MCQ Questions for Class 12 Medical Physics Magnetism And Matter Quiz 14

Which of the following is not suitable for the core of the electromagnets ?

0%
Cu-Ni alloy
0%
Soft iron
0%
Steel
0%
Air

What creates a magnetic field? More than one answer may be correct.

0%
a stationary object with electric charge
0%
a moving object with electric charge
0%
a stationary conductor carrying electric current
0%
a difference in electric potential
0%
a charged capacitor disconnected from a battery and at rest

The magnetic lines of forces :

0%
Interested each other
0%
Do not intersect each other
0%
Parallel to each other
0%
Perpendicular to each other

A thin sheet of the conductor, when allowed to oscillate in a magnetic field normal to the sheet, then the motion is -

0%
damped due to air friction
0%
damped due to eddy currents
0%
accelerated due to eddy currents
0%
not effected by induced currents

Two short magnets with pole strengths of $900Am$ and $100Am$ are placed with their axes in the same vertical line, with similar poles facing each other. Each magnet has a length of $1 cm$ . When separation between the nearer poles is $1cm$ , the weight of the upper magnet is supported by the repulsive force between the magnets. If $g=10ms^{-2}$ , then the mass of upper magnet is ( $x \times10^{-4}$ )

0%
$1.6 kg$
0%
$2.4 kg$
0%
$5.5 kg$
0%
$4.8 kg$

Explanation

$A = \dfrac{\mu_o}{4\pi} \dfrac{m_1m_2}{r^2}$ = Force exerted between two mag. poles

$m_1 = 900, m_2=100,r=0.01$ .

Total force of repulsion =

$F_T =A(1-\dfrac{2}{2^2} + \dfrac{1}{3^2})$

[near poles are like 1cm apart, repulsion, next nearest 2 pairs atract 2cm apart, the furthest pair repulse3 cm apart]

$F_T = Mg$

$F_t = 10^{-7}\times 900\times \dfrac{100}{(0.01)^2} \times (\dfrac{5.5}{9}) = Mg = 10M$

$5.5 \times 10 = 10M$

Therefore,

$M = 5.5kg$

Two charges repel each other with a force F when they are at certain distance apart. If the charges are halved and the distance between them is also halved, then force between the charges now is

0%
$\displaystyle \frac{F}{16}$
0%
$\displaystyle \frac{F}{4}$
0%
$\displaystyle \frac{F}{8}$
0%
F

Find the interaction force of two coils with magnetic moments

$p_{1m}=4.0\:mA$ -

$m^2$ and

$p_{2m}=6.0\:mA$ -

$m^2$ and collinear axes if the separation between the coils is equal to

$l=20\:cm$ which exceeds considerably their linear dimensions.

0%
$9$
0%
$5$
0%
$8$
0%
$10$

Explanation

$\displaystyle F=P_{2m}\dfrac{\partial}{\partial l}\left[\dfrac{\mu_0}{4\pi}\dfrac{3\vec{p_{1m}}\cdot\vec{r}\vec{r}-\vec{p_{1m}}r^2}{r^5}\right]$

$\displaystyle=P_{2m}\dfrac{\partial}{\partial l}\left[\dfrac{\mu_0}{4\pi}\dfrac{p_{1m}}{l^3}\right]=\frac{-3}{2}\dfrac{\mu_0p_{1m}p_{2m}}{\pi l^4}=9\:nN$

The tangent drawn at any point on the magnetic lines of force gives direction of mangetic field at that point.

