Magnetism and Matter

Get insights from 91 questions on Magnetism and Matter, answered by students, alumni, and experts. You may also ask and answer any question you like about Magnetism and Matter

91
Questions
0
Discussions
0
Active Users
0
Followers

Have a question about Magnetism and Matter? Ask our experts

Ask Your Question

Keep it short & simple. Type complete word. Avoid abusive language.

New answer posted•a year ago

Magnetism and Matter Class 12th

5.25 The magnetic moment vectors and associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:

$μ_{s} = - (\frac{e}{m}) s \dots \dots . . (1)$

$μ_{l} = - (\frac{e}{2 m}) l$ …….(2)

Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

5.25 Of the two, the relation, the relation (2) is in accordance with classical physics. It follows easily from the definitions of

$μ_{l}$ = IA = $\frac{e}{T}) π r^{2}$ ( …. (i)

I = mvr = m (2 $π r^{2} / T)$ …… (ii)

where r is the radius of the circular orbit which the electron of mass m and charge (-e) completes in time T.

Dividing (i) by (i

Upvote

New answer posted•a year ago

Magnetism and Matter Class 12th

5.24 A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetizing current of 1.2 A?

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

5.24 Mean radius of a Rowland ring, r = 15 cm = 0.15 m

Number of turns on a ferromagnetic core, n = 3500

Relative permeability of core material, $μ_{r}$ = 800

Magnetizing current, I = 1.2 A

The magnetic field is given by the relation,

B = $\frac{4.5 \times 0.98 \times 4.2}{0.64 \times 2.8}$ , where $μ_{0}$ = Permeability of free space = 4 $π \times 10^{- 7}$ T m $A^{- 1}$

B = $\frac{800 \times 4 π \times 10^{- 7} \times 1.2 \times 3500}{2 π \times 0.15}$ = 4.48 T

Ther

Upvote

New answer posted•a year ago

Magnetism and Matter Class 12th

5.23 A sample of paramagnetic salt contains 2.0 × 10²⁴ atomic dipoles each of dipole moment 1.5 × 10^–23 J T^–1. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)

9 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

5.23 Number of atomic dipoles, n = 2 $\times 10^{24}$

Dipole moment of each atomic dipole, M = 1.5 $\times 10^{- 23}$ J/T

Magnetic field, $B_{1}$ = 0.64 T

The sample is cooled to the temperature, $T_{1}$ = 4.2 K

Total dipole moment of the atomic dipole, $M_{t o t}$ = n $\times$ M

= 2 $\times 10^{24}$ $\times 1.5 \times 10^{- 23} = 30 J / T$

Magnetic saturation is achieved at 15 %

Hence, effective dipole moment, = 15 %

Upvote

New answer posted•a year ago

Magnetism and Matter Class 12th

5.22 A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me = 9.11 × 10^–31kg).

[Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

20 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

5.22 Energy of an electron beam, E = 18 keV = 18 $\times 10^{3}$ eV

Charge of an electron, e = 1.6 $\times$ $10^{- 19}$ C

Total energy of the electron beam, $E_{T}$ = E $\times e$ = 18 $\times 10^{- 3}$ $\times$ 1.6 $\times$ $10^{- 19}$

Magnetic field, B = 0.04 G

Mass of an electron, $m_{e}$ = 9.11 $\times 10^{- 31}$ kg

Distance up to which the electron beam travels, d = 30 cm = 0.3 m

The kinetic energy of an el

Upvote

New answer posted•a year ago

Magnetism and Matter Class 12th

5.21 A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10^–2T. If the dipole comes $B_{2}$ to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field?

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

5.21 Magnitude of one of the magnetic fields, $B_{1}$ = 1.2 $\times 10^{- 2}$ T

Let the magnitude of other magnetic field be = $B_{2}$ $\times 10^{- 3}$

Angle between two fields, $θ$ = 60 $°$

At stable equilibrium, the angle between the dipole and the field , $B_{1}$ = $θ_{1}$ = $15 °$

Angle between the dipole and the field $B_{2}$ , $θ_{2}$ = $θ$ - $θ_{1}$ = 60 $°$ - $15 °$ = $45 °$

At rotational eq

Upvote

New question posted•a year ago

Magnetism and Matter Class 12th

5.21 A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10–2 T. If the dipole comes <math> <msub> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field?

3 Views|Posted a year ago

Asked by Payal Gupta

New answer posted•a year ago

Magnetism and Matter Class 12th

5.20 A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.

(a) Determine the horizontal component of the earth’s magnetic field at the location.

(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

5.20 Number of turns in the circular coil, n = 30

Radius of the circular coil, r = 12 cm = 0.12 m

Current in the coil, I = 0.35 A

Angle of dip, $β$ = 45 $°$

The magnetic field due to current I at a distance r is given by

B = $\frac{μ_{0} 2 π n I}{4 π r}$ , where $B_{1}$

$μ_{0}$ = Permeability of free space = 4 $π \times 10^{- 7}$ T m $A^{- 1}$

Then B = $\frac{4 π \times 10^{- 7} \times 2 π \times 30 \times 0.35}{4 π \times 0.12} = 5.50 \times 10^{- 5}$ T

The compass needle

Upvote

New answer posted•a year ago

Magnetism and Matter Class 12th

5.19 A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

12 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

5.19 Number of horizontal wires in the telephone cable, n = 4

Current in each wire, I = 1.0 A

The Earth’s magnetic field at a location, H = 0.39 G = 0.39 $\times 10^{- 4}$ T

Angle of dip at the location, $β$ = 35 $°$

Angle of declination, $θ \approx$ 0 $°$

For a point 4.0 cm below the cable, r = 4.0 cm = 0.04 m

The horizontal component

Upvote

New answer posted•a year ago

Magnetism and Matter Class 12th

5.18 A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

5.18 Current in the wire, I = 2.5 A

Angle of dip at the given location of earth, $β$ = 0 $°$

Earth’s magnetic field, H = 0.33 G = 0.33 $\times 10^{- 4}$ T

The horizontal component of earth’s magnetic field is given as:

$H_{H}$ = H $\cos ? β$ = 0.33 $\times 10^{- 4} \cos ? 0 °$ = 0.33 $\times 10^{- 4}$ T

The magnetic field at the neutral point at a distance R from the cable is given

Upvote

New answer posted•a year ago

Magnetism and Matter Class 12th

5.17 Answer the following questions:

(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetization curve of a ferromagnet.

(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetization, which piece will dissipate greater heat energy?

(c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.

(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?

(e) A certain region of space is to be shielded from magnetic fields. Suggest a method.

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

5.17 (a) The hysteresis curve of a ferromagnetic material is given below.

(b)The dissipated heat energy is directly proportional to the area of a hysteresis loop. A carbon steel piece has a greater hysteresis curve area. Hence, it dissipates greater heat energy.

(c) The value of magnetisation i

Upvote

Magnetism and Matter

Questions

Discussions

Active Users

Followers