Magnetism and Matter Test 23

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Magnetism and Matter Test 23

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Question 1
1 / -0

Permanent magnet has properties retentivity and coercivity respectively as:

A
high-high

B
low-low

C
low-high

D
high-low

Solution

The material for a permanent magnet should have high retentivity (so that magnet is strong) and high coercivity (so that magnetism is not wiped out by strong magnetic fields).
Question 2
1 / -0

A current carrying loop is placed in a uniform magnetic field in four different orientations I, II, III and IV as shown in figure. Arrange them in decreasing order of potential energy.

A
I > III > II > IV

B
I > II > III > IV

C
I > IV > II > III

D
III > IV > I > II

Solution

$$\textbf{Hint: Use the formula of the potential energy of a current-carrying loop placed in a uniform magnetic field}$$

$$\textbf{Step1: Formula of potential energy}$$
As we know that, potential energy of a magnet in a magnetic field
$$U=-m\cdot B$$
$$=-mB\cos { \theta }$$
where, $$ m=$$ magnetic dipole moment of the magnet
$$B=$$ magnetic field

$$\textbf{Step2: Calculate potential energy in different cases}$$
Case I $$\theta ={ 180 }^{ o }$$
$$\therefore { U }_{ 1 }=-mB\cdot \cos { { 180 }^{ o } }$$
$$ =mB$$ $$\left[ \because \cos { { 180 }^{ o } } =-1 \right]$$
Case II $$ \theta ={ 90 }^{ o }$$
$${ U }_{ 2 }=0$$ $$\left[ \because \cos { { 90 }^{ o } } =0 \right]$$
Case III $$ \theta$$ is acute angle
$$ \theta \in \left( 0,{ 90 }^{ o } \right) $$
$$\therefore \cos { \theta } =$$ positive
Thus, $${ U }_{ 3 }=$$ negative
Case IV $$\theta$$ is obtuse
$$ \theta \in \left( { 90 }^{ o },{ 180 }^{ o } \right)$$
$$ \therefore \cos { \theta } \in \left( 0,-1 \right)$$
Thus, $$ { U }_{ 4 }=$$ positive
Therefore, decreasing order of PE is
I > IV > II > III
$$\textbf{Hence option C correct}$$
Question 3
1 / -0

Magnetic susceptibility of diamagnetic materials is of the order of (SI units) :

A
$$+ 10^{5}$$

B
$$+10^{-4}$$ to $$+10^{-2}$$

C
$$+10^{-5}$$

D
$$-10^{-5}$$

Solution

Magnetic susceptibility of diamagnetic substance is negative and close to zero. So, from the given option (d) is correct.
Question 4
1 / -0

The magnetic moment $$ ( \mu ) $$ of a revolving electron around the nucleus varies with principal quantum number $$n$$ as :

A
$$ \mu \propto 1/n $$

B
$$ \mu \propto 1/n^2 $$

C
$$ \mu \propto n $$

D
$$ \mu \propto n^2 $$

Solution

Orbital magnetic moments of an electron
$$ \mu_1 = \dfrac {neh} {4 \pi me} \therefore \mu_1 \propto n $$
Question 5
1 / -0

A. Ferromagnetic materials will lose their magnetism if heated above a point known as the Curie temperature.
B. You would have to re-magnetize the magnet again, either in a solenoid or with another permanent magnet, in order to restore the magnetism.
C. If you heat a magnet up a little bit, it will lose some of its magnetism, but on returning to room temperature full magnetism can be restored.
Which among the following is/are true?

A
Only A

B
Both A and B

C
Only C

D
All A, B and C

Solution

All the three statements are true that the ferromagnetic materials lose their magnetism on heating above the Curie temperature,
The magnet has to re-magnetized with another magnet or solenoid to restore its magnetism and the magnetism of the magnet is fully recovered when it is kept at room temperature after heating.
Question 6
1 / -0

A bar magnet is allowed to fall vertically through a copper coil placed in a horizontal plane. The magnet falls with a net acceleration

A
$$= g$$

B
Zero

C
$$ < g$$

D
$$> g$$

Solution

As the magnet will fall the flux through the coil will increase ,according to lenz law, the coil will induce current in order to oppose the flux ,That current will reduce the acceleration of magnet so $$a_{magnet}=g-a_{current}$$
Question 7
1 / -0

The magnetic moment of electron due to orbital motion is proportional to
($$n =$$ principal quantum number)

A
$$\dfrac {1}{n^{2}}$$

B
$$\dfrac {1}{n}$$

C
$$n^{2}$$

D
$$n$$

Solution

The magnetic moment of electron can be expressed as
$$\mu=\dfrac{-neh}{4\pi m_e}$$ ------ (1)
From the equation we can say that magnetic moment $$\mu$$ is directly proportional to the principal quantum number.
Question 8
1 / -0

0.8 J work is done in rotating a magnet by $$60^o$$, placed parallel to a uniform magnetic field. How much work is done in rotating it $$30^o$$ further?

