Magnetism and Matter Test 24

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Magnetism and Matter Test 24

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Question 1
1 / -0

A circular coil of $$100$$ turns has an effective radius of $$0.05 m$$ and carries a current of $$0.1 A$$. How much work is required to turn it in an external magnetic field of $$1.5 { Wb }/{ { m }^{ 2 } }$$ through $${ 180 }^{ o }$$ about its axis perpendicular to the magnetic field? The plane of the coil is initially perpendicular to the magnetic field :

A
$$0.456 J$$

B
$$2.65 J$$

C
$$0.2355 J$$

D
$$3.95 J$$

Solution

The potential energy of circular loop in the magnetic field $$=-MB \cos { \theta } $$, where $$\theta $$ is the angle between normal to plane of coil and magnetic field $$B$$.
Initial potential energy,
$${ U }_{ 1 }=-NiAB\cos { { 0 }^{ o } } =-NiAB$$
When coil is turned through $${ 180 }^{ o }$$, therefore final potential energy,
$${ U }_{ f }=-NiAB\cos { { 180 }^{ o } } =NiAB$$
$$\therefore $$ Required work, $$W=$$ gain in potential energy
$$={ U }_{ i }-{ U }_{ i }=NiAB-\left( -NiAB \right) =2NiAB$$
Here, $$N=100$$, $$i=0.1 A$$, $$B=1.5{ Wb }/{ { m }^{ 2 } }$$ and radius $$r=0.05 m$$
$$\therefore A=\pi { r }^{ 2 }=3.14\times { \left( 0.05 \right) }^{ 2 }=7.85\times { 10 }^{ -3 }{ m }^{ 2 }$$
$$\therefore $$ Work, $$W=2\times 100\times 0.1\times 7.85\times { 10 }^{ -3 }\times 1.5$$
$$\Rightarrow W=0.2355 J$$
Question 2
1 / -0

Which of the following is represented by the area enclosed by a hysteresis loop (B-H curve) ?

A
Permeability

B
Retentivity

C
Heat energy lost per unit volume in the sample

D
Susceptibility

Solution

Area enclosed by B-H curve represents energy lost. If the area of hysterosis loop is less energy loss is low whereas if the area of hysteresis loop is large energy loss is high.
Question 3
1 / -0

Curie temperature is defined as the temperature above which.

A
A paramagnetic material becomes ferromagnetic

B
A ferromagnetic material becomes paramagnetic

C
A ferromagnetic material becomes diamagnetic

D
A diamagnetic material becomes ferromagnetic

Solution

The Curie temperature $$T_C$$ is defined as the temperature at which the ferromagnetic material becomes paramagnetic.
Question 4
1 / -0

The magnetic lines of force inside a bar magnet.

A
Are from S-pole to N-pole

B
Are from N-pole to S-pole

C
Depends on teh cross-sectional area of magnet

D
Do not exist

Solution
Question 5
1 / -0

A ferromagnetic material heated above its curie temperature. Which one is a correct statement?

A
Ferromagnetic domains are perfectly arranged

B
Ferromagnetic domains become random

C
Ferromagnetic domains are not influenced

D
Ferromagnetic material changes into diamagnetic material

Solution

Beyond curie temperature, ferromagnetic material turns into paramagnetic material, as if ferromagnetic domains become random.
Question 6
1 / -0

The areas under the I-H hysteresis loop and B-H hysteresis loop are denoted by $$A_1$$ and $$A_2$$, then the ratio of $$\frac{A_2}{A_2}$$ is given by

A
$$\mu_0$$

B
$$\dfrac {1}{\mu_0}$$

C
$$1$$

D
$$\sqrt{\mu_0}$$

Solution

The relation between B and H is given by
$$B= \mu_0 (H+I)$$
$$db=\mu_0 dH+\mu_0 dl$$
$$\oint H\cdot dB=\mu _0\oint H\cdot dH +\mu _0\oint H\cdot dl$$
Also $$\oint H\cdot dH=0$$
$$\oint H\cdot dB=\mu _0\oint H\cdot dl$$
Area of B-H loop $$=\mu_0 \times area \, of \, l-H \, loop$$
$$A_2 =\mu_0 A_1$$
$$\Rightarrow \frac {A_1}{A_2} =\mu_0$$
Question 7
1 / -0

Horizontal component of the earth's field is $$3 \times 10^{-5} \, T$$ and of dip is $$53^o$$. Magnetic field of the earth at that place is

A
$$25 \times 10^{-5} \, T$$

B
$$5 \times 10^{-5} \, T$$

C
$$85 \times 10^{10} \, T$$

D
$$9.9 \times 10^{-8} \, T$$

Solution

Horizontal component of the earths magnetic fields is given by
$$B_H=B \, cos \delta $$
$$\Rightarrow B=\frac {B_H}{cos\,\delta }=\frac {3\times 10^{-5}}{cos\, 53^o}$$
$$ = 5 \times 10^{-5}\, T$$
Question 8
1 / -0

Choose the correct statement.

