Magnetism and Matter Test -15

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Magnetism and Matter Test -15

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Question 1
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β rays deflected in

A
Gravitational; field

B
Only in magnetic field

C
Only in electrical field

D
In magnetic and electrical field both

Solution

A magnetic field (only one pole is shown) affects radioactive rays differently depending on the type of ray.

Alpha rays (heavy, positively charged particles) are deflected slightly in one direction.

Beta rays (light, negatively charged electrons) are deflected strongly in the opposite direction.
Question 2
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Demagnetisation of magnets can be done by:

A
Rough handling

B
Heating

C
Magnetising in the opposite direction

D
All of above

Solution

1. Magnet can be demagnetised by heating past the Curie point, applying a strong oppsing magnetic field, applying alternating current, or hammering the metal.

2. By dropping it from height several times. Hammering it, passing electric current, forcibly keeping it near to the like poles of other strong magnets.
Question 3
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If a diamagnetic substance is brought near north or South Pole of a bar magnet is:

A
Attracted by the poles

B
Repelled by the pole

C
Repelled by the North Pole and attracted by the South Pole

D
Attracted by the North Pole and repelled by the South Pole

Solution

Diamagnetic materials create an induced magnetic field in a direction opposite to an externally applied magnetic field, and are repelled by the applied magnetic field.

Hence, if a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is repelled by the both the poles.
Question 4
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The only property possessed by ferromagnetic substances is

A
Hysteresis

B
Magnetic Moment

C
Directional property

D
Attracting magnetic substance

Solution

Hysteresis Effect.

The magnetization of ferromagnetic substances due to a varying magnetic field lags behind the field. This effect is called hysteresis, and the term is used to describe any system in whose response depends not only on its current state, but also upon its past history.
Question 5
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Substance in which the magnetic moment of a single atom is not 0 is known as

A
Diamagnetism

B
Ferromagnetism

C
Paramagnetism

D
Ferrimagnetisms

Solution

The property of Paramagnetism is found in those substances whose atoms have an excess of electrons spinning in the same direction. Hence atoms of paramagnetic substances have a net non-zero magnetic moment of their own.
Question 6
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The magnetic moment of diamagnetic atom is

A
Much greater than one

B
1

C
Between 0 and 1

D
Equal to 0

Solution

Atoms that have paired electrons have magnetic moment zero. Because diamagnetism is the intrinsic property of every material and it is generated due to the mutual interaction between the applied magnetic field and orbital motion of electrons.

So, the magnetic moment of a diamagnetic atom is equal to zero.
Question 7
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A superconductor exhibits perfect

A
Ferrimagnetisms

B
Ferromagnetism

C
Paramagnetism

D
Diamagnetism

Solution

Diamagnetism is a quantum mechanical effect that occurs in all materials; when it is the only contribution to the magnetism, the material is called diamagnetic. A superconductor exhibits perfect Diamagnetism.
Question 8
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Which one of the following is non-magnetic substance?

A
Iron

B
Cobalt

C
Nickel

D
Glass

Solution

The substances which are not attracted by a magnet are called non-magnetic substances.

For example, wood, paper, water, copper, glass, gold, silver, etc., are non-magnetic substances.
Question 9
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At time $$t = 0$$ magnetic field of $$1000$$ Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to $$500$$ Gauss, in nest $$5s$$, then induced EMF in the loop is:

A
$$36\mu V$$

B
$$28\mu V$$

C
$$56\mu V$$

D
$$48\mu V$$

Solution

Area of loop $$AEBCFDA =$$ area of rectangle $$ABCD \ -$$ Area of triangle $$ABE\ -$$ Area of triangle $$DCF$$

$$\therefore$$ Required area $$=16cm \times 4 cm - \dfrac{1}{2}\times 4cm \times 2cm - \dfrac{1}{2}\times 4cm \times 2cm$$

$$= 64cm^2 - 4cm^2 - 4cm^2 = 56cm^2$$

Magnetic flux passing through an area is given by

$$\displaystyle \phi = \int B.dA = \int BdA \cos \theta$$.

Here, $$\theta = 0$$ and $$B$$ is constant throughout the area.
$$\therefore \phi_{initial} = B_{initial}.A = 1000 \, Gauss \times 56cm^2$$

$$= 56000 \, gauss\, cm^2$$

$$= 5.6\times 10^{-4}Wb$$

$$\begin{bmatrix} 1 \, Gauss = 10^{-4}T \, and \, 1cm^2 = 10^{-4}m^2\\ \therefore 1\, Gauss\, cm^2 = 10^{-8}Tm^2 = 10^{-8}Wb \end{bmatrix}$$

Also, $$\phi _{final} = B_{final}. A = 500\, Gauss \times 56cm^2$$

$$= 28000\, Gauss\, cm^2 = 2.8\times 10^{-4}Wb$$

Since the flux changes linearly,
$$\in = \left|\dfrac{d\phi}{dt}\right| = \dfrac{|\phi_{final} - \phi_{initial}|}{\Delta t} = \dfrac{(5.6-2.8)\times 10^{-4}}{5}V$$

$$= 56\times 10^{-6}V = 56\mu v$$
Question 10
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The figure gives experimentally measured B vs. H variation in a ferromagnetic material. The retentivity, co-ercivity and saturation, respectively, of the material are :

A
$$1.5 T, 50 A/m $$ and $$1.0T$$

B
$$150 A/m, 1.0 T$$ and $$1.5 T$$

C
$$1.5 T, 50 A/m$$ and $$1.0 T$$

D
$$1.0 T, 50 A/m$$ and $$1.5 T$$

Solution

The retentivity measures the induced magnetic field at which applied field is zero.

