MCQ Questions for Class 12 Medical Physics Magnetism And Matter Quiz 11

The magnetic field of earth at the equator is approximately

$4\times 10^{-5}\ T$ . The radius of earth is

$6.4\times 10^{6}\ m$ . Then the dipole moment of the earth will be near the order of

0%
$10^{23}\ A\ m^{2}$
0%
$10^{20}\ A\ m^{2}$
0%
$10^{16}\ A\ m^{2}$
0%
$10^{10}\ A\ m^{2}$

Explanation

$B=\cfrac {\mu_om}{4 \pi d^3}= \cfrac {4 \times 10^{-7}}{4 \pi (6.4 \times 10^6)^3}= \approx 10^{23} Am^2$

The coercivity of a small magnet where the ferromagnet gets demagnetized is

$3\times 10^{3}\ Am^{-1}$ . The current required to be passed in a solenoid of length

$10\ cm$ and number of turns

$100$ , so that the magnet gets demagnetized when inside the solenoid, is:

0%
$3A$
0%
$6A$
0%
$30\ mA$
0%
$60\ mA$

Explanation

Coercivity refers to the intensity of magnetic fields

$(B)$ where the magnet gets demagnetised

For solenoid we have,

$B=\mu _{0}nI$

(

$\mu _{0}=$ absolute permeability,

$n=$ no. of turns per unit length,

$B=$ Magnetic field in solenoid , I= current through each turn)

$\Rightarrow \mu _{0}H= \mu _{0}nI$ (

$H=$ intensity of magnetic field)

$\Rightarrow I=\dfrac{H}{n}$

$\Rightarrow I=\dfrac{3\times 10^{3}A/m}{\dfrac{100}{0.1 m}}=3A$

So current needed to be passed through solenoid

$=3A$

A wire of length L metres carrying a current I amperes is bent in the form of circle. The magnitude of its magnetic moment is _________ in MKS units.

0%
$\dfrac{IL^5}{4\pi}$ .
0%
$\dfrac{IL^2}{4\pi}$ .
0%
$\dfrac{IL^3}{4\pi}$ .
0%
$\dfrac{IL^4}{4\pi}$ .

Out of soft iron and steel which has more coercivity, less retentivity.

0%
iron
0%
steel
0%
both are equal
0%
can't say

The susceptibility of a diamagnetic substance is

0%
$\infty$
0%
zero
0%
Small but negative
0%
Small but positive

Some metals get magnetised by orientation of atomic magnetic moments in external magnetic field. What are they called ?

0%
just magnets
0%
Paramagnetics
0%
dimagnetics
0%
ferromagnetic

Consider the following statements I and II.
(I) Paramagnetism is explained by domain theory
(II) Susceptibility of a diamagnetic substance is independent of temperature.
Identify the correct choice from the following.

0%
Both I and II are correct
0%
Both I and II are incorrect
0%
I is correct and II is incorrect.
0%
I is incorrect and II is correct

Domain formation is the necessary features of

0%
Diamagnetism
0%
Paramagnetism
0%
Ferromagnetism
0%
All of these

Susceptibility is positive and small for

0%
Paramagnetic substances
0%
Ferromagnetic substances
0%
Non-magnetic substances
0%
Diamagnetic substances

The graph of X versus T for a paramagnetic substance will look like

The relative permeability is represented by

$\mu_r$ and the susceptibility by

$\chi$ for a magnetic substance. Then for a paramagnetic substance.

0%
$\mu_r> 1, \chi >0$
0%
$\mu_r> 1, \chi < 0$
0%
$\mu_r > 1, \chi >0$
0%
$\mu_r < 1, \chi < 0$

Explanation

Susceptence

$(\chi) = \dfrac{B-B_0}{B_0}$

Where

$B_0$ is external applied magnetic field and

$B$ is magnetic field in presence of paramagnetic substance.

Since dipoles in paramagnetic substance align in direction of applied magnetic filed, therefore

$B > B_0$

And

$B$ is bit more than

$B_0$

$\Rightarrow \chi > 0$

And also

$\mu_r = \chi + 1$

$\Rightarrow \mu_r > 0 + 1$

$\Rightarrow \mu_r > 1$

Therefore option A is correct.

Which of the following statements is incorrect about hysteresis?

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The effect is common to all ferromagnetic substances.
0%
The hysteresis loop area is proportional to the thermal energy developed per unit volume of the material.
0%
The hysteresis loop area is independent of the thermal energy developed per unit volume of the material.
0%
The shape of the hysteresis loop is characteristic of the material.

