Magnetism and Matter Test -8

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Magnetism and Matter Test -8

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Weekly Quiz Competition

Question 1
1 / -0

A cylinder of iron is placed so that it is free to rotate around its axis. Initially the cylinder is at rest, and a magnetic field is applied to the cylinder so that it is magnetized in a direction parallel to its axis. If the direction of the external field is suddenly reversed, the direction of magnetization will also reverse and the cylinder will begin rotating around its axis because

A
Conservation of energy

B
Conservation of parity

C
Conservation of mass

D
Conservation of angular momentum

Solution

Conservation of angular momentum and spin angular momentum is of the same nature as the angular momentum of rotating bodies
Question 2
1 / -0

A short bar magnet placed with its axis at 30 degree with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10⁻² J. What is the magnitude of magnetic moment of the magnet?

A
0.30 J/T

B
0.26 J/T

C
0.36 J/T

D
0.33 J/T

Solution

Torque, τ = mB sin θ, where m is magnetic moment, B is magnetic field and θ is angle between magnetic moment and magnetic field.
4.5 × 10⁻²= 0.25 × m × 0.5
m = 9/25 J/T = 0.36 J/T
Question 3
1 / -0

A closely wound solenoid of 800 turns and area of cross section 2.5 × 10⁻⁴m² carries a current of 3.0 A. What is its associated magnetic moment?

A
0.6 J/T

B
0.8 J/T

C
0.5 J/T

D
0.4 J/T

Solution

m = NIA = 800 × 3 × 2.5 × 10⁻⁴= 0.6 J/T
Question 4
1 / -0

A closely wound solenoid of 800 turns and area of cross section 2.5 × 10⁻⁴m² carries a current of 3.0 A.It is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied.Magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field

A
0.075J

B
0.080J

C
0.085J

D
0.070J

Solution

m = NIA = 0.6 J/T
torque = mBsinθ = 0.6 × 0.25 × 0.5 = 0.075J
Question 5
1 / -0

A bar magnet of magnetic moment 1.5 J/T lies perpendicular to the direction of a uniform magnetic field of 0.22 T. What is the torque acting on it?

A
0.23J

B
0.33 J

C
0.38J

D
0.43J

Solution

torque = mBsinθ
= 1.5 × .22 × 1 = 0.33J
Question 6
1 / -0

A bar magnet of magnetic moment 1.5 J/T lies parallel to the direction of a uniform magnetic field of 0.22 T. What is the torque acting on it?

A
-0.3J

B
0.3J

C
0.23J

D
0.0J

Solution

torque =mBsinθ
Since magnetic moment and magnetic field are parallel, angle between them is zero. Hence torque acting is zero.
Question 7
1 / -0

A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10⁻⁴m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. What is the magnetic moment associated with the solenoid?

A
1.18 Am²

B
1.38 Am²

C
1.08 Am²

D
1.28 Am²

Solution

m = NIA
=2000 × 1.6 × 10⁻⁴× 4
=1.28 Am²
Question 8
1 / -0

A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10⁻⁴m² , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. If a uniform horizontal magnetic field of 7.5 × 10⁻² T is set up at an angle of 3030 with the axis of the solenoid,force and torque on the solenoid are

A
0.0N,0.058 Nm

B
0.5N,0.078 Nm

C
0.0N,0.048 Nm

D
0.5N,0.068 Nm

Solution

m = NIA = 1.28 Am²
torque = mBsinθ = 1.28 × 7.5 × 10⁻²× 0.5 = 0.048 Nm
Since field is uniform, net force = zero
Question 9
1 / -0

How will the force between the two poles change if the two poles are connected by an iron rod keeping the distance between two magnetic poles constant ?

A
Increases

B
Decreases

C
Remains same

D
None of these

Solution

correct answer is option A
An iron rod placed between the poles of a magnet provides a better path than air for the magnetic flux. The flux density in iron is greater than it is in air; therefore, iron is said to have a high permeability. Hence the force between the poles increases. Also the Iron rod gets magnetised with north pole at the end in contact with south pole and south pole at the end in contact with the north pole of the bar magnet.
Question 10
1 / -0

Lenz's law is a consequence of the law of conservation of

A
$$energy$$

B
$$momentum$$

C
$$angular \ momentum$$

D
$$charge \ and \ mass$$

Solution

according to Lenz's law an induced electromotive force (emf) always gives rise to a current whose magnetic field opposes the original change in magnetic flux.
It is a common way of understanding how electromagnetic circuits obey newton's third law and the conservation of energy.
Question 11
1 / -0

A bar magnet for dipole moment $$10^4\ JT^{-1}$$ is free to rotate around a horizontal plane. A horizontal magnetic field $$4\times 10^{-5}\ T$$ exists on the space. Find the work done in rotating the magnet slowly from a direction parallel to the field to a direction $$60^0$$ from the field.

A
$$0.1\ J$$

B
$$0.2\ J$$

C
$$0.4\ J$$

D
$$0.5\ J$$

Solution

$$W=MB(\cos 0-\cos 60)$$
$$W=\dfrac{1}{2}MB$$
$$W=\dfrac{1}{2}\times4\times10^{-5}\times10^4$$
$$W=0.2\ J$$
Question 12
1 / -0

When the current through the electromagnet of a relay reaches a particular value

A
It breaks the circuit

B
It open the circuit by pulling in an iron contact

C
It closes the circuit by pulling in an iron contact

D
Both A or C

Solution

They are ways of switching using a low current to use an electromagnet to close or open a spring steel contact.They are often used to isolate a user from a high voltage that needs to be switched, using low current, low voltage to operate the relay, rather than having a user come in direct proximity to the higher voltage that needs to be switched.When the current through the electromagnet of a relay reaches a particular value it either breaks the circuit by repllening or closes the circuit by pulling in an iron contact.
Question 13
1 / -0