A
Polar molecules have permanent electric dipole moment

B
$$C{ O }_{ 2 }$$ molecule is a polar molecule

C
$${ H }_{ 2 }O$$ is a non-polar molecule

D
The dipole field at large distances falls of as $$\dfrac { 1 }{ { r }^{ 2 } } $$

E
The dipole moment is a scalar quantity

Solution

The total charge on a molecule is zero, the nature of chemical bonds is such that the positive and negative charges do not completely overlap in most molecules. Such molecules are said to be polar because they possess a permanent dipole moment. A good example is the dipole moment of the water molecule.
Question 9
1 / -0

The earth's magnetic field at some place on magnetic equator of the earth is $$0.5\times { 10 }^{ -4 }T$$. Consider the radius of the earth at that place as $$6400 km$$. Then, magnetic dipole moment of the earth is ___________ $${ Am }^{ 2 }$$. $$\left( { \mu }_{ 0 }=4\pi \times { 10 }^{ -7 }Tm{ A }^{ -1 } \right) $$

A
$$1.31\times { 10 }^{ 23 }$$

B
$$1.05\times { 10 }^{ 23 }$$

C
$$1.15\times { 10 }^{ 23 }$$

D
$$1.62\times { 10 }^{ 23 }$$

Solution

Given,
$${ R }_{ e }=6400 km=6.4\times { 10 }^{ +6 }m$$
$${ B }_{ e }=0.5\times { 10 }^{ -4 }T$$
$${ \mu }_{ 0 }=4\pi \times { 10 }^{ -7 }Tm{ A }^{ -1 }$$
$$M=$$?
The magnetic field at the equator
$$B=\dfrac { { \mu }_{ 0 } }{ 4\pi } \cdot \dfrac { i }{ { R }_{ e } }$$
and $$M=iA$$
$$M=\dfrac { 4\pi { R }_{ e }\cdot B }{ { \mu }_{ 0 } } \times \pi { R }_{ e }^{ 2 }$$
$$=\dfrac { 4\pi \times 64\times { 10 }^{ 5 }\times 5\times { 10 }^{ -5 }\times 3.14\times 4096\times { 10 }^{ 10 } }{ 4\pi \times { 10 }^{ -7 } }$$
$$=130\times { 10 }^{ 7 }\times 3.14\times 4096\times { 10 }^{ 10 }$$
$$=1673508.57\times { 10 }^{ 17 }$$
$$=1.67\times { 10 }^{ 23 }{ Am }^{ 2 }$$
Question 10
1 / -0

Which of these devices uses a magnet or electromagnet to make it work?

A
An electric bell

B
A compact disc

C
A microchip

D
All of these

Solution

An electric bell is a mechanical bell that functions by means of an electromagnet. When an electric current is applied, it produces a repetitive buzzing or clanging sound. Electric bells have been widely used at railroad crossings, in telephones, fire and burglar alarms, as school bells, doorbells, and alarms in industrial plants, since the late 1800s, but they are now being widely replaced with electronic sounders. It consists of coils of insulated wire wound round iron rods. When an electric current flows through the coils, the rods became magnetic and attract a piece of iron attached to a clapper. The clapper hits the bell and makes it ring.

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Magnetism and Matter Test 24

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Magnetism and Matter Test 24

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Question 1

$$0.456 J$$

$$2.65 J$$

$$0.2355 J$$

$$3.95 J$$

Solution

Question 2

Permeability

Retentivity

Heat energy lost per unit volume in the sample

Susceptibility

Solution

Question 3

A paramagnetic material becomes ferromagnetic

A ferromagnetic material becomes paramagnetic

A ferromagnetic material becomes diamagnetic

A diamagnetic material becomes ferromagnetic

Solution

Question 4

Are from S-pole to N-pole

Are from N-pole to S-pole

Depends on teh cross-sectional area of magnet

Do not exist

Solution

Question 5

Ferromagnetic domains are perfectly arranged

Ferromagnetic domains become random

Ferromagnetic domains are not influenced

Ferromagnetic material changes into diamagnetic material

Solution

Question 6

$$\mu_0$$

$$\dfrac {1}{\mu_0}$$

$$1$$

$$\sqrt{\mu_0}$$

Solution

Question 7

$$25 \times 10^{-5} \, T$$

$$5 \times 10^{-5} \, T$$

$$85 \times 10^{10} \, T$$

$$9.9 \times 10^{-8} \, T$$

Solution

Question 8

Polar molecules have permanent electric dipole moment

$$C{ O }_{ 2 }$$ molecule is a polar molecule

$${ H }_{ 2 }O$$ is a non-polar molecule

The dipole field at large distances falls of as $$\dfrac { 1 }{ { r }^{ 2 } } $$

The dipole moment is a scalar quantity

Solution

Question 9

$$1.31\times { 10 }^{ 23 }$$

$$1.05\times { 10 }^{ 23 }$$

$$1.15\times { 10 }^{ 23 }$$

$$1.62\times { 10 }^{ 23 }$$

Solution

Question 10

An electric bell

A compact disc

A microchip

All of these

Solution

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