A
$$0.8\times 10^7 erg$$

B
$$0.8 erg$$

C
$$8 J$$

D
$$0.4 J$$

Solution

Work done, $$W=MB (\cos {\theta}_1 - \cos {\theta}_2)$$
When the magnet is rotated from $$0^o$$ to $$60^o$$, then work done is 0.8 J
$$0.8 = MB(\cos 0^o - \cos 60^o) = \dfrac{MB}{2}$$
$$MB=0.8\times 2 = 1.6 N-m$$
In order to rotate the magnet through an angle of $$30^o$$, i.e., from $$60^o$$ to $$90^o$$, the work done is
$$W' = MB(\cos 60^o - \cos 90^o)$$
$$=MB(\dfrac{1}{2}-0)=\dfrac{MB}{2}$$
$$W'=\dfrac{1.6}{2}=0.8 J$$
$$=0.8\times 10^7 erg$$
Question 9
1 / -0

Curie-Weiss law is obeyed by iron.

A
At curie temperature only

B
At all temperatures

C
Below curie temperature

D
Above curie temperature

Solution

$$\textbf{Hint:}$$ Below curie temperature ferromagnetic substance behave as paramagnetic .

$$\textbf{Step 1-Curie -Weiss law}$$
The Curie-Weiss Law relates the susceptibility of ferromagnets like iron to their temperature.
$$\chi=\dfrac{C}{T-T_c}$$
Here C is the Curie constant and dependent only on the material.
and $$T_c$$ is Curie temperature.

$$\textbf{Step 2-Behaviour of iron}$$
Iron is a ferromagnetic material so it obeys curie-weiss law above curie temperature.
Below curie temperature ferromagnetic material changes to paramagnetic material and then it obeys curie law.
Question 10
1 / -0

Two identical coaxial coils $$P$$ and $$Q$$ carrying equal amount of current in the same direction are brought nearer. The current in

A
$$P$$ increases while in $$Q$$ decreases

B
$$Q$$ increases while in $$P$$ decreases

C
Both $$P$$ and $$Q$$ increases

D
Both $$P$$ and $$Q$$ decreases

E
Both $$P$$ and $$Q$$ remains constant

Solution

Explanation:
Lenz law:
$$\bullet$$ Lenz’s law states that the induced electromotive force with different polarities induces a current whose magnetic field opposes the change in magnetic flux through the loop in order to ensure that original flux is maintained through the loop when current flows in it.
$$\bullet$$When the coils P and Q are brought nearer the magnetic flux linked with each coil will increase and so the induced current will try to decrease the flux and hence current in both P and Q decreases.
Hence option $$D$$ is correct.

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Magnetism and Matter Test 23

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Magnetism and Matter Test 23

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Question 1

high-high

low-low

low-high

high-low

Solution

Question 2

I > III > II > IV

I > II > III > IV

I > IV > II > III

III > IV > I > II

Solution

Question 3

$$+ 10^{5}$$

$$+10^{-4}$$ to $$+10^{-2}$$

$$+10^{-5}$$

$$-10^{-5}$$

Solution

Question 4

$$ \mu \propto 1/n $$

$$ \mu \propto 1/n^2 $$

$$ \mu \propto n $$

$$ \mu \propto n^2 $$

Solution

Question 5

Only A

Both A and B

Only C

All A, B and C

Solution

Question 6

$$= g$$

Zero

$$ < g$$

$$> g$$

Solution

Question 7

$$\dfrac {1}{n^{2}}$$

$$\dfrac {1}{n}$$

$$n^{2}$$

$$n$$

Solution

Question 8

$$0.8\times 10^7 erg$$

$$0.8 erg$$

$$8 J$$

$$0.4 J$$

Solution

Question 9

At curie temperature only

At all temperatures

Below curie temperature

Above curie temperature

Solution

Question 10

$$P$$ increases while in $$Q$$ decreases

$$Q$$ increases while in $$P$$ decreases

Both $$P$$ and $$Q$$ increases

Both $$P$$ and $$Q$$ decreases

Both $$P$$ and $$Q$$ remains constant

Solution

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