Time period of a bar magnet oscillating in earth's magnetic field is T. Its magnetic moment is M. Now, a bar magnet of magnetic moment

$2$ M is brought near it. The new time period will. be

0%
$T$
0%
$\dfrac{T}{\sqrt3}$
0%
$T\sqrt3$
0%
$3T$

Explanation

we know,

Time Period of a magnet in a magnetic field is

$=2\pi\sqrt{\dfrac{I}{M.B}}$

now,

$M_i=M$ and

$M_f=3m$

$T_i =2\pi\sqrt{\dfrac{I}{M.B}}$

$T_f=2\pi\sqrt{\dfrac{I}{3.M.B}}$

$=\dfrac{1}{\sqrt{3}}\times 2\pi\sqrt{\dfrac{I}{M.B}}$

$T_f=\dfrac{1}{\sqrt 3}\ T_i$

Hence, Option

$\textbf B$ is correct.

The given figure represents a material which is

0%
Paramagnetic
0%
Diamagnetic
0%
Ferromagnetic
0%
Ferrimagnetic

A frog is levitated in magnetic field produced by current in a vertical solenoid below the frog is:

0%
Diamagnetic
0%
Paramagnetic
0%
Ferromagnetic
0%
non-magnetic

Two small bar magnets are placed in a line at certain distance

$d$ part. If the length of each magnet is negligible compared to

$d$ , the force between them will be inversely proportional to:

0%
$d$
0%
${d}^{2}$
0%
${d}^{3}$
0%
${d}^{4}$

Explanation

For bar magnet

$F=\dfrac{\mu_0}{4\pi}\left(\dfrac{6MM'}{d^4}\right)$ in axial position.

$\implies F\alpha \dfrac 1{d^4}$

The figure shows the various positions (labelled by subscripts) of small magnetised needles P and Q. The arrows show the direction of their magnetic movement.configuration corresponds to the lowest potential energy among all the configurations shown

0%
$PQ_3$
0%
$PQ_4$
0%
$PQ_5$
0%
$PQ_6$

Explanation

Given, two small magnetic needle P and Q respectively,

So, At

$pQ_6$ when all the movements are in the parallel direction potential will be minium.

987855_1041730_ans_d076a2109d4c4d31bc70a61e899cd9db.png

A conductor behave as a super conductor

0%
Above critical temperature
0%
At critical temperature
0%
At ${100}^{o}C$
0%
At boiling point of that metal

Force between two identical bar magnets whose centres are

$r$ metre apart is

$4.8N$ , when their axes are in the same line. If separation is increased to

$2r$ , the force between them is reduced to.

0%
$0.3\ N$
0%
$0.6\ N$
0%
$2.4\ N$
0%
$1.2\ N$

Explanation

Force acting between the bar magnets

$F\propto \dfrac{1}{r^2}$

where

$r$ is the separation between the two bar magnets.

Thus if the separation between them is increased by

$2$ times, the force between them gets reduced by

$4$ times.

So, the new force becomes

$F' = \dfrac{F}{4} = \dfrac{4.8}{4} = 1.2 \ N$

A small rod of bismuth is suspended freely between the poles of a strong electromagnet. It is found to arrange itself at right angles to the magnetic field. This observation establishes that bismuth is :

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Diamagnetic
0%
Paramagnetic
0%
Ferro-magnetic
0%
Antiferro-magnetic

As a result of a change in the magnetic flux linked to the closed loop shown in the figure, an e.m.f. . V volt is induced in the loop. The work done(joules) in taking a charge Q coulomb once along the loop is:

0%
QV
0%
Zero
0%
$2$ QV
0%
QV $/2$

The wing span of an aeroplane is 20 meter. it is flying in a field.where the vertical component of magnetic field of earth is

$5 \times 10^-5$ tesla with velocity

$360mh^1$ . The potential difference product between the blades is:

0%
0.15 V
0%
0.16 V
0%
0.20 V
0%
0.30 V

Explanation

Given,

Wing span, $L=20\,m$

Magnetic Field $\overset{\scriptscriptstyle\rightharpoonup}{B}=\ 5\times {{10}^{-5}}\ T$

$1\ mile=1.6\ kilometer$

Velocity, $v=360\ miles/h=\ 360\times 1.6\ =576\ km/s$

$v=\ 160\ m/s$

Potential Difference, $V=BLv=5\times {{10}^{-5}}\times 20\times 160=0.16\ V$

Potential difference is $0.16\,V$

A magnetising force of

$360 Am^{-1}$ produces a magnetic flux density of

$0.6 T$ in a ferromagnetic material. The susceptibility of the material is:

0%
$1625$
0%
$1329$
0%
$2105$
0%
$1914$

Explanation

Permeability of material,

$\mu = \frac{B}{\mu } = \frac{{0.6}}{{360}} = 1.67 \times {10^{ - 3}}{T^{ - 1}}\,Am$

$\mu = {\mu _0}\left( {1 + {X_m}} \right)$

$\therefore$ Susceptibility,

${X_m} = \frac{\mu }{{{\mu _0}}} - 1 = \frac{{1.67 \times {{10}^{ - 3}}}}{{47 \times {{10}^{ - 7}}}} - 1 = 1329$

Hence, Option

$B$ is correct.