So, Retentivity $$= OA = 1.0T$$

Co-ercivity is the resistance of a ferromagnetic material to become demagnetized $$= OC = 50 A/m$$.
$$\therefore $$ Co-ercivity $$= 50 A/m$$

At saturation magnetization; graph becomes horizontal
So, saturation magnetization $$= M_s = 1.5 T$$

option (D) is correct.
Question 11
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The corecivity of a small magnet where the ferromagnet gets demagnetized is $$3 \times 10^3 Am^{-1}$$. The current required to be passed in a solenoid of length $$10 cm$$ and number of turns $$100$$, so that the magnet gets demagnetized when inside the solenoid is:

A
$$3A$$

B
$$6A$$

C
$$30 mA$$

D
$$60 mA$$

Solution

$$\mu_o H = \mu_o n i$$
$$n = \dfrac{N}{L} = \dfrac{100}{0.1}$$
$$3 \times 10^3 = \dfrac{100}{0.1} i$$
$$i= 3A $$
Question 12
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A cylindrical conducting rod is kept with its axis along positive z-axis, where a uniform magnetic field exists parallel to z-axis. The current induced in the cylinder is

A
zero

B
clockwise as seen from +z axis

C
anti-clockwise as seen from +z axis

D
opposite to the direction of magnetic field.

Solution

The induced emf $$e$$ in a conducting rod of length $$l$$, moving with a velocity $$v$$, in a uniform magnetic field, is given by,
$$e=Bvl\sin\theta$$,
where $$\theta=$$ angle between $$\vec v$$ and $$\vec B$$,
in the given problem, as the rod is static i.e. $$v=0$$ and parallel to the magnetic field $$\theta=0$$ therefore induced emf in rod will be, $$e=0$$,
hence, $$i=0$$ (induced current)
Question 13
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A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III & IV, arrange them in the decreasing order of Potential Energy.

A
$$|>|||>||>$$

B
$$|>||>|||>$$

C
$$|>|\mathrm{V}>||>$$ III

D
$$|||>|\mathrm{V}>|>$$ II

Solution
Question 14
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The magnetic susceptibility is negative for:

A
diamagnetic material only

B
paramagnetic material only

C
ferromagnetic material only

D
paramagnetic and ferromagnetic materials

Solution

Hint: Magnetic susceptibility is a proportionality constant that determines the degree of magnetization of a material with respect to the applied magnetic field.

Solution:

Step1: The magnetic susceptibility denotes whether a material is repelled out or attracted of a magnetic field. It is negative only for diamagnetic material and it is positive for paramagnetic material and ferromagnetic material has large positive magnetic susceptibility. Hence option ‘A’ is correct.
Question 15
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The magnetic moment of a diamagnetic atom is

A
much greater than one

B
one

C
between zero and one

D
equal to zero

Solution

The magnetic momentum of a diamagnetic atom is equal to zero.
Question 16
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A short bar magnet of magnetic moment $$0.4 J {T}^{-1}$$ is place in a uniform magnetic field of $$0.16 T$$. The magnet is stable equilibrium when the potential energy is

A
$$-0.064 J$$

B
Zero

C
$$-0.082 J$$

D
$$0.064$$

Solution

For stable equilibrium
$$U = -MB$$
$$= -\left(0.4\right) \left(0.16\right)$$
$$= -0.064 J$$.
Question 17
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A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. the energy required to rotate it by $$60^o$$ is $$W$$, Now the torque required to keep the magnet in this new position is.

A
$$\dfrac{2W}{\sqrt{3}}$$

B
$$\dfrac{W}{\sqrt{3}}$$

C
$$\sqrt{3} W$$

D
$$\dfrac{\sqrt{3}}{2} W$$

Solution

Hint:-Use potential energy formula to calculate the energy required to rotate the magnet and hence calculate torque.

Step 1: Calculate energy required to rotate magnet by $$60^{\circ}$$.
Energy required or work done in moving from $$\theta=0^{\circ}$$ to $$\theta=60^{\circ}$$ is given by
$$W=-mBcos 60^{\circ}-(- mBcos 0)=\dfrac{1}{2}mB$$

Step 2: Calculate the torque required to keep the magnet in the new position.
The torque acting on a bar magnet kept in a magnetic field $$\vec{B}$$ is given by
$$\tau=\vec{m}\times \vec{B}=mBsin\theta$$

$$\tau=mBsin60^{\circ}$$
$$\tau=\dfrac{\sqrt3}{2}mB=\sqrt3W$$

$$\textbf{Hence option C correct}$$
Question 18
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A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from.

A
The lattice structure of the material of the rod

B
The current source

C
The induced electric field due to the changing magnetic field

D
The magnetic field

Solution

Here when electromagnet is connected to battery, emf source the magnetic field is produced which pushes rod upwards and so gravitational potential energy is stored.
This stored energy will require work which will come from current source or battery which is connected to electromagnet.
So option 2 is correct answer