**Design of Electromagnets MCQs ( Design Of Electrical Machines ) MCQs – ****Design Of Electrical Machines MCQs**

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# Types of Electromagnets

**1. How many types of electromagnets are present?****a) 2**

b) 3

c) 4

d) 5**Answer: a****Explanation: There are 2 types of electromagnets present. They are i) Tractive type and ii) Portative type.****2. What is the other name of the Tractive electromagnet and what is the means of movement of the armature?**

a) solenoidal, electrical movement**b) solenoidal, mechanical movement**

c) traction, electrical movement

d) traction, mechanical movement**Answer: b****Explanation: The other name for the tractive electromagnet is solenoids. They are designed to produce mechanical.****3. What is the supply given to the tractive electromagnets?**

a) only dc supply

b) only ac supply

c) ac and dc supply**d) ac or dc supply****Answer: d****Explanation: The other name for the tractive electromagnet is solenoids. The tractive electromagnets are operated either from ac or dc supply.****4. Among the following what are the applications of the tractive electromagnets?**

a) track switches

b) electric bells

c) buzzers**d) track switches, bells, buzzers****Answer: d****Explanation: The tractive electromagnets have a large number of applications. They are made use of in the track switches, electric bells and buzzers.****5. How does the portative electromagnet work as?****a) holding magnet**

b) connecting magnet

c) repulsion magnets

d) attraction magnets**Answer: a****Explanation: The second type of electromagnet is the portative electromagnet. They usually function as a holding magnet.****6. What type of supply is being provided to the portative electromagnet?**

a) only ac supply**b) only dc supply**

c) ac and dc supply

d) ac or dc supply**Answer: b****Explanation: Portative electromagnets are one type of electromagnet, which function as a holding magnet. They operated usually from dc supply only.****7. How many most commonly used electromagnets are present?**

a) 2**b) 3**

c) 4

d) 5**Answer: b****Explanation: There are 3 most commonly used electromagnets are present. They are I) Flat-faced armature type, II) Horse shoe type, III) Flat-faced plunger type.****8. What type of magnet is made use of to produce large force through a relatively small distance?****a) flat-faced armature type**

b) horse shoe type

c) flat-faced plunger type

d) flat-faced plunger type and horse shoe type**Answer: a****Explanation: There are 3 types of most commonly used electromagnets. The flat faced armature type electromagnet is made use of to produce large force through a relatively small distance.****9. What material is the flat faced armature type made of?**

a) hard steel**b) cast steel**

c) cast iron

d) soft steel**Answer: b****Explanation: The flat faced armature type is made up of cast steel. It is used for lifting scrap iron, sheet iron and iron castings.****10. How are the air gaps arranged in the flat faced armature type?****a) magnetic in series and mechanical in parallel**

b) magnetic in series and parallel

c) mechanical in series and parallel

d) magnetic in parallel and mechanical in series**Answer: a****Explanation: The flat faced armature types have 2 air gaps within them. They are magnetic in series and are mechanical in parallel and hence produce a holding surface of large effective area.****11. Which among the following are the application of portative electromagnets?**

a) lifting magnets

b) magnetic clutches

c) magnetic chucks**d) lifting magnets, magnetic clutches, magnetic chucks****Answer: d****Explanation: The portative electromagnets generally function as holding magnets. The lifting magnets, magnetic chucks, magnetic clutches are all applications of portative electromagnets.****12. What is the relation between force and the air gap length in the flat-faced armature type?**

a) force is directly proportional to the air gap length

b) force is indirectly proportional to the air gap length**c) force is directly proportional to the square of the air gap length**

d) force is indirectly proportional to the square of the air gap length**Answer: c****Explanation: The force is directly proportional to the square of the air gap length. This condition exists under ideal conditions wherein the effects of saturation and magnetic leakage are negligible.****13. Horse shoe is usually employed for the small magnets.****a) true**

b) false**Answer: a****Explanation: The horse shoe is usually employed for the small magnets. It is because of the mechanical adaptability and the ease with which it can be constructed.****14. How many air gaps are present in the flat-faced plunger type?****a) 1**

b) 2

c) 3

d) 4**Answer: a****Explanation: The magnetic circuit in the flat faced plunger type is usually short and heavy. It has only one air gap present.****15. What is the relation of the force and the air gap length in the flat faced plunger type?**

a) force is directly proportional to the air gap length

b) force is indirectly proportional to the air gap length**c) force is directly proportional to the square of the air gap length**

d) force is indirectly proportional to the square of the air gap length**Answer: c****Explanation: Force is directly proportional to the square of the air gap length. The characteristics are similar to that of the flat-faced armature type.**