The force between two magnetic poles is

$F$ in air. If the pole is doubled that second pole is tripled and the distance between the poles is halved the force between the poles becomes and increase by

0%
$24\ F$ and $2300\%$
0%
$24\ F$ and $2400\%$
0%
$\dfrac {3}{2}\ F$ and $150\%$
0%
$2\ F$ and $100\%$

Temperature of a conductor increases by

$5^\circ C$ on passing electric current for some time. The increases inits temperature when double current is passed through the same conductor for the same time is _______

$^\circ C$ .

0%
$10$
0%
$12$
0%
$18$
0%
$20$

The armature of dc motor has 20Ω resistance .It draws current of 1.5 ampere when run by 220 volts dc supply. The value of back e.m.f. induced in it will be

0%
190 V
0%
150 V
0%
170 V
0%
120 V

Explanation

We know that,

$i=\cfrac{E-e}{R}$

where

$E=emf$ and

$R=$ Resistance

$1.5=\cfrac{220-e}{20}\\e=190V$

When a diamagnetic substance is placed near a magnet then it is

0%
Attracted
0%
Repelled
0%
No effect
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None of these

The force between two poles is reduced to P newton. when their separation in increased to x times, it is increased to Q newton, when their separation is made 1/xth of their original value.

0%
$Q=XP$
0%
${ Q=X }^{ 2 }P$
0%
${ Q=X }^{ 3 }P4$
0%
${ Q=X }^{ 4 }P$

Copper is :

0%
Paramagnetic
0%
Diamagnetic
0%
Ferromagnetic
0%
None of these

Which of the followings is not property of magnetic lines of force.

0%
The tangent at any point on the magnetic lines of pole given the direction of the magnetic field at that point.
0%
No two magnetic lines of pole can intersect each other.
0%
Magnetic lines of force are crowded where the magnetic field is strong and far from each other where field is weak.
0%
They are closed continuous curves. They start from south pole and end on north pole.

a force acting on a magnetic pole of

$0.07 A m$ is

$31.5 N$ . the magnetic field at that point is

0%
$450$
0%
$400$
0%
$350$
0%
$300$

Explanation

Given,

$F=31.5N$

$M=0.07Am$

The force acting on a magnetic pole is given by

$F=MB$

$B=\dfrac{F}{M}$

$B=\dfrac{31.5}{0.07}=450T$

The correct option is A.

The magnetic dipole moment of a small bar magnet is

$1.6 Am^{2}$ . It is placed in magnetic meridian with its north pole pointing toward geographic south of the earth. If the neutral point is obtained at

$20$ cm from the centre of the magnet then, the horizontal component of earth's magnetic field is:

0%
$10^{-5}T$
0%
$4\times 10^{-5}T$
0%
$4\times 10^{+5}T$
0%
$4\times 10^{-7}T$

Explanation

at the neutral point, the net magnetic field in zero.

The field on a axial line (Bax) due to the magnetic

dipole is given by at a distance

$r=0.2m$

$Bax=\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$ (M : Magnetic moment =

$1.6Am^2$ )

$\Rightarrow Bax=\dfrac{10^{-7}\times 2\times 1.6}{(0.2)^3}=4\times 10^{-5}T$

at the neutral point,

$Bax = B_H$

$\Rightarrow B_H=4\times 10^{-5}T$

1068298_1166289_ans_7df70be85c054c5d810a938c8a11221c.png

Magnetic field at point 'P'due to following current distribution is

0%
$\dfrac { \mu _ { 0 } I } { \pi r ^ { 2 } } \sqrt { r ^ { 2 } - a ^ { 2 } }$
0%
$\dfrac{{{\mu _0}i}}{{2\pi a}}$
0%
$\dfrac { \mu _ { 0 } I } { 2 \pi r ^ { 2 } } \sqrt { r ^ { 2 } - a ^ { 2 } }$
0%
$\dfrac { \mu _ { 0 } l a } { 2 \pi r ^ { 2 } }$

Explanation

Hence, option

$(B)$ is correct answer.
1380535_1242248_ans_df3ebfb020704044a0bdea721b46af71.jpg

For protecting a sensitive equipment from the external magnetic field, it should be

0%
Placed inside an aluminium cane
0%
Placed inside an iron cane
0%
Wrapped with insulation around it when passing current through it
0%
Surrounded with fine copper sheet

Atomic number of an element is

$26$ . The element shows:

0%
Ferromagnetism
0%
Diamagnetism
0%
Paramagnetism
0%
None of these

Explanation

Iron is magnetic in its α form. The α form occurs below a special temperature called the Curie point, which is 770 °C. Iron is paramagnetic above this temperature and only weakly attracted to a magnetic field. Magnetic materials consist of atoms with partially-filled electron shells.

