816 + Interview Questions in MAGNETISM IN GENERAL AWARENESS Page 3 InterviewSolution

101.	Analogy:1. Natural magnet: Magnetic stones. Artificial magnet: _______2. Magnetic substance : Attracted by magnets Non magnetic substance : _____3. Repel to each other: Like poles. Attract to each other : ______4. Demagnetisation : _________ Making magnets: Rubbing with one end to another end without changing direction.5 Electromagnet: Magnetic crane. Ordinary magnet: _______
Answer» 1. Bar magnet 2. Not attracted by magnets 3. Unlike poles 4. Hit with hammer 5. Mobile phone covers.

101.

Analogy:1. Natural magnet: Magnetic stones. Artificial magnet: _____2. Magnetic substance : Attracted by magnets Non magnetic substance : _3. Repel to each other: Like poles. Attract to each other : 4. Demagnetisation : _ Making magnets: Rubbing with one end to another end without changing direction.5 Electromagnet: Magnetic crane. Ordinary magnet: ___

Answer»

1. Bar magnet

2. Not attracted by magnets

3. Unlike poles

4. Hit with hammer

5. Mobile phone covers.

Discussion

102.	Distinguish between natural and artificial magnets.
Answer» 1. Natural Magnets: These are found in nature. Have irregular shapes and dimensions. The strength of a natural magnet is well determined and difficult to change. These are permanent magnets. They have a less usage. 2. Artificial Magnets: These are man-made magnets. They can be made in different shapes and dimensions. Artificial magnets can be made with required and specific strength. Their properties are time bound. They have a vast usage in day-to-day life.

102.

Distinguish between natural and artificial magnets.

Answer»

1. Natural Magnets:

These are found in nature.
Have irregular shapes and dimensions.
The strength of a natural magnet is well determined and difficult to change.
These are permanent magnets.
They have a less usage.

2. Artificial Magnets:

These are man-made magnets.
They can be made in different shapes and dimensions.
Artificial magnets can be made with required and specific strength.
Their properties are time bound.
They have a vast usage in day-to-day life.

Discussion

103.	Match the following :I.Natural magnets(a)Levitating propellerII.Artificial magnets(b)Wooden spoonIII.Magnetic substance(c)HeatingIv.Non magnetic substance(d)Bar magnetv.Demagnetization(e)Pin holdersvI.Demagnetizaiton(f)Magnetite
Answer» I – f II – d III – e IV – b V – c VI – a.

103.

Match the following :I.Natural magnets(a)Levitating propellerII.Artificial magnets(b)Wooden spoonIII.Magnetic substance(c)HeatingIv.Non magnetic substance(d)Bar magnetv.Demagnetization(e)Pin holdersvI.Demagnetizaiton(f)Magnetite

Answer»

I – f

II – d

III – e

IV – b

V – c

VI – a.

Discussion

104.	Why are the artificial magnets preferred over the natural magnets ?
Answer» Artificial magnets are preferred over natural magnets because natural magnets are weak and often irregular in shape, they can readily be magnetised and demagnetised by turning the current on or off in the coil.

Discussion

105.	A magnet is placed in iron powder and then taken out, then maximum iron powder is atA. Some away from north poleB. Some away from south poleC. The middle of the magnetD. The end of the magnet
Answer» Correct Answer - D

Discussion

106.	A magnet is placed in iron poweder and the taken out , them maximum iron powder is atA. some distance away from north poleB. some distance away from south poleC. the middle of the magnetD. the ends of the magnet
Answer» Correct Answer - D Magnetic pole strength is stronger at end part of magnet , so maximum iron powder is collected at the end points of magnet .