0%
True
0%
False

An electric dipole of moment

$\vec P$ is placed at the origin along the

$x-$ axis. The angle made by electric field with

$x-$ axis at point

$P$ , whose position vector makes an angle

$\theta$ with x-axis, is (where

$\tan \alpha = \dfrac{1}{2}\tan \theta$ )

0%
$\alpha$
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$\theta$
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$\alpha +\theta$
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$2\theta + \alpha$

A small, magnetized sphere A of pole strength $m$ is counter-poised by a pan in a balance. Another magnetised sphere B of the same mass as A is placed below A so that their centres are at a distance $1cm$ . If A and B are unlike poles of same strength and to restore counter the weight in the pan is increased by $500 gm$ , the pole strength of each sphere is:

0%
$1.4 Am$
0%
$9.8 Am$
0%
$0.7 Am$
0%
$70Am$

Explanation

Let pole strengths be

$P_{1}=P_{2}=P$

Force between two magnetic poles is given by:

$F_{1}=\dfrac{\mu _{0}P_{1}P_{2}}{4\pi d^{2}}$
where

$d= 0.01m$

Force due to added weight is

$F_{2}=mg$

where

$m=0.5 kg; g=10m/s^2$

Since the counter is balanced,

$F_{1}=F_{2}$

$\cfrac{\mu _{0}P^2}{4\pi d^{2}}=mg$

$\cfrac{10^{-7}P^2}{0.01^2}=0.5 \times 10$

$P=70.7 Am$

The figure shows an isosceles triangle wire frame with apex angle equal to

$\pi /2$ . The
frame starts entering into the region of uniform magnetic field

$B$ with constant velocity

$v$ at

$t= 0$ . The longest side of the frame is perpendicular to the direction of velocity.
If i is the instantaneous current through the frame then choose the alternative showing
the correct variation of i with time

A rectangle has sides of lengths 6 cm and 8 cm. Two magnetic poles each of pole strength 'm' are placed at the ends of one side of smaller length and magnetic force acting between them is 'F'. If they are moved to new positions of same rectangle so that they are diagonally opposite. Find final force acting on each pole.

0%
$\displaystyle \frac{18 F}{50}$
0%
$\displaystyle \frac{18 F}{10}$
0%
$\displaystyle \frac{18 F}{40}$
0%
$\displaystyle \frac{18 F}{80}$

A superconductor has

$T_c(0)=100 K$ . When a magnetic field of 7.5 T is applied, its

$T_c$ decreases to 75 K. For this material one can definitely say that when

0%
$B=5T, T_c(B)=80 K$
0%
$B=5T, 75K < T_c(B) < 100 K$
0%
$B=10T, 75K < T_c(B) < 100 K$
0%
$B=10 T, T_c=70 K$

Explanation

We can observe the graph and state the answer. At

$B = 7.5T, Tc(B)=75K$

=> For a

$B>7.5T$ Tc(B) will be further lower, and B<7.5T will give

$Tc(B) .75K$

But exact values cannot be found due to lack of information.

The diagram shows a magnet being placed in a magnetic field. What will happens to the bar magnet?

0%
The bar magnet does not react to the magnetic field
0%
The bar magnet moves in the direction of the field
0%
The bar magnet moves in the opposite direction of the field
0%
The bar magnet rotates clockwise by $90^o$

Explanation

Net force on magnet will be zero but there is torque due to which magnet rotate and stop when magnet is along field line i.e. after

$90°$ rotation clockwise.

The coercivity of a magnet is

$3\times 10^{3}Am^{-1}$ . What current should be passed through a solenoid of length

$10\ cm$ and number of turns

$50$ that the magnet is demagnetized when inserted in it?

0%
$3A$
0%
$6A$
0%
$9A$
0%
$12A$

Explanation

Length of solenoid

$L = 10 \ cm = 0.1 \ m$
Number of turns

$N = 50$
So, number of turns per unit length

$n = \dfrac{N}{L} = \dfrac{50}{0.1} = 500$
Current to be passed

$i = \dfrac{H}{n}$

$\implies \ i = \dfrac{3\times 10^3}{500} = 6 \ A$

Two strong bar magnets A and B are placed on a horizontal table such that the axis of B intersects A at mid-point as shown in figure. B is fixed while A is free to move.There are no magnetic fields except those due to A and B. Then A will

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rotate but not move toward or away from B.
0%
rotate and move toward B.
0%
move toward B and not rotate.
0%
rotate and move away from B.
0%
move way from B and not rotate.