[Ar]

$3d^64s^2$ ..

The variation of magnetic susceptibility (x) with temperature for a diamagnetic substance is best represented by

If a solution of ferromagnetic material is poured into a U-tube and one arm of this is placed between the poles of a strong magnet with the meniscus in line with the field, then the level of the solution will:

0%
Rise
0%
Fall
0%
Rise till the liquid comes out
0%
Remain unchanged

The basic magnetization curve for a ferromagnetic material is shown in figure. Then,the value of relative permeability is highest for the point

0%
P
0%
Q
0%
R
0%
S

Explanation

Hint:- Use relationship between magnetising field and magnetic filed.

Explanation :-

As we know that magnetic field is given by

$B=\mu_o\mu_r H$

$\mu_r=\dfrac B{\mu_o H}$

$\mu_o=constant$ So ,

$\mu_r \propto \dfrac BH$ =Slope of B-H curve

Hence

$\mu_r$ is greatest for which slope is maximum and slope is maximum for PQ so

$\mu_r$ is greatest for PQ

In the given figure, the electron enters into the magnetic field. It deflects in x direction .
Two parallel beams of electrons moving in the same direction produce a mutual force

0%
+ ve X direction
0%
- ve X direction
0%
+ ve Y direction
0%
- ve Y direction

A charge particle of charge q = 2 C has velocity

$v=100{ ms }^{ -1 }$ . When it passes point A adn has velocity in the direction shown. The strength of magnetic field at the point B due to this moving charge is (r = 2m)

0%
$2.5\mu T$
0%
$5.0\mu T$
0%
$2.0\mu T$
0%
None of these

Explanation

Given :charge(q)=2C,v=100m/s, r=2m,

$\theta$ =30v=100ms−1

Solution: We know the formula of magnetic field on moving charge is :

$B=\frac{\mu _{0}M}{4\pi r^{3}}$

And as we know the know the formula of magnetic moment in terms of q &v

$M=q\left ( \vec{v}\times \vec{r} \right )$

$B=\frac{\mu _{0}q\left ( vr sin 30^{0} \right )}{4\pi r^{3}}=\frac{10^{-7} \times2\times100\times1 }{4\times 2} =2.5\mu T$

Hence the correct answer is A

A wire carrying

$I$ is shaped as shown. Section

$AB$ is a quarter circle of radius

$r$ . The magnetic field at

$C$ is directed:

0%
Along the bisector of the angle $ACB$ , away from $AB$
0%
Along the bisector of the angle $ACB$ , towards $AB$
0%
Perpendicular to the plane of the paper directed into the paper
0%
At an angle $\pi /4$ to the plane of the paper

Explanation

Use Right hand palm rule or Maxwell?s Cork screw rule.

Perpendicular to the plane of the paper directed into the paper.

Hence,

Option

$C$ is correct answer.

A coil of surface area

$100{cm}^{2}$ having

$50$ turns is held perpendiculars to the magnetic field intensity

$0.02Wb{m}^{-2}$ . The resistance of the coil is

$2\Omega$ . If it is removed from the magnetic field in

$1s$ , the induced charge in the coil is ______

0%
$5C$
0%
$0.5C$
0%
$0.05C$
0%
$0.005C$

Explanation

$\begin{array}{l} \Delta \phi =\dfrac { { 100 } }{ { 10000 } } \times 50\times 0.02 \\ =\dfrac { { 10 } }{ { 1000 } } =\dfrac { 1 }{ { 100 } } \\ \Delta Q=\dfrac { { \Delta \phi } }{ R } \\ =\dfrac { 1 }{ { 100\times } } =0.05C \\ Hence, \\ option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

A thin straight vertical conductor has

$10amp$ current flows vertically upwards. It is present at a place where

${B}_{H}=4\times {10}^{-6}T$ . At which point magnetic feild will be maximum

0%
$0.5m$ on south of conductor
0%
$0.5m$ on west of conductor
0%
$0.5m$ on east of conductor
0%
$0.5m$ on north-east of conductor

Explanation

Given,

$B_H=4\times 10^{-6}T$

$I=10A$

The magnetic field at point

$r$ from the vertical conductor,

$B=\dfrac{\mu_0 I}{2\pi r}$

$r=\dfrac{\mu_0 I}{2\pi B_H}=\dfrac{4\pi \times 10^{-7}\times 10}{2\pi \times 4\times10^{-6} }=0.5m$ on west of the conductor for the maximum magnetic field.