# Construction of Electromagnets

**1. What type of core does the electromagnetic consist of?**

a) paramagnetic

b) diamagnetic**c) ferromagnetic**

d) paramagnetic and diamagnetic**Answer: c****Explanation: Electromagnets consists of a ferromagnetic core. It carries the flux and a winding which produces a flux when excited by an external source.****2. What material is used for the construction of core of electromagnets?****a) soft magnetic materials**

b) hard magnetic materials

c) non-magnetic materials

d) anti-ferromagnetic materials**Answer: a****Explanation: The electromagnets consist of ferromagnetic core. The soft magnetic materials are made use of in the construction of core of electromagnets.****3. What are the ferromagnetic elements used in the electromagnets?**

a) iron

b) nickel

c) cobalt**d) iron, nickel, cobalt****Answer: d****Explanation: The electromagnets consist of ferromagnetic core. The ferromagnetic materials include iron. nickel and cobalt.****4. What are the non-magnetic materials being used in the electromagnets?**

a) silicon

b) molybdenum**c) silicon, chromium, molybdenum**

d) chromium**Answer: c****Explanation: The electromagnets consists of the ferromagnetic core. Some times the non-ferro-magnetic materials are being made use of like silicon, chromium and molybdenum.**

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Design Of Current Transformers MCQs

**5. Coils are being made use of in electromagnets as an exciting source for production of magnetic field.****a) true**

b) false**Answer: a****Explanation: The coils are being made use of in electromagnets. They are used as an exciting source for the production of magnetic field.****6. What is the insulation material being used in the electromagnets?****a) paper**

b) wood

c) brass

d) copper**Answer: a****Explanation: Insulation material is being used in between the coils of the electromagnet to provide insulation. The material used in the electromagnet is paper.****7. What is the conductor material being used in the electromagnet?****a) copper**

b) zinc

c) bronze

d) aluminum**Answer: a****Explanation: Conductor materials are being used in the electromagnet to conduct the current in the electromagnet. Copper is the material used in the electromagnet.****8. What type of conductors are being used in the coils made of heavy wire?**

a) circular

b) rounded

c) conical**d) rectangular****Answer: d****Explanation: The cross-section of coils is generally rectangular and the cross-section of conductors is usually rounded. The coils made of heavy wires, rectangular conductors with rounded corners are used.****9. The coil insulation used in the electromagnets is of sheet form.****a) true**

b) false**Answer: a****Explanation: Insulation material used in the machine is usually paper. The insulation is arranged in the form of sheets.****10. What materials are used along with paper for insulation?**

a) varnish

b) glass

c) synthetic resin**d) varnish, glass, synthetic resin****Answer: d****Explanation: The insulation used in the electromagnets is paper. It is being insulated along with the varnish, glass and synthetic resin and treated into the form of sheets to form proper insulation.**

# Design of Magnet Coils

**1. What is the formula for the mean diameter of the magnet coils?****a) mean diameter = inside diameter of coil + outer diameter of coil / 2**

b) mean diameter = inside diameter of coil â€“ outer diameter of coil / 2

c) mean diameter = inside diameter of coil * outer diameter of coil / 2

d) mean diameter = inside diameter of coil / outer diameter of coil / 2**Answer: a****Explanation: First the inner diameter of coil is calculated. Secondly, the outer diameter of coil is calculated. On substitution, we finally get the mean diameter.****2. What is the formula for the outside diameter of the magnet coils?**

a) outside diameter = mean diameter + 2*depth of winding**b) outside diameter = mean diameter + depth of winding**

c) outside diameter = mean diameter â€“ 2*depth of winding

d) outside diameter = mean diameter â€“ depth of winding**Answer: b****Explanation: The mean diameter is found out from its respective formula. Next, the depth of the winding is calculated and on substitution gives the outside diameter.****3. What is the formula for depth of winding of the magnet coils?****a) depth of winding = mean diameter of coil â€“ inner diameter**

b) depth of winding = mean diameter of coil + inner diameter

c) depth of winding = mean diameter of coil â€“ 2* inner diameter

d) depth of winding = mean diameter of coil + 2*inner diameter**Answer: a****Explanation: The mean diameter of coil is calculated first from its respective formula. The inner diameter is next calculated and on substitution gives the depth of winding.****4. What is the formula of the cross winding area of the magnet coils?**

a) cross winding area = axial length of coil + depth of winding

b) cross winding area = axial length of coil â€“ depth of winding**c) cross winding area = axial length of coil * depth of winding**

d) cross winding area = axial length of coil / depth of winding**Answer: c****Explanation: First the axial length of coil is calculated. Next, the depth of winding is calculated and on substitution gives the cross winding area of the magnet coils.****5. What is the formula for the length of mean turn of magnet coils?****a) length of mean turns = 3.14 * (inside diameter of coil + depth of windings)**

b) length of mean turns = 3.14 / (inside diameter of coil + depth of windings)