A bar magnet is 10 cm long is kept with its north (N)-pole pointing north. A neutral point is formed at a distance of 15 cm from each pole:Given the horizontal component of earth's field is 0.4 Gauss, the pole strength of the magnet is

0%
9 A-m
0%
6.75 A-m
0%
27 A-m
0%
1.35 A-m

Explanation

Length of magnet = 10 cm = 10

$\times$ 10

$^{-2}$ m,
r = 15

$\times$ 10

$^{-2}$ m

$op \, = \, \overline {225 \, - \, 25} \, = \, \overline {200} \, cm$
Since , at the neutral point , magnetic field due to the magnet is equal to

$B_H$

$B_H \, = \,\displaystyle \frac{\mu _O}{4 \pi } \, \cdot \, \displaystyle \frac{M}{(OP^2 \, + \, AO^2)^{3/2}}$

$0.4 \, \times \, 10^{-4} \, = \, 10^{-7} \, \times \, \displaystyle \frac {M}{(200 \, \times \, 10^{-4} \, + \, 25 \, \times \, 10^{-4})^ {3/2}}$

$\displaystyle \frac {0.4 \, \times \, 10^{-4}}{10^{-7}} \, \times \, (225 \, \times \, 10^{-4})^{3/2} \, = \, M$

$0.4 \, \times \,10^{-6} \, 225 ^{3/2} \, = \, M$

$M \, = \, 1.35 \, A-m$

For diamagnetic materials, magnetic susceptibility is

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Small and negative
0%
Small and positive
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Large and negative
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Large and positive

In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields

$B_1$ (solid line) and

$B_2$ (dashed line). If

$B_2$ is larger than

$B_1$ , which of the following graphs shows the correct variation of R and T in these fields?

Explanation

Resistance usually varies directly with temperature.

But when considering superconductors, the least temperature is Tc(0).

In presence of B, Tc decreases. But that doesn't say anything about general R and T relation.

Only info we have is Tc(

$B_2$ ) < Tc(

$B_1$ ) if

$B_2>B_1$ . (from observing the graph)

The period of osillation of a freely suspended bar magnet is 4 second. If it is cut into two equal parts in length, then the time period of each part will be

0%
$4 sec$
0%
$2 sec$
0%
$0.5 sec$
0%
$0.25 sec$

Explanation

We know that, Torque on bar

$T = \vec{\mu} \times \vec{B}$

$I \alpha = -\mu B \sin \theta$

$I \alpha = - \mu B \theta$

$\sin \theta \approx$ (for very small

$\theta$ )

$I \dfrac{d^2 \theta}{dt^2} + \mu B \theta = 0$

$\dfrac{d^2 \theta}{dt^2} + \left(\dfrac{\mu B}{I} \right) \theta = 0$

from S.H.M

$w^2 = \dfrac{\mu B }{I} \Rightarrow T = 2 \pi \sqrt{\dfrac{I}{\mu B}}$

$T \alpha \sqrt{\dfrac{I}{\mu}}$

$\dfrac{\mu_1}{\mu_2} = \dfrac{2 m (l_1)}{2m (l_2)} = \dfrac{2m (l)}{2m (1/2)} = 2$

$\dfrac{I_1}{I_2} = \dfrac{\rho l_1 b_1 (l_1^2 + b^2_1)/12}{\rho l_2 b_2 (l_2^2 + b^2_2)/12}$

$\dfrac{I_1}{I_2} = \left(\dfrac{l_1}{l_2}\right)\left(\dfrac{b_1}{b_2}\right) \left(\dfrac{l_1^2}{l_2^2}\right) \left(\dfrac{1 + (b_1/l_1)^2}{1 +(b_2/l_2)^2}\right)$

