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H Harsh Kankaria
We know,    is in expected from classical physics. Now, Magnetic moment associated with the orbital motion of the electron is:  = Current x Area covered by orbit = I x A =  And, l = angular momentum = mvr = (m is the mass of the electron having charge (-e), r is the radius of the orbit of by the electron around the nucleus and T is the time period.) Dividing these two equations:     , which is...

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H Harsh Kankaria
Given, Radius of ring, r = 15cm = 0.15m Number of turns in the ring, n = 3500 Relative permeability of the ferromagnetic core,  = 800 Current in the Rowland ring, I = 1.2A We know, Magnetic Field due to a circular coil, B  =  ∴ B =  = 4.48T Therefore, the magnetic field B in the core for a magnetising current is 4.48 T

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H Harsh Kankaria
Given, Magnetic field,  = 0.64 T Temperature,  = 4.2K And, saturation = 15% Hence, Effective dipole moment,  = 15% of Total dipole moment   = 0.15 x (no. of atomic dipole × individual dipole moment)   = = 4.5 Now,  Magnetic field,  = 0.98 T and Temperature,  = 2.8 K Let be the new dipole moment. We know that according to Curie’s Law,  ∴ The ratio of magnetic dipole moments    Therefore, the...

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D Devendra Khairwa
The energy of electron beam   =   18 eV                    We can write:-                                                            so                                                      We are given horizontal magentic field :          B  = 0.40 G  Also,                                                                                          We obtain,                                       ...

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H Harsh Kankaria
Given, The magnitude of the first magnetic field, B1 = 1.2 × 10–2 T The angle between the magnetic field directions,  = 60° The angle between the dipole and the magnetic field  is  = 15° Let B2 be the magnitude of the second magnetic field and M be the magnetic dipole moment Therefore, the angle between the dipole and the magnetic field B2 is  = = 45° Now, at rotational equilibrium, The torque...

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H Harsh Kankaria
Given, Number of turns in the coil, n = 30 Radius of coil, r = 12cm = 0.12m Current in the coil, I = 0.35A The angle of dip,  = 45o We know, Magnetic fields due to current carrying coils, B =      Now, Horizontal component of the earth’s magnetic field = Bsin   (Hint: Take sin45o as 0.7)
Number of long straight horizontal wires = 4 The current carried by each wire = 1A  earth’s magnetic field at the place = 0.39 G the angle of dip = 350 magnetic field due to infinite current-carrying straight wire r=4cm =0.04 m magnetic field due to such 4 wires The horizontal component of the earth's magnetic field the horizontal component of the earth's magnetic field At the point below...

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H Harsh Kankaria
Given, Current in the cable, I = 2.5 A Earth’s magnetic field at the location, H = 0.33 G = 0.33 × 10-4T The angle of dip,  = 0 Let the distance of the line of the neutral point from the horizontal cable = r m. The magnetic field at the neutral point due to current carrying cable is:  , We know, Horizontal component of earth’s magnetic field,  = Also, at neutral points,   ⇒ =     Required...

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H Harsh Kankaria
The region can be surrounded by a coil made of soft iron to shield from magnetic fields.

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H Harsh Kankaria
Ceramic, a ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer.

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H Harsh Kankaria
Ferromagnets have a record of memory of the magnetisation cycle. Hence it can be used to store memories. 

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H Harsh Kankaria
Material that has a greater area of hysteresis loop will dissipate more heat energy. Hence after going through repeated cycles of magnetization, a carbon steel piece dissipates greater heat energy than a soft iron piece, as the carbon steel piece has a greater hysteresis curve area.

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H Harsh Kankaria
According to the graph between B (external magnetic field) and H (magnetic intensity) in ferromagnetic materials, magnetization persists even when the external field is removed. This shows the irreversibility of magnetization in a ferromagnet.

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H Harsh Kankaria
Yes, the maximum possible magnetisation of a paramagnetic sample will be of the same order of magnitude as the magnetisation of a ferromagnet for very strong magnetic fields.

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H Harsh Kankaria
Since the permeability of ferromagnetic material is always greater than one, the magnetic field lines are always nearly normal to the surface of ferromagnetic materials at every point. 

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H Harsh Kankaria
We know that the permeability of ferromagnetic materials is inversely proportional to the applied magnetic field. Therefore it is more for a lower field.

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H Harsh Kankaria
A toroid using bismuth for its core will have slightly less magnetic field than a toroid with an empty core because bismuth is a diamagnetic substance.

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D Devendra Khairwa
Magnetic moment M is    Earth's field = 0.42 G. The magnetic field due to magnet at equitorial line is given by :-                                                 or                                                   or                                                                                                                      

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H Harsh Kankaria
The magnetism in a diamagnetic substance is due to induced dipole moment. So the random thermal motion of the atoms does not affect it which is dependent on temperature. Hence diamagnetism is almost independent of temperature.
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