Magnetism and Matter Test 21

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Magnetism and Matter Test 21

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Weekly Quiz Competition

Question 1
1 / -0

Electricity and magnetism are mutually

A
Complementary

B
Opposite

C
Equal and opposite

D
Strongest forces

Solution

Electricity and magnetism are complementary to each other.

Changing magnetic field produces electric current that is it produces electric field. And moving charges that is changing electric field produces magnetic effect like moving charges produces magnetic field.

Answer-(A)
Question 2
1 / -0

An electron in a circular orbit of radius .05 nm performs $$10^{16}$$ revolutions per second. The magnetic moment due to this rotation of electron is (in $$Am^2$$)

A
$$2.16\times 10^{-23}$$

B
$$3.21 \times 10^{-22}$$

C
$$3.21 \times 10^{-24}$$

D
$$1.26\times 10^{-23}$$

Solution

The magnetic moment of an electron revolving in a circular path is $$iA=\dfrac{q}{T}A$$
$$=q\nu \pi r^2$$
$$=1.6\times 10^{-19}\times 10^{16}\times 3.14\times (0.05\times 10^{-9})^2$$
$$=1.26\times 10^{-23}Am^2$$
Question 3
1 / -0

In a current carrying loop of area $$A$$ a current $$I$$ is flowing. loop is placed inside a magnetic field H in upward direction. Find the potential energy of current loop inside the magnetic filed.

A
$$I\times A\times H$$

B
$$I^2\times A \times H$$

C
$$\dfrac{I}{A} \times H$$

D
None of these

Solution

Magnetic moment of the loop $$\mu = I\times A$$
Potential energy of the loop inside the magnetic field $$P.E = \mu\times H$$
$$\implies$$ $$P.E = I \times A\times H$$
Question 4
1 / -0

The magnetic moment of a current carrying loop is $$\mu =6.25\times { 10 }^{ -4 } A{ m }^{ 2 }$$ and a magnetic field of value $$0.4 \ T$$ is applied in upward direction perpendicular to the plane of loop.Find the potential energy of loop in magnetic field in joules. (The direction of current in loop is in clockwise direction.)

A
$$-1.25\times { 10 }^{ -4 }\quad J$$

B
$$-2.51\times { 10 }^{ -4 }\quad J$$

C
$$2.51\times { 10 }^{ -4 }\quad J$$

D
$$1.25\times { 10 }^{ -4 }\quad J$$

Solution

As the direction of the current in the loop is clockwise, thus magnetic moment points in the downward direction.
$$\therefore$$ Magnetic moment $$\mu = 6.25\times 10^{-4} Am^2$$ (downwards)
Magnetic field $$B = 0.4 T$$ (upwards)
Potential energy of loop $$P.E = -\mu . B$$
As $$\mu$$ and $$B$$ are anti-parallel, thus $$\mu.B = -\mu B$$
$$\implies$$ $$P.E = \mu B$$
$$\therefore$$ $$P.E = 6.25\times 10^{-4} \times 0.4 = 2.5\times 10^{-4}$$ $$J$$
Question 5
1 / -0

Two current carrying loops having $$N_1$$ turns and $$N_2$$ turns respectively both carrying a current equal to $$I$$ in the same direction, are placed inside a magnetic field $$B$$. If the radii of both loops are in the ratio $$1: 3$$ then what will be the ratio of the potential energy of loops in that magnetic field?

A
$$\dfrac{N_1}{N_2}$$

B
$$\dfrac{2N_1}{N_2}$$

C
$$\dfrac{N_1}{3N_2}$$

D
$$\dfrac{N_1}{9N_2}$$

Solution

Given : $$\dfrac{R_1}{R_2} = \dfrac{1}{3}$$
Magnetic moment of the loop $$\mu = N I A$$
where $$A = \pi R^2$$
$$\therefore$$ $$\mu = \pi N I R^2$$
Potential energy of loop in the magnetic field $$P.E = \mu B$$
$$\therefore$$ $$P.E =\pi NIR^2 B$$
We get $$P.E \propto NR^2$$
$$\implies$$ $$\dfrac{P.E_1}{P.E_2} = \dfrac{N_1}{N_2}\dfrac{R_1^2}{R_2^2}$$
Or $$\dfrac{P.E_1}{P.E_2} = \dfrac{N_1}{N_2}\dfrac{1}{9} = \dfrac{N_1}{9N_2}$$
Question 6
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When a superconductor is placed in a weak external magnetic field H, and cooled below its transition temperature,

A
the magnetic field is ejected

B
the magnetic field get magnetized

C
the magnetic field strength increased

D
none of the above

Solution

When a superconductor is placed in a weak external magnetic field $$H$$, and cooled below its transition temperature, the magnetic field is ejected from it and this phenomenon is called effect Meissner effect. The ability of this expulsion effect of the magnetic field is determined by the nature of equilibrium formed by the neutralization within the unit cell of a superconductor.
Question 7
1 / -0