$\dfrac{l_1}{l_2} = 2$

$b_1 = b_2$ and

$l_1 >> b_1$

So,

$\dfrac{l}{b} \approx 0$

$\therefore \dfrac{I_1 }{I_2} = (2) (1) (2)^2 = 8$

$T \alpha \sqrt{\dfrac{I}{\mu}}$

$\dfrac{T_1}{T_2} = \sqrt{\dfrac{I_1}{I_2}} \sqrt{\dfrac{\mu_2}{\mu_1}}$

$\dfrac{T_1}{T_2} = \sqrt{8} \sqrt{\dfrac{1}{2}}$

$\left(\because \dfrac{\mu_1}{\mu_2} = 2\right)$

$\dfrac{T_1}{T_2} = \sqrt{4} \Rightarrow T_2 = \dfrac{T_1}{2}$

$T_2 = \dfrac{4}{2} = 2 \sec$

$\therefore$ option B is correct

1719423_731707_ans_a8d4a9289d3b43afbcdf8b3aee191f07.png

A rectangle with its dimensions is as shown. A magnetic mono pole of pole strength '

$m$ ' is placed at the point '

$A$ ' and another pole of strength '

$5m$ ' is placed at the point '

$B$ '. The magnetic force acting between them is '

$F$ '. If magnetic pole at the point '

$B$ ' is moved to the point '

$C$ ', find final magnetic force acting between them.

0%
${F}/{2}$
0%
${F}/{3}$
0%
${F}/{4}$
0%
${F}/{17}$

A domain in ferromagnetic iron in the form of cube is having

$5\times 10^{10}$ atoms. If the side length of this domain is

$1.5\mu m$ and each atom has a dipole moment of

$8\times 10^{-24}$ A

$m^2$ , then magnetisation of domain is?

0%
$11.8\times 10^5$ A $m^{-1}$
0%
$1.18\times 10^4$ A $m^{-1}$
0%
$11.8\times 10^4$ A $m^{-1}$
0%
$1.18\times 10^5$ A $m^{-1}$

Explanation

The volume of the cubic domain is

$V=(1.5\times 10^{-6}m)^3=3.38\times 10^{-18}m^3=3.38\times 10^{-12}cm^3$
Number of atoms in domain (N)

$=5\times 10^{10}$ atoms
Since each iron atom has a dipole moment(m)

$=8\times 10^{-24}$ A

$m^2$

$m_{max}=$ total number of dipole moment for all atoms

$=N\times m$

$=5\times 10^{10}\times 8\times 10^{-24}$

$=40\times 10^{-14}=4\times 10^{-13}$ A

$m^2$
Now the consequent magnetisation

$M_{max}=\dfrac{m_{max}}{Domain volume}=\dfrac{4\times 10^{-13}Am^2}{3.38\times 10^{-18}m^3}$

$=1.18\times 10^5$ A

$m^{-1}$ .

Magnetic lines of force always cross each other.

0%
True
0%
False

Let

${ B }_{ P }$ and

${ B }_{ Q }$ be the magnetic field produced by the wire P and Q which are placed symmetrically in a rectangular loop ABCD as shown in figure. Current in wire P is I directed inward and in Q is 2 I directed outwards.
if

$\displaystyle \int _{ A }^{ B } \vec { { B }_{ Q } }.\vec { d\ell } =2{ \mu }_{ 0 }$ tesla meter,

$\displaystyle \int _{ D }^{ A }\vec { { B }_{ P } }.\vec { d\ell } =-2{ \mu }_{ 0 }$ tesla meter &

$\displaystyle \int _{ A }^{ B }\vec { { B }_{ P } }.\vec { d\ell } =-{ \mu }_{ 0 }$
tesla meter the value of I will be:

0%
8 A
0%
4 A
0%
5 A
0%
6 A

Aston mass spectrograph is a device to measure the atomic masses of isotopes with great precision. Its principle consists in applying.