A current carrying wire in the neighbourhood produces -

A
No field

B
Electric field only

C
Magnetic field only

D
Electric and magnetic fields

Solution

Current means flow of electron in particular direction it does not charge the conductor hence outside wire electric field is zero and only magnetic field exists.
Question 14
1 / -0

A bar magnet has coercivity $$4\times10^3 Am^{-1}$$. It is placed in a solenoid having $$40$$ turns per cm and is $$1.2m$$ long. The current carried by the solenoid to demagnetize the bar magnet is :

A
$$6A$$

B
$$8A$$

C
$$5.78A$$

D
none of these

Solution

$$H=4*10^3A/m$$
$$B=H*l=4*10^3A/m*1.2=48*10^2A$$
$$\mu_0=4\pi*10^{-7} Tm/a$$
$$48*10^2=\mu_o NI=4\pi*10^{-7}*40*I$$
$$I=\dfrac{48*10^2}{4\pi*10^{-7}*40}=0.095*10^9A$$
Question 15
1 / -0

If a bar magnet in magnetic moment $$m$$ is deflected from an angle $$\theta$$ in a uniform magnetic field of induction $$B$$, the work done in reversing the direction is

A
$$MB\sin\theta$$

B
$$MB$$

C
$$MB(1-\cos\theta)$$

D
$$MB\cos\theta$$

Solution

Now lets assume that the bar magnetic is initially placed parallel to the field the initial potential energy will be $$U_i=-\overrightarrow M.\overrightarrow B=-MB\cos 0=-MB$$
Now when it is deflected by an angle $$\theta$$ then $$U_f=-\overrightarrow M.\overrightarrow B=-MB\cos \theta$$
Now we know that the work done = $$-\Delta U$$
$$\Delta U=U_i-U_f$$
$$\Delta U=-MB(1-\cos \theta)$$
therefore, work done $$=MB(1-\cos \theta)$$.
Question 16
1 / -0

What is the direction of magnetic field at the centre of a coil carrying clockwise current ?

A
No direction

B
normal to the axis of coil inwards.

C
Along the axis of coil inwards.

D
Along the axis of coil outwards.

Solution

A solenoid is a long coil of wire wrapped in many turns. The magnetic field within a solenoid depends upon the current and density of turns. Its direction depends on the direction of the current. When a current passes through it, it creates a nearly uniform magnetic field inside or along the axis of the solenoid. Outside the coil or solenoid, that is, at the ends and beyond the magnetic field is small and appears to diverge.
When the current flows through the solenoid in the clockwise direction, then the magnetic lines of force inside or center of the coil will be along the axis inwards according to Fleming's right hand rule.
Hence Magnetic field at the centre of a coil carrying a clockwise current will be along the axis of coil inwards.
Question 17
1 / -0

At Curie temperature the ferromagnetic materials get converted into.

A
non-magnetic material.

B
para magnetic material.

C
dia magnetic material.

D
all of the above.

Solution

Above curie temperature ferromagnetic material behaves like paramganetic materials. There magnetic susceptibilty starts varying linearly with temperature. Their $$I$$ becomes too small above curies temperature.
Therefore, nature of ferrromagnetic materials becomes paramagnetic.
Question 18
1 / -0

A straight wire lying in a horizontal plane carries a current from north to south. What will be the direction of magnetic field at a point just underneath it?

A
Towards north.

B
Towards east.

C
Towards west.

D
Towards south.

Solution

When electric current flows through a conductor, it creates a magnetic field around it. The magnetic field lines form concentric circles around the conductor. The direction of the magnetic field is perpendicular to the wire and is in the direction of wrapped fingers of the right hand holding the conductor and the thumb pointing upwards indicates the direction of the current. The direction of the magnetic field can be deduced by holding the current carrying conductor in the right hand with the thumb extended it will be pointing in the direction of the current flow from north to south.
The position of the fingers laid across the conductor will now point in the direction of the magnetic lines of force as shown in the figure.
Hence, the direction of the magnetic field will be towards east, just beneath the conductor.
If the direction of the current flowing through the conductor is reversed, the right hand will need to be placed onto the other side of the conductor with the thumb pointing in the new direction of the current flow.

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Magnetism and Matter Test -8

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Question 1

Conservation of energy

Conservation of parity

Conservation of mass

Conservation of angular momentum

Solution

Question 2

0.30 J/T

0.26 J/T

0.36 J/T

0.33 J/T

Solution

Question 3

0.6 J/T

0.8 J/T

0.5 J/T

0.4 J/T

Solution

Question 4

0.075J

0.080J

0.085J

0.070J

Solution

Question 5

0.23J

0.33 J

0.38J

0.43J

Solution

Question 6

-0.3J

0.3J

0.23J

0.0J

Solution

Question 7

1.18 Am2

1.38 Am2

1.08 Am2

1.28 Am2

Solution

Question 8

0.0N,0.058 Nm

0.5N,0.078 Nm

0.0N,0.048 Nm

0.5N,0.068 Nm

Solution

Question 9

Increases

Decreases

Remains same

None of these

Solution

Question 10

$$energy$$

$$momentum$$

$$angular \ momentum$$

$$charge \ and \ mass$$

Solution

Question 11

$$0.1\ J$$

$$0.2\ J$$

$$0.4\ J$$

$$0.5\ J$$

Solution

Question 12

It breaks the circuit

It open the circuit by pulling in an iron contact

It closes the circuit by pulling in an iron contact

Both A or C

Solution

1.18 Am²

1.38 Am²

1.08 Am²

1.28 Am²