Superconducting metal in superconducting state relative permeability of

A
zero

B
one

C
negative

D
more than one

Solution

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Question 8
1 / -0

A current carrying loop of radius $$2 \ cm$$ has $$4 A$$ current flowing in an anti-clockwise direction. The plane of the loop makes an angle of $$\theta$$ with the direction of the magnetic field which is in an upward direction. the value of the magnetic field is $$0.7 \ Tesla$$.if potential energy of loop in magnetic field is $$\approx 0.01 \ J$$, then value of $$\theta $$ is:

A
$$45^0$$

B
$$60^0$$

C
$$30^0$$

D
None of these

Solution

Potential energy $$=\mu B=\mu B\cos { \theta } $$ where $$\mu=$$magnetic moment
$$\mu =iA$$ where $$A$$= Area of loop
$${ p }_{ E }=iA\cos { \theta } \times B=0.01$$
$$0.7\times 4\times \pi \times 4\times { 10 }^{ -4 }\cos { \theta } =0.01$$
$$\\ \cos { \theta } =\dfrac { 0.01\times { 10 }^{ -4 } }{ 16\pi \times 0.7 } =\dfrac { 100 }{ 50\times 0.7 } =\dfrac { 2 }{ 0.7 } $$
$$\cos { \theta } >1\quad $$ is not possible.
So option C is correct.
Question 9
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If the magnetizing field on a ferromagnetic material is increased, its permeability

A
Decreased

B
Increased

C
Is unaffected

D
May be increased or decreased

Solution

The magnetic permeability of a substance is defined as
$$\mu=\dfrac{B}{H}$$
where $$B$$ is the established magnetic field inside the material,
$$H$$ is the applied external magnetic field.
When the external applied magnetic field $$H$$ is increased, the established magnetic field for a ferromagnetic material remains the same. Thus the magnetic permeability decreases.
Question 10
1 / -0

A charged particle $$(charge = q; mass = m)$$ is rotating in a circle of radius $$'R'$$ with uniform speed $$'V'$$. Ratio of its magnetic moment $$(\mu)$$ to the angular momentum $$(L)$$ is

A
$$\dfrac {q}{2m}$$

B
$$\dfrac {q}{m}$$

C
$$\dfrac {q}{4m}$$

D
$$\dfrac {2q}{m}$$

Solution

Period of revolution of charged particle $$q$$ ,
$$T=\dfrac{\text{circumference}}{\text{ velocity}}$$
or $$T=2\pi R/v$$
Circulating current , $$I=q/T=q/(2\pi R/v)=qv/2\pi R$$
Therefore magnetic moment of charged particle ,
$$M=IA$$ (where $$A= $$area of circular path)
$$M=(qv/2\pi R)(\pi R^{2})=(qvR/2)$$ ..........eq1
Now , angular momentum of particle of mass $$m$$ ,
$$L=mvR$$ .............eq2
Dividivg eq1 by eq2 ,
$$\dfrac{M}{L}=\dfrac{qvR/2}{mvR}=q/2m$$

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Magnetism and Matter Test 21

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Question 1

Complementary

Opposite

Equal and opposite

Strongest forces

Solution

Question 2

$$2.16\times 10^{-23}$$

$$3.21 \times 10^{-22}$$

$$3.21 \times 10^{-24}$$

$$1.26\times 10^{-23}$$

Solution

Question 3

$$I\times A\times H$$

$$I^2\times A \times H$$

$$\dfrac{I}{A} \times H$$

None of these

Solution

Question 4

$$-1.25\times { 10 }^{ -4 }\quad J$$

$$-2.51\times { 10 }^{ -4 }\quad J$$

$$2.51\times { 10 }^{ -4 }\quad J$$

$$1.25\times { 10 }^{ -4 }\quad J$$

Solution

Question 5

$$\dfrac{N_1}{N_2}$$

$$\dfrac{2N_1}{N_2}$$

$$\dfrac{N_1}{3N_2}$$

$$\dfrac{N_1}{9N_2}$$

Solution

Question 6

the magnetic field is ejected

the magnetic field get magnetized

the magnetic field strength increased

none of the above

Solution

Question 7

zero

one

negative

more than one

Solution

Question 8

$$45^0$$

$$60^0$$

$$30^0$$

None of these

Solution

Question 9

Decreased

Increased

Is unaffected

May be increased or decreased

Solution

Question 10

$$\dfrac {q}{2m}$$

$$\dfrac {q}{m}$$

$$\dfrac {q}{4m}$$

$$\dfrac {2q}{m}$$

Solution

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