0%
Successfully first the magnetic field and then the electric field at right angle to each other
0%
Electric and magnetic fields simultaneously in perpendicular direction
0%
Election and magnetic fields simultaneously and in parallel direction
0%
Successfully first the electric field and then the magnetic field at right angle to each other

$M_2$ = magnetization of a paramagnetic sample, B = external magnetic field, T = absolute temperature, C = curie constant then according to Curie's law in magnetism, the correct relation is

0%
$M_z = \dfrac{T}{CB}$
0%
$M_z = \dfrac{CB}{T}$
0%
$C = \dfrac{M_z}{T}$
0%
$C = \dfrac{T^2}{M_zB}$

Explanation

According to Curie's law:-

$X\propto\dfrac{1}{T}$ , where

$X$ =magnetic susceptibility and

$T=$ temperature

$\implies X=\dfrac{C}{T}$ , where

$C$ =Curie's constant

Magnetization,

$M_z=XB$

$\implies M_z=\dfrac{CB}{T}$

Hence, answer is option-(B).

Two small magnets X and Y of dipole moments

$M_1$ and

$M_2$ are fixed perpendicular to each other with their north poles in contact. This arrangement is placed on a floating body so as to move freely in earth's magnetic field as shown in figure. Find the ratio of magnetic moment.

0%
$1:\sqrt3$
0%
$2:\sqrt3$
0%
$\sqrt3:2$
0%
$\sqrt3:1$

A current / flows along the length of an infinitely long, straight, thin-walled pipe. Then

0%
The magnetic field at all points inside the pipe is the same, but not zero
0%
The magnetic field at any point inside the pipe is zero
0%
The magnetic field is zero only on the axis of the pipe
0%
The magnetic filed is different point inside the pipe.

Which of the statement is/are true about MAGLEV Technology?
It is Magnetic Suspension technology to keep object suspended with no support.
This technology is used in planes and ships.
Select the correct answer:

0%
Only 1
0%
Only 2
0%
Both 1 and 2
0%
None of the above

The area enclosed by a hysteresis loop is a measure of

0%
Retentivity
0%
Susceptility
0%
Permeability
0%
Energy loss per cycle

Explanation

$\textbf{Correct option : option D}$

$\textbf{Explanation:}$

Ferromagnetic material form hysteresis loop. The magnetic field in the ferromagnetic material changes is represented by this . Due to magnetization and demagnetization of material loss of energy happens in the form of heat and this is called hysteresis loop and this loss of energy per cycle is measured by area enclosed by a hysteresis loop.

Write the dimension of the magnetic flux in terms of mass, time, length and charge.

0%
$ML^3T^{-1}Q^{-1}$ .
0%
$ML^2T^{-1}Q^{-1}$ .
0%
$ML^4T^{-1}Q^{-1}$ .
0%
$ML^5T^{-1}Q^{-1}$ .

There are two current carrying planar coils each made from identical wires of length L.

$C_1$ is circular (radius R) and

$C_2$ is square(side a). They are so constructed that they have same frequency of oscillation when they are place in the same uniform magnetic field

$\vec{B}$ and carry the same current. Find a in terms of R.

0%
$a=R$
0%
$a=2R$
0%
$a=3R$
0%
$2a=5R$

Explanation

For coil

$C_1$ :

$n_1=\dfrac{L}{2\pi R}$ ,

$m_1=n_1iA_1=\left(\dfrac{L}{2\pi R}\right)$ (i)

$(\pi R^2)=\dfrac{LiR}{2}$

$I_1$ (moment of inertia of the coil

$C_1$ about an axis through its diameter)

$=\dfrac{1}{2}MR^2$
Also,

$\omega_1=\sqrt{\dfrac{m_1B}{I_1}}\left(as T_1=2\pi\sqrt{\dfrac{I_1}{mB}}\right)$
For the coil

$C_2$ :
As

$n_2=\dfrac{L}{4a}, m_2=n_2iA_2=\left(\dfrac{L}{4a}\right)(i)(a^2)=\dfrac{Lia}{4}$

$I_2$ (moment of inertia of coil

$C_2$ about an axis through its centre and parallel to one of its sides)

$=\dfrac{1}{12}Ma^2$
Also,

$\omega_2=\sqrt{\dfrac{m_2B}{I_2}}$
Since

$\omega_1=\omega_2, \dfrac{m_1}{I_1}=\dfrac{m_2}{I_2}$
or

$\dfrac{LiR/2}{\dfrac{1}{2}MR^2}=\dfrac{Lia/4}{\dfrac{1}{12}Ma^2}$ or

$\dfrac{Li}{MR}=\dfrac{3Li}{Ma}$
Hence,

$a=3R$ .