The correct option is B.

The magnetic susceptibility is negative for

0%
Paramagnetic materials only
0%
Diamagnetic material only
0%
Ferromagnetic materials only
0%
paramagnetic and ferromagnetic materials

A compass needle made of pure iron (with density

$8000 kg/m^3$ ) has a length 5 cm, width 1.0 mm and thickness

$0.50$ mm. The magnitude of magnetic dipole moment of an iron is

$\mu Fe = 2\times 10^{-23}$ JiT, If magnetisation of needle is equivalent to the alignment of 10% of the atoms in the needle, what is the magnitude of the needle's magnetic dipole moment

$\overrightarrow { \mu }$ ? (mass of iron per mole

$= 0.05 kg/mole, N_A = 6 x 10^{23}$ )

0%
$2.4 \times 10^{-3} J/T$
0%
$4.8 \times 10^{-3} J/T$
0%
$7.2 \times 10^{-3} J/T$
0%
$9.6 \times 10^{-3} J/T$

Explanation

Given: $\rho=8000kg/m^3$

$volume,V=5*10^{-2}*10^{-3}*0.5*10^{-3} m^3=2.5*10^{-8}m^3$

$\mu Fe=2*10^{-23}J/T$

$1 mole=50gm=6*10^{23} atoms$

To find: needle's magnetic dipole moment, $\mu=?$

Solution: mass of compass needle,

$M=\rho*V$

$M=8000*2.5*10^{-8}$

$M=2*10^{-4}Kg$

$M=0.2gm$

now, we have

no of atoms in

$50gm=6*10^{23} atoms$

$==>$ no of atoms in

$0.2gm=\dfrac{0.2*6*10^{23}}{50} atoms$

$==>$

$N=0.024*10^{23}atoms$

now, we know that

$\mu=\mu Fe*N$

$==>$

$\mu=2*10^{-23}*0.024*10^{23}$

$\mu=4.8*10^{-3}J/T$

Temperature above which a ferromagnetic substance becomes paramagnetic is called

0%
Critical temperature
0%
Boyle's temperature
0%
Debye's temperature
0%
Curie temperature

The mass of a specimen of a ferromagnetic material is

$0.6kg$ and its density is

$7.8\times{10}^{3}kg/{m}^{3}$ . If the area of hysteresis loop of alternating magnetising field of frequency

$50Hz$ is

$0.722MKS$ units then the hysteresis loss per second will be-

0%
$277.7\times{10}^{-5}$ Joule
0%
$277.7\times{10}^{-6}$ Joule
0%
$277.7\times{10}^{-4}$ Joule
0%
$277.7\times{10}^{-7}$ Joule

Explanation

Given: m=0.6 kg, density= 7.8

$\times 10^{3}$ frequency=50Hz, A=0.722

Solution: As we know the formula of hysteresis loss per second :

$Volume\times\nu \times A=\frac{0.6}{7.8\times10^{3}}\times50 \times0.722$

W=277.7

$\times 10^{-5}$

Hence the correct option is A.

The susceptibility of a paramagnetic substance was found for different temperatures and a graph of

$\chi$ against

$\frac{1}{T}$ was plotted. From the graph, it was found that what

$\chi =0.5, \frac{1}{T}=5 \times 10^{-3}/K$ . What is the curie constant for the substance ?

0%
50 K
0%
75 K
0%
100 K
0%
125 K

Explanation

Curie's law :

$\chi = \dfrac{c}{T}$

So,

$0.5 =c\times 5\times 10^{-3}$

$\implies \ c = 100 \ K$

Magnetic field is not associated with

0%
a change in uniform motion.
0%
an accelerated charge.
0%
a deaccelerated charge.
0%
a stationary charge.

Explanation

Magnetic field is not associated with

$a\,stationary\,charg e.$

Hence

$,$ option

$(D)$ is correct

$.$

Magnetic dipole moment is a:

0%
Scalar quantity
0%
Vector quantity
0%
Constant quantity
0%
None of these

CBSE Questions for Class 12 Medical Physics Magnetism And Matter Quiz 11 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Magnetism And Matter Quiz 11 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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