c) length of mean turns = 3.14 * (inside diameter of coil * depth of windings)

d) length of mean turns = 3.14 + (inside diameter of coil + depth of windings)**Answer: a****Explanation: The inside diameter of the coil is first calculated. Next, the depth of windings is next calculated and on substitution gives the length of mean turns.****6. What is the formula for the total heat dissipating surface of the magnet coils?**

a) total heat dissipating surface = length of mean turn * depth of winding * axial length of coil

b) total heat dissipating surface = length of mean turn * depth of winding + axial length of coil**c) total heat dissipating surface = 2 * length of mean turn * (depth of winding + axial length of coil)**

d) total heat dissipating surface = 2 * length of mean turn * depth of winding * axial length of coil**Answer: c****Explanation: The length of mean turn is calculated first. Next, the depth of winding and axial length of coil is next calculated and on substitution gives the total heat dissipating surface.****7. What is the formula for the outer cylindrical heat dissipating surface of the magnet coils?**

a) outer cylindrical heat dissipating surface = 3.14 * outer diameter of coil + axial length of coil

b) outer cylindrical heat dissipating surface = 3.14 + outer diameter of coil + axial length of coil

c) outer cylindrical heat dissipating surface = 3.14 / outer diameter of coil + axial length of coil**d) outer cylindrical heat dissipating surface = 3.14 * outer diameter of coil * axial length of coil****Answer: d****Explanation: The outer diameter of the coil is first calculated. Next, the axial length of the coil is next calculated and on substitution gives the outer cylindrical heat dissipating surface of the magnet coils.****8. What is the formula of the inner cylindrical heat dissipating surface?**

a) inner cylindrical heat dissipating surface = length of mean turn * axial length of coil**b) inner cylindrical heat dissipating surface = 2 *length of mean turn * axial length of coil**

c) inner cylindrical heat dissipating surface = length of mean turn / axial length of coil

d) inner cylindrical heat dissipating surface =1 / length of mean turn * axial length of coil**Answer: b****Explanation: The length of mean turn is first calculated. Next, the axial length of coil is calculated and on substitution gives the inner cylindrical heat dissipating surface.****9. What is the ambient temperature of the magnet coils?**

a) 10Â°C

b) 15Â°C**c) 20Â°C**

d) 25Â°C**Answer: c****Explanation: The temperature is one of the factors which is used in the efficient operation of the magnet coils. The ambient temperature of the magnet coils is 20Â°C.****10. What is the formula for the area of the conductors of the magnet coils?**

a) area of the conductors = mmf per coil * resistivity of conductor * length of mean turn * terminal voltage

b) area of the conductors = mmf per coil / resistivity of conductor * length of mean turn * terminal voltage**c) area of the conductors = mmf per coil * resistivity of conductor * length of mean turn / terminal voltage**

d) area of the conductors = mmf per coil * resistivity of conductor / length of mean turn * terminal voltage**Answer: c****Explanation: For calculating the area of the conductors, first the mmf per coil is calculated along with the resistivity of conductors. The length of mean turn and terminal voltage is calculated and on substitution gives the area of the conductors.****11. What is the value of the resistivity temperature coefficient of copper?**

a) 0.017 ohm per m per mm^{2}

b) 0.0173 ohm per m per mm^{2}**c) 0.01734 ohm per m per mm ^{2}**

d) 0.0175 ohm per m per mm

^{2}

**Answer: c**

**Explanation: The resistivity temperature coefficient of copper is first calculated at a temperature of 20Â°C. The resistivity temperature coefficient of copper is 0.01734 ohm per m per mm**

^{2}.**12. What is the value of the resistance temperature coefficient of copper?**

**a) 0.00393 per Â°C**

b) 0.0040 per Â°C

c) 0.00383 per Â°C

d) 0.00373 per Â°C

**Answer: a**

**Explanation: The resistance temperature coefficient of copper is calculated at a temperature of 20Â°C. The resistance temperature coefficient of copper is 0.00393 per Â°C.**

**13. What is the formula for total number of turns in the magnet coils?**

a) total number of turns = mmf per coil * current

**b) total number of turns = mmf per coil / current**

c) total number of turns = mmf per coil â€“ current

d) total number of turns = mmf per coil + current

**Answer: b**

**Explanation: The mmf per coil is first calculated. Next, the current flowing through the coils is measured and on substitution gives the total number of turns.**

**14. What is the formula for the total winding area?**

a) total winding area = number of turns * area of each conductor * space factor

b) total winding area = number of turns / area of each conductor * space factor

**c) total winding area = number of turns * area of each conductor / space factor**

d) total winding area = 1/number of turns * area of each conductor * space factor

**Answer: c**

**Explanation: First the number of turns is calculated along with the area of each conductor. Next, the space factor is calculated and on substitution gives the total winding area.**