According to Curie's law, the magnetic susceptibility of a substance at an absolute temperature T is proportional to

0%
T
0%
$1/T^2$
0%
$T^2$
0%
$1/T$

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at

$22^0$ with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be

$0.35 G$ . Determine the magnitude of the earth's magnetic field at the place.

0%
0.12
0%
1.23
0%
4.56
0%
0.38

Explanation

$We\quad horizontal\quad component\quad of\quad earth\quad magnetic\quad field\quad H=\quad R\quad cos\quad \theta \\ Where\quad \theta =angle\quad of\quad dip=\quad { 22 }^{ 0 }\quad and\quad the\quad value\quad of\quad H=\quad 0.35\quad G\\ So\quad putting\quad the\quad values\quad respectively\quad R=\frac { H }{ cos\theta } =\quad \frac { 0.35 }{ cos\quad 22 } =\frac { 0.35 }{ 0.927 } =0.377=0.38\quad G\quad$

The two I-H graph A and B in the adjacent figure are for

0%
A paramagnetic and a ferromagnetic substance, respectively
0%
A diamagnetic and a paramagnetic substance, respectively
0%
Steel and soft iron, respectively
0%
Soft iron and steel, respectively

Statement-1: The core of the transformer is laminated to avoid loss of energy.
Statement-2: A laminated metal sheet placed in a changing magnetic field has lower eddy current.

0%
Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
0%
Statement-! is true, statement-2 is false.
0%
Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
0%
Statement-1 is false, statement-2 is true.

Explanation

The core of transformer is laminated to avoid loss of energy This statement is true and also a laminated metal sheet placed in a changing

$\vec{B}$ has lower eddy current.

A domain in ferromagnetic iron is in the form of a cube of side length

$2$

$\mu m$ then the number of iron atoms in the domain are (Molecular mass of iron

$=55$ g

$mol^{-1}$ and density

$=7.92$ g

$cm^{-3}$ )

0%
$6.92\times 10^{12}$ atoms
0%
$6.92\times 10^{11}$ atoms
0%
$6.92\times 10^{10}$ atoms
0%
$6.92\times 10^{13}$ atoms

Explanation

The volume of the cubic domain

$V=(2\mu m)^3=(2\times 10^{-6}m)^3$

$=8\times 10^{-18}m^3=8\times 10^{-12}cm^3$
and mass

$=$ volume

$\times$ density

$=8\times 10^{-12}cm^3\times 7.9g cm^{-3}$

$=63.2\times 10^{-12}$ g
Now the Avagadro number

$(6.023\times 10^{23})$ of iron atoms have a mass of

$55$ g. Hence the number of atoms in the domain are.

$N=\dfrac{63.2\times 10^{-12}\times 6.023\times 10^{23}}{55}=6.92\times 10^{11}$ atoms.

Three magnets of the same length but moments M,

$2$ M and

$3$ M are arranged in the forms of an equilateral triangle with opposite pole nearer, the resultant magnetic moment of the arrangement is?

0%
$6$ M
0%
Zero
0%
$\sqrt{3}$ M
0%
$\dfrac{\sqrt{3}}{2}$ M

Explanation

$\vec M=(M-2M\cos 60-3M \cos 60)\hat i+(2M\cos 30-3M\cos 30)$

$\vec M=\dfrac {-3}{2}M\hat i-\dfrac {\sqrt 3}{2}M$

$M=|\vec M|=\sqrt {\left (\dfrac {3}{2}\right)^2 +\left (\dfrac {\sqrt 3}{2}\right)^2}M=\sqrt 3 M$

Magnetic moment

$=\sqrt 3M$

A magnetic needle suspended by a silk thread is vibrating in the earth's magnetic field. If temperature of the needle is increased to

$1000^0C$ , then

0%
The time period decreases
0%
The time period remains unchanged
0%
The time period increases
0%
The needle stops vibrating

Explanation

Time period T is directly proportional to the inverse of the root of magnetic moment M as

$T \propto \dfrac{1}{\sqrt{M}}$

Since magnetic moment decreases with increase in temperature so the time period will increase.

