SSC JE Magnetic-Circuit Solved Questions (2018-2009)

Ques.71. Magnetic recording tape is most commonly made from _________ (SSC-2016)

  1. Ferric oxide
  2. The small particle of iron
  3. Silicon-iron
  4. Any of the above

Answer.1. Ferric oxide

Explanation:-

Magnetic tape is a thin strip of plastic, coated on one side with a film of tiny permanent magnets. Each of these permanent magnets has a north magnetic pole at one end and a south magnetic pole at the other. During recording, a strong magnetic field can swap the poles of these particles, altering the tape in a way that can be detected during playback.  Magnetic recording can either be analog, as in the audio recording of signals on magnetic tape or dismal recording, as used in the storage of information of data on magnetic disks and tapes for computer applications. 

There are three major types of magnetic solids, ferromagnetic, antiferromagnetic, and ferrimagnetic, distinguished by the ordering of their individual magnetic atoms. Iron, cobalt, nickel, and chromium dioxide are ferromagnetic solids, meaning that each atom’s magnetic dipole is aligned with those of its neighbors. Chromium is an antiferromagnetic solid, in which the atoms’ magnetic dipoles point in alternating or gradually varying directions and completely cancel one another. Ferrite (γ-iron oxide) and barium ferrite are ferrimagnetic solids, meaning that its atoms’ magnetic dipoles alternate back and forth but, because they have different strength, don’t cancel one another completely.

Permanent magnets

Both ferromagnetic and ferrimagnetic materials can be used to make permanent magnets. Common plastic sheet magnets usually contain barium ferrites, which are ferrimagnetic substances. Most plastic magnetic recording tapes, however, are coated with iron or other ferromagnetic materials. There isn’t any profound reason for these choices; they’re the consequences of technical and practical concerns.

Magnetic recording media use a different type of hard magnetic material—tiny elongated particles of the γ-iron oxide, chromium dioxide, cobalt-coated iron oxide, or pure iron. Each of these particles is so small that it contains only a single magnetic domain. It has a north pole at one end and a south pole at the other, because that arrangement permits the opposite poles of adjacent magnetic atoms to be as close together as possible and minimizes the particle’s total magnetic potential energy. With its magnetization pointing along with its cigar shape, the particle is in a stable magnetic equilibrium: if something disturbs its magnetization direction, it will experience restoring effects that return its magnetization to that equilibrium direction. The most widely used magnetic recording materials are γ-Fe2O3 (Gamma ferric oxide) and cobalt-doped γ-Fe2O3

Magnetic tape

Each of these particles is a tiny permanent magnet that can’t be demagnetized. It’s always magnetized in one direction or the other along its length. However, the particle’s magnetic poles can be interchanged by exposing it to a strong magnetic field. The particle itself won’t flip, just its magnetic poles. This behavior makes such a particle ideal for magnetic recording. A magnetic recorder can choose a particle’s magnetic orientation during recording and expect it to remain that way for an extremely long time.

 

Ques.72. The materials having low retentivity are suitable for making______ (SSC-2016)

  1. Permanent magnet
  2. Temporary Magnet
  3. Any of the above
  4. None of the above

Answer.2. Temporary Magnet

Explanation:-

The value of the intensity of magnetization of a material when the magnetizing field is reduced to zero is called retentivity or residual magnetism of the material. If the material has high retentivity it will remain magnetized in the absence of the magnetizing field.

For making temporary magnets the material should possess low retentivity such as electromagnets.

 

Ques.73. The relative permeability of a material is 0.95. The material is (SSC-2016)

  1. Diamagnetic
  2. Paramagnetic
  3. Ferromagnetic
  4. Ferrimagnetic

Answer.1. Diamagnetic

Explanation:-

Permeability is a measure of how easy it is to establish the flux in a material. Ferromagnetic materials have high permeability and hence low Reluctance, while nonmagnetic materials have low permeability and high Reluctance.

Diamagnetic materials are materials with relative permeabilities slightly smaller than 1 (μr < 1). This class includes important materials such as mercury, gold, silver, copper, lead, silicon, and water. The relative permeability of most diamagnetic materials varies between 0.9999 and 0.99999 (susceptibility varies between (-10-5 and -104), and for most applications, they may be assumed to be nonmagnetic.

 

 

Ques.74. A wire bent into a semi-circle in the center and straight at both ends is placed in a uniform magnetic field B pointing out of the page shown in the figure. If the wire carries the current I, the force on each straight section xy and zw is (SSC-2016)

  1. IB(Current × Magnetic field)
  2. IB(Length × Magnetic field)
  3. BIL
  4. Zero

Answer.3. BIL

Explanation:-

The magnetic force on a current

If a current is placed in a region of the magnetic field, it will experience a magnetic force. In Figure, a magnetic field is established out of the page and wire carries a current from left to right, perpendicular to the magnetic field. The magnitude of the force is proportional to the current I, the magnetic field magnitude B and the length L of the wire that is in the magnetic field.

current carrying condcutor

Mathematically

F ∝ BIL

F =kBIL

where k is a constant of proportionality. This constant can be made to equal one by proper choice of the unit of the magnetic field. We can make k = 1 by saying that when the force on 1 m of wire carrying a current of 1 A is 1 N. then the magnitude of the magnetic field is defined to be 1 Tesla (so 1 T = 1 N A−1 m−1). So the force on the current-carrying wire is

F =BIL

Remember, though, that the magnetic field we at right angles to the wire. If there is an angle between them then:

The force on a length L of the wire is given by

F = BIL sinθ

where θ is the angle between the current and the direction of the magnetic field.

