Magnetism and Matter Test 43

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Magnetism and Matter Test 43

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Question 1
1 / -0

The area enclosed by a hysteresis loop is a measure of

A
Retentivity

B
Susceptility

C
Permeability

D
Energy loss per cycle

Solution

$$\textbf{Correct option : option D}$$

$$\textbf{Explanation:}$$

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.
Question 2
1 / -0

Susceptibility of ferromagnetic substance is

A
> 1

B
< 1

C
0

D
1

Solution

Latex form:

$$\textbf{Correct option : option A}$$
Question 3
1 / -0

Figure shows a circular area of radius $$R$$ where a uniform magnetic field $$\vec { B }$$ is going into the plane of paper and increasing in magnitude at a constant rate. In rate case, which of the following graphs, drawn schematically, correctly shown the variation of the induced electric field $$E(r)$$?

A

B

C

D
Question 4
1 / -0

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

A
T

B
$$1/T^2$$

C
$$T^2$$

D
$$1/T$$

Solution
Question 5
1 / -0

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.

A
0.12

B
1.23

C
4.56

D
0.38

Solution

$$We\quad horizontal\quad component\quad of\quad earth\quad magnetic\quad field\quad H=\quad R\quad cos\quad \theta \\ Where\quad \theta =angle\quad of\quad dip=\quad { 22 }^{ 0 }\quad and\quad the\quad value\quad of\quad H=\quad 0.35\quad G\\ So\quad putting\quad the\quad values\quad respectively\quad R=\frac { H }{ cos\theta } =\quad \frac { 0.35 }{ cos\quad 22 } =\frac { 0.35 }{ 0.927 } =0.377=0.38\quad G\quad $$
Question 6
1 / -0

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.

A
60 mm

B
30 mm

C
2350 mm

D
2380 mm

Solution
Question 7
1 / -0

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

A
A paramagnetic and a ferromagnetic substance, respectively

B
A diamagnetic and a paramagnetic substance, respectively

C
Steel and soft iron, respectively

D
Soft iron and steel, respectively
Question 8
1 / -0

The magnetic field in a region is given by B=$$\dfrac{B_0}{L}$$y(k) where L is a fixed length. A conducting rod of length L lies along the Y-axis between the origin and the point (0.L,0). If the rod moves with a velocity v=$$v_0$$i, find the emf induced between thee ends of the rod.

A
$$\dfrac{B_0v_0l}{2}$$

B
$${B_0v_0l}$$

C
$$\dfrac{B_0v_0l}{4}$$

D
$$\dfrac{B_0v_0l}{6}$$

Solution

$$B=\dfrac{B_0}{L}gk$$
$$\Rightarrow $$ Length $$=L$$
$$\Rightarrow $$ Velocity $$=v=v_0\hat {i}$$
$$\Rightarrow \triangle \phi _{ B }=Bvl\cos \theta$$
$$=B_0\;v_o\;l\;\cos 60°$$
$$=\dfrac{B_0v_0l}{2}$$
$$\therefore \varepsilon =\dfrac { \triangle \phi _{ B } }{ \triangle t } =\dfrac{B_0v_0l}{2}$$
Hence, the answer is $$\dfrac{B_0v_0l}{2}.$$
Question 9
1 / -0

A train is moving from south to north with a velocity of $$90$$ km/h. The vertical component of earth's magnetic induction is $$0.4\times { 10 }^{ -4 }\quad Wb/{ m }^{ 2 }.$$ If the distance between the two rails is $$1 m,$$ what is the induced e.m.f. in its axle?

A
$$1mV$$

B
$$0.1mV$$

C
$$10mV$$

D
$$100mV$$

Solution

By taraday's law,
Einduced - $$ \frac{-d\phi }{dt}$$
$$ \phi = \vec{B}.\vec{A} = |\vec{B}| |\vec{A}| cos\theta $$
Here $$ \theta = 0$$ so,
$$ \phi = B.A$$
$$ \dfrac{-d\phi }{dt} = \dfrac{dB.A}{dt} = \dfrac{-BdA}{dt}-\dfrac{AdB}{dt}$$
As B is constant, $$ \dfrac{dB}{dt} = 0$$
$$ = \dfrac{d\phi }{dt} = \dfrac{-B.dA}{dt}$$
Here, $$ \dfrac{dA}{dt} = \dfrac{dlb}{dt} = \dfrac{ldb}{dt}+\dfrac{bdl}{dt}$$
b(breadth) of the train is not
changing it a length that is going
into the field is varying with speed
90km/h(25m/s),so.
$$ \dfrac{dA}{dt} = l(0)+bv = (1)(25) = 25$$
so,$$ \dfrac{d\phi }{dt} = -(0.4) \times 10^{-4} \times 25 = 10^{-3}v$$
Einduced =$$ 10^{-3}v = 1 mv$$
Option - A is correct.
Question 10
1 / -0

In a hydrogen atom, an electron of mass m and charge e revolves in an orbit of radius r making n revolutions per second. If the mass of hydrogen nucleus is M, the magnetic moment associated with the orbital motion of electron is :

A
$$\cfrac { \pi ne{ r }^{ 2 }m }{ M+m } $$

B
$$\pi ne{ r }^{ 2 }$$

C
$$\cfrac { \pi ne{ r }^{ 2 } }{ m } $$

D
$$\cfrac { \pi ne{ r }^{ 2 }m }{ M } $$

Solution

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