Protons and singly iorized atoms of

$U^235 & U^238$ are passed in turn (which means one after the other and not at the same time) through a velocity selector and then enter a uniform magnetic field. The protons describe semicircles of radius 10 mm. The separation between the ions of $$U^235 and U^238 after describing semicircles is given by.

0%
60 mm
0%
30 mm
0%
2350 mm
0%
2380 mm

A thin conducting rod of length

$l$ is moved such that its end

$B$ moves along the X-axis while end

$A$ moves along the Y-axis. A uniform magnetic field

$B = {B_0}\hat k$ exists in the region. At some instant, the velocity of end

$B$ is

$v$ and the rod makes an angle of

$\theta = 60$ with the X-axis as shown in the figure. Then, at this instant

0%
angular speed of rod $AB$ is $\omega = \dfrac{{2v}}{{\sqrt 3 l}}$
0%
angular speed of rod $AB$ is $\omega = \dfrac{{\sqrt 3 v}}{{2l}}$
0%
e.m.f. induced in rod $AB$ is $Blv\sqrt 3$
0%
e.m.f. induced in rod $AB$ is $Blv/2\sqrt 3$

Explanation

$\begin{array}{l}OB = x = l\cos \theta \\\dfrac{{dx}}{{dt}} = - l\sin \theta \left( {\dfrac{{d\theta }}{{dt}}} \right)\\\dfrac{{d\theta }}{{dt}} = - \dfrac{1}{{l\sin \theta }}\left( {\dfrac{{dx}}{{dt}}} \right)\\ \Rightarrow \omega = - \dfrac{1}{{l\sin \theta }} \times {v_B}\\ \Rightarrow \omega = - \dfrac{1}{{l\sin 60}} \times v\\ \Rightarrow \omega = - \dfrac{v}{{l \times \sqrt 3 /2}} = \dfrac{{ - 2v}}{{\sqrt 3 l}}\\so,\omega = \dfrac{{2v}}{{\sqrt 3 l}}\end{array}$

A coil of radius 200mm is to produce a field of 0.4 G In its center with a current of 0.25. How many turns must be there in the coil?

0%
61
0%
51
0%
41
0%
63

A vertical circular coil of one turn and radius 9.42 cm is placed with its plane in the magnetic meridian and short magnetic needle is pivoted at the centre of the coil so that it can freely rotate in the horizontal plane. If a current of 6A is passed through the coil, then the needle deflects by

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

0%
$30^ \circ$
0%
$45^ \circ$
0%
$60^ \circ$
0%
$90^ \circ$

Explanation

The needle at the centre gets deflected and comes to an equilibrium position under the action of two perpendicular field

$:$

one due to horizontal component of earth and the other due to field

$(B)$ set up the coil due to current

$.$

$B = {B_H}\tan \theta$

and

$B = \dfrac{{{\mu _0}ni}}{{2r}}$

$n = no.of\,terms = 1,i = 6A,r = 9.42cm\,and\,{B_{^H}} = 4 \times {10^{ - 5}}T$

therefore

$,$

$B = {B_H}\tan \theta = \dfrac{{{\mu _0}ni}}{{2r}}$

$\tan \theta = \dfrac{{{\mu _0}ni}}{{2r{B_H}}} = \dfrac{{4\pi \times {{10}^{ - 7}} \times 1 \times 6}}{{2 \times 9.42 \times {{10}^{ - 2}} \times 4 \times {{10}^{ - 5}}}} = 1$

$\theta = {45^0}$

hence,

option

$(B)$ is correct answer.