 

Ques.75. The material used for the magnetic circuit where the high value of flux density required is (SSC-2016)

  1. Cast iron
  2. Soft steel
  3. Ferro Cobalt
  4. Gray cast iron

Answer.3. Ferro Cobalt

Explanation:-

Residual induction: Residual induction is the residual flux density remaining in a saturated magnetic material after the magnetizing force has been withdrawn. It represents the maximum flux output from the magnet that occurs at zero air gap.

or

The residual induction is any magnetic induction that remains in a magnetic material after removal of an applied magnetic field

Remanence: Remanence is defined as the magnetic induction that remains in a magnetic circuit after the withdrawal of an applied magnetizing force.

Coercive Force – The amount of reverse magnetic field must be applied to a magnetic material to make the magnetic flux return to zero.

Properties of the material of a permanent magnet :

(1) It should have high retentivity so that it remains magnetized in the absence of the magnetizing field.

(2) It should have a high saturation magnetization.

(3) It should have high coercivity so that it does not get demagnetized easily.

  • Hard magnetic materials have large hysteresis loss due to a large hysteresis loop area.
  • In these materials, the domain wall movement is difficult because of the presence of impurities and crystal imperfections and it is irreversible.
  • The coercivity and retentivity are large.

Hence, these materials cannot be easily magnetized and demagnetizer

Hard magnetic materials are used to produce permanent magnet. Hysteresis losses are of no significance here as no repeated reversals of magnetization is involved in a permanent magnet. The permanent magnets must have high residual induction B. and large coercive field Hc. The area of the hysteresis loop between Br and Hc represents the energy required to demagnetize a permanent magnet. The maximum value of this area (= BrHr) called the energy product, must be as large as possible for permanent magnets.

permanent magnet B H curve

As steel and alnico have high coercivity, therefore, they are used for making permanent magnets. The metals in Alnico magnets are indicated by the pairs of letters making up the name, Aluminum, Nickel, and Cobalt.

 

Ques 76. The magnitude of AT required to establish a given value of flux in the air gap will be much greater than that required for iron part of a magnetic circuit because (SSC-2015)

  1. Air in Gas
  2. Air is a good conductor of magnetic flux
  3. Air has the lowest relative permeability
  4. Iron has the lowest permeability

Answer.3. Air has the lowest relative permeability

Explanation:-

The reluctance of a uniform magnetic circuit can be calculated as:

R = l/μo μr A

R ∝ 1/μo μr

Where l = length

μo = permeability of vacuum, equal to 4π × 10−7 

μr  =  relative magnetic permeability of the material

A =  cross-sectional area of the circuit

The reluctance of air > Reluctance of iron

So air has the lowest relative permeability.

 

Ques 77. The unit for permeability is (SSC-2015)

  1. Wb/Atm
  2. At/m
  3. Wb
  4. At/Wb

Answer.1. Wb/Atm

Explanation:-

In electromagnetism, permeability is the measure of the ability of a material to support the formation of a magnetic field within itself. Hence, it is the degree of magnetization that a material obtains in response to an applied magnetic field.

The reluctance in the magnetic circuit

R = l/µA

Where l = length in meter

µ = Permeability of the material —————1

Also, Reluctance in the magnetic circuit is the ratio of m.m.f and flux and the SI unit is amp.turn/Weber……..2

From equation 1 permeability of the material is given as

µ = l/aR

= meter/(meter2 x amp.turns/Weber)

= wb/Atm

 

Ques 78. A bar of Iron 1 cm2 in cross-section has 10-4 wb of magnetic flux in it. If μr = 2000. What is the magnetic field intensity in the bar? (SSC-2015)

  1. 796 x  103 AT/m
  2. 398 AT/m
  3. 398 x  10-4 AT/m
  4. 398 x  104 AT/m

Answer.2. 398 AT/m

Explanation:-

Given area 1 cm2 = 10-4 m2

Flux Φ = 10-4 wb

Relative permeability μr = 2000

the magnetic permeability of free space μo = 4π × 10−7 N A−2

solution11

 

Ques 79. If the coefficient of coupling between two coils is increased, mutual inductance between the coils (SSC-2015)

  1. Changes depend on current only
  2. Increased
  3. Decreased
  4. Remain unchanged

Answer.2. Increased

Explanation:-

Mutual inductance between two coils are

M = K√L1L2

Where K is the coefficient of coupling and L1 & L2 are inductance between the coil

Therefore from the above equation, we can say that the mutual inductance is directly proportional to the coefficient of coupling i.e M ∝ K hence if the coefficient of coupling increased the mutual inductance will also increase.

 

Ques 80. The B-H curve is used to find the MMF of this section of the magnetic circuit. The section is: (SSC-2014)

  1. Vaccum
  2. Iron part
  3. Air gap
  4. Both iron part and air gap

Answer.2. Iron part

Explanation:-

The level of magnetic flux established in a ferromagnetic core is a direct function of the permeability of the material. Ferromagnetic materials have a very high level of permeability, while nonmagnetic materials such as air, vacuum, and would have very low levels of permeability.

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