Magnetism and Matter Test 39

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Magnetism and Matter Test 39

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Question 1
1 / -0

The most appropriate magnetization $$ M $$ versus magnetising field $$ H $$ curve for a paramagnetic substance is

A
$$ A $$

B
$$ B $$

C
$$ C $$

D
$$ D $$

Solution

For paramagnetic substance magnetization $$ M $$ proportional to magnetising field $$ H $$, and $$ M $$ is positive. Hence option $$A$$ is correct.
Question 2
1 / -0

For ferromagnetic material, the relative permeability $$ ( \mu_r) $$ , versus magnetic intensity $$ (H) $$ has the following shape

A

B

C

D

Solution

$$ \mu_r=1+\dfrac{1}{H}; $$ as we know $$ I $$ dependent on $$ H $$, initially value of $$ \dfrac{I}{H}$$ is smaller so value of $$ \mu $$ increases with $$H $$ but slowly but with further increases of $$H$$ value of $$\dfrac{I}{H} $$ also increases i.e. $$ \mu $$ increases speedily. When material fully magnetised $$I$$ becomes constant then with the increase of $$H$$ ($$\dfrac{1}{H} $$ decreases) $$ \mu $$ decreases, This is confirm field with the option (d).
Question 3
1 / -0

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

A
$$ P $$

B
$$ Q $$

C
$$ R $$

D
$$ S $$

Solution

$$ B= \mu_0\mu_rH \Rightarrow \mu_r \propto \dfrac{B}{H}$$ = slope $$ B-H $$ curve According to the given graph, slope of the graph is highest point $$ Q$$.
Question 4
1 / -0

The variation of magnetic susceptibility $$ (\chi) $$ with absolut temperature $$ T $$ for a ferromagnetic material is

A

B

C

D

Solution

Susceptibility of a ferromagnetic substance falls with rise of temperature $$ \left( \chi =\dfrac { c }{ T-{ T }_{ c } } \right) $$ and the substance becomes paramagnetic above curie temperature, so magnetic susceptibility becomes very small above curie temperature.
Question 5
1 / -0

Magnetic susceptibility is negative and very less for :

A
ferromagnetic substances

B
paramagnetic substances

C
diamagnetic substance

D
all of the above

Solution

Magnetic susceptibility is negative and very less for diamagnetic substances.
Question 6
1 / -0

If a conducting rod moves with a constant velocity $$ v $$ in a magnetic field, emf is induced between both its ends if:

A
$$ v $$ and $$ B $$ are parallel

B
$$ v $$ is perpendicular to $$ \overrightarrow{B} $$

C
$$ v$$ and $$ B $$ are in opposite direction

D
All of the above

Solution

The emf will only be induced if $$ v $$ is perpendicular to $$ \overrightarrow{B} $$
Velocity $$(\overrightarrow{v})$$ is perpendicular to magnetic field $$(\overrightarrow{B})$$
Question 7
1 / -0

The cause of dimegnetism is :

A
orbital motion of electrons

B
spin motion of electrons

C
paired electrons

D
none of the above

Solution

Due to orbital motion of electrons.
Question 8
1 / -0

The phase difference between induced emf and magnetic flux linked with a coil rotating in a uniform magnetic field is:

A
$$ \frac{\pi}{4} $$

B
$$ \frac{\pi}{2} $$

C
$$ \frac{\pi}{3} $$

D
$$ \pi $$

Solution

For a coil rotating in uniform magnetic field, the induced emf is always perpendicular to the magnetic field. So, the angle between them is $$ 90^o $$.
Question 9
1 / -0

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

A
Cu-Ni alloy

B
Soft iron

C
Steel

D
Air

Solution

Air because with it only magnetic field doesn't increase.
Question 10
1 / -0

An electric motor is a device to convert electrical energy into mechanical energy. The motor shown below has a rectangular coil $$(15\ cm \times 10\ cm)$$ with 100 turns placed in a uniform magnetic fleld $$B=2.5\ T$$. When a current is passed through the coil, lt completes $$50$$ revolutions in one second. the power output of the motor is $$1.5\ kW$$. The current rating of the coil should be

A
1 A

B
2 A

C
3 A

D
4 A

Solution

Let $$I$$ be the current through the coil of 100 turns. Let the magnetic moment $$\vec{M}$$ make an angle $$\theta $$ at an instant of time.
Then, torque on the coil
$$|\vec{\tau }|=MB\sin\theta $$
Work done by the torque in each half revolution $$=\dfrac{W}{2} $$, where $$W$$ is the work done in one full revolution.
$$\therefore \dfrac{W}{2}=\int_{0}^{\pi }MBsin\theta d\theta =IANB\int_{0}^{\pi }\sin\theta d\theta $$
$$=2IANB$$
$$\therefore $$work output per revolution $$=W=4INAB$$
Hence work output per second$$=Power=4INABf$$
$$\therefore 1.5\times 10^{3}=4INABf$$
$$I=\dfrac{1.5\times 10^{3}}{4\times 100\times 150\times 10^{-4}\times 2.5\times 50}$$
$$\therefore I=\dfrac{15\times 10^{2}}{15\times 50}=2\ A$$