre long, sitting in a flux density equal to one Gauss and carrying a current of 1 abampere experiences a force equal to 1 dyne.

Thus

$1G=\dfrac{1dyne}{(1cm*10A)}=\dfrac{10^{-5}N}{(10^{-2}m*10A)}=\dfrac{10^{-4}N}{Am}=10^{-4}T$

A conducting rod AB moves parallel to X-axis in a uniform magnetic field, pointing in the positive X-direction. The end A of the rod gets

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positively charged
0%
negatively charged
0%
neutral
0%
first positively charged and then negatively charged

The formation of a dipole is due to two equal and dissimilar point charges placed at a

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short distance
0%
long distance
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above each other
0%
None of these

If the current is doubled, the deflection is also doubled in

0%
a tangent galvanometer
0%
a moving-coil galvanometer
0%
both
0%
None of these

Ampere rule is used to find the

0%
direction of current
0%
direction of magnetic field
0%
direction of motion of the conductor
0%
magnitude of current

Explanation

Ampere's law states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop.

Mathematically

$\mu_0I_{enc }=\int \vec{B}.d\vec{l}$

This is used to find the magnetic field due to a current carrying conductor.

A magnetic field exerts no force on:

0%
A magnet
0%
An unmagnetised iron bar
0%
A moving charge
0%
Stationary charge

Scale used in moving coil galvanometer is :

0%
function scale
0%
linear scale
0%
exponential scale
0%
none of these

An electron of charge

$e$ and mass

$m_e$ moves in a circular path of radius

$r$ in a uniform magnetic field

$B$ .
The Kinetic Energy of the electron can be described by the expression

0%
$\dfrac{1}{2}m_ev^2$
0%
$evBr$
0%
$\dfrac{1}{2}evBr$
0%
$2evBr$
0%
$\sqrt{\dfrac{1}{2}evBr}$

Explanation

Using

$eB r = mv$

$\implies v = \dfrac{eBr}{m}$

$\therefore$ kinetic energy of the electron

$K.E = \dfrac{1}{2}m_ev^2$ where

$v = \dfrac{eBr}{m}$

A charged particle moves in a magnetic field. The only force influencing the particle is the force caused by the magnetic field.
During the particle's movement in the magnetic field, what will NOT change?

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the particle's velocity
0%
the particle's acceleration
0%
the particle's speed
0%
the particle's momentum
0%
the particle's position

In ballistic galvanometer, the frame on which the coil is wound is non-metallic to:

0%
Avoid the production of induced emf
0%
Avoid the production of eddy currents
0%
Increase the production of eddy currents
0%
Increase the production of induced emf

A current carrying conductor.

0%
Experience a force when it is in magnetic field
0%
Does not experience a force when it is in magnetic field
0%
Experiences the force only when the field is electromagnetic in nature
0%
None of the above

Explanation

As per right hand thumb rule, when a current carrying conductor is in magnetic field, it experiences force as

$F=IlB\sin\theta$

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 2 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Moving Charges And Magnetism Quiz 2 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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