The area of hysteresis loop of a material is
equivalent to 250 Joule. When 10kg material is magnetised by an alternating
field of 50Hz then energy lost in one hour will be if the density of material
is 7.5 $gm/c{m^3}$

0%
$6 \times {10^4}$ Joule
0%
$6 \times {10^4}$ Erg
0%
$3 \times {10^2}$ Joule
0%
$3 \times {10^2}$ Erg

Two concentric circular loops of radii

$R$ and

$2R$ carry currents of

$2i$ and

$i$ respectively in opposite sense (ie. Clockwise in one coil and center clockwise in the other oil). The resultant magnetic field at their common center is

0%
$\mu _{0}\dfrac {i}{4R}$
0%
$\mu _{0}\dfrac {5i}{4R}$
0%
$\mu _{0}\dfrac {3i}{4R}$
0%
` $\mu _{0}\dfrac {i}{2R}$

The radius of a circular wire is 0.5 m and the current is 10 A. The magnitude of magnetic field at the centre of the circular wire is :

0%
${ 12.57\times 10 }^{ -6 }T$
0%
${ 12.57\times 10 }^{ -5 }T$
0%
${ 12.57\times 10 }^{ -4 }T$
0%
${ 12.57\times 10 }^{ -3 }T$

B-H curve (Hysteresis loops) of samples of Iron , sample 'P' and sample 'Q' , are given below. Find the correct statement.

0%
Both 'P' and 'Q' are suitable for making permanent magnet.
0%
Both 'P' and 'Q' are suitable for making electromagnets.
0%
'P' is suitable for making permanent magnet and 'Q' for electromagnet.
0%
'P' is suitable for making electromagnet and 'Q' for permanent electromagnet.

Which of the following magnetic line alignment is correct for ferrimagnetism?

0%
$\uparrow, \uparrow, \uparrow, \downarrow, \downarrow$
0%
$\uparrow, \uparrow, \uparrow, \uparrow$
0%
$\uparrow, \downarrow, \uparrow, \downarrow$
0%
$\downarrow, \downarrow, \downarrow, \downarrow$

At a place earth's magnetic field,

$5\times 10^{-5}\,Wb/m^2$ is acting perpendicular to a coil of radius R=5 cm. If

$\dfrac{\mu_0}{4\pi}=10^{-7}$ , then how much current is induced in circular loop?

0%
0.2 A
0%
0 A
0%
4 A
0%
40 A

Every iron atom in a ferromagnetic domain in iron has a magnetic dipole moment equal to

$9.27 \times 10^{-24} \ A-m^2$ . A ferromagnetic domain in iron has the shape of a cube of side

$1 \mu m$ . The maximum dipole moment occurs when all the dipoles are aligned. The molar mass of iron is

$55 \ g$ and its specific gravity is

$7.9$ . The magnetisation of the domain is:

0%
$8.0 \times 10^5 \ A/m$
0%
$8.0 \times 10^8 \ A/m$
0%
$8.0 \times 10^{11} \ A/m$
0%
$8.0 \times 10^{14} \ A/m$

Explanation

Dipole moment,

$p=9.27\times 10^{-24} A-m^2$

$a=10^{-6}m$

$m=55g$

$N_A=6.023\times 10^{24} atom$

Volume of the domain,

$V=a^3$

$V=10^{-12}cm=10^{-18}m$

Density,

$\rho=7.9 g/cm^3$

Mass of domain,

$M=V\times \rho$

$M=10^{-12}\times 7.9$

$M=7.9\times 10^{-12}g$

The number of atom in domain,

$n=M\dfrac{N_A}{m}$

$n=7.9\times 10^{-12}\times \dfrac{6.023\times 10^{24}}{55g}$

$n=8.65\times 10^{10}atom$

The total dipole moment,

$P=np$

$P=8.65\times 10^{10}\times 9.27\times 10^{-24}$

$P=8\times 10^{-13}A-m^2$

The magnetization of the domain,

$M=\dfrac{P}{V}$

$M=\dfrac{8\times 10^{-13}}{10^{-18}}$

$M=8\times 10^5A/m$

The correct option is A.

CBSE Questions for Class 12 Medical Physics Magnetism And Matter Quiz 14 - 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 14 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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