Discussion

107.	Asseriton (A): A magnetic suspended freely in an uniform magnetic field experiences no net force, but a torque that tends to algn the magnet along the field when it is deflected form equilibrium position Reason (R ): Net force `mB-mB-0`, but the forces on north and south poles being equal, unlike and parallel make up a couple that tends to align the magnet, along the field.A. Both `A` and `R` are true and `R` is the correct explanation of `A`.B. Both `A` and `R` are true and `R` is not correct explanation of `A`.C. `A` is true, But `R` is falseD. `A` is false, But `R` is true
Answer» Correct Answer - A

Discussion

108.	A bar magnet is suspended in a uniform magnetic field in a position such that it experiences maxiumum torque. The angle through which it must be rotated from this position such that it experiences half of the maximum torque. The angle through which it must be rotated from this positon such that it experiences half of the maxium troque isA. `60^(@)`B. `30^(@)`C. `45^(@)`D. `37^(@)`
Answer» Correct Answer - A `tau_(max)=MB`

Discussion

109.	The electron in hydrogen atom moves with a speed of `2.2xx10^(6)m//s` in an orbit of radius `5.3xx10^(-11)cm`. Find the magnetic moment of the orbiting electron.A. `8.3 xx 10^(-23)A-m^(2)`B. `9.3 xx 10^(-24)A-m^(2)`C. `7.2 xx 10^(-24)A-m^(2)`D. `6 xx 10^(-24)A-m^(2)`
Answer» Correct Answer - B Frequency of revolution , ` f=(v)/( 2pir )` The moving charge is equivalent to a current loop , given by `i=f xx e or i=(ev)/( 2pir )` If A be the area of the orbit , then the magnetic moment of the orbiting electron is , `M=iA=((ev)/( 2pi r))(pir^(2))=(evr)/(2)` Putting the values , we have `M=((1.6 xx 10^(-19))(2.2 xx 10^(6))(5.3 xx 10^(-11)))/(2)` `=9.3 xx 10^(-24)A-m^(2)`

Discussion

110.	The electron in hydrogen atom moves with a speed of `2.2xx10^(6)m//s` in an orbit of radius `5.3xx10^(-11)cm`. Find the magnetic moment of the orbiting electron.A. `8.27 xx 10^(-26) Am^(2)`B. `9.27 xx 10^(-27) Am^(2)`C. `9.3 xx 10^(-26)Am^(2)`D. `8.8 xx 10^(-27)Am^(2)`
Answer» Correct Answer - C Frequency of revolution , ` f=(v)/( 2pir )` The moving charge is equivalent to a current loop , given by `l=f xx e or l=(ev)/( 2pir )` If A be the area of the orbit , then the magnetic moment of the orbiting electron is , `M=lA=((ev)/( 2pi r))(pir^(2))=(evr)/(2)` Putting the values , we get `M=((1.6 xx 10^(-19))(2.2 xx 10^(6))(5.3 xx 10^(-11) xx 10^(-2)))/(2)` `=9.3 xx 10^(-26)A-m^(2)`

Discussion

111.	A closlly wound coil of 1000 turns and cross sectional area `2xx10^(-4) m^(2)` carries a current of 1.0 A.The magnetic moment of the coil isA. `0.1Am^(2)`B. `0.2 AM^(2)`C. `0.4 Am^(2)`D. `0.6 Am^(2)`
Answer» Correct Answer - B `M=nIA=1000xx1xx2xx10^(-4)=0.2 Am^(2)`

Discussion

112.	A magnetic needle is made to vibrate in uniform field `H`, then its time period is `T`. If it vibrates in the field of intensity `4H`, its time period will beA. 2TB. `T//2`C. `2//T`D. T
Answer» Correct Answer - B

Discussion

113.	A straight solenoid of length `50cm` has `1000` turns and a mean cross sectional area of `2xx10^-4m^2`. It is placed with its axis at `30^@` with a uniform magnetic field of `0*31T`. Find the torque acting on the solenoid when a current of `2A` is passed through it.
Answer» Correct Answer - `0.064 Nm`

Discussion

114.	A closely wound solenoid of `200` turns and area of cross-section `1.5xx10^(-4)m^(2)` carries a current of `2.0 A` It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field `5xx10^(-2)T`, making an angle of `30^(@)` with the axis of the solenoid. The torque on the solenoid will be
Answer» Correct Answer - `5 xx 10^(-2) J T^(-1) , 5 xx 10^(-3) J 2 xx 10^(-2) J `

Discussion

115.	The magnetic moment per unit volume of the material is termed as ……
Answer» The magnetic moment per unit volume of the material is termed as intensity of magnetisation.

Discussion

116.	In diamagnetic materials the net magnetic moment of atoms is ………
Answer» In diamagnetic materials the net magnetic moment of atoms is zero.

Discussion

117.	A freely suspended magnet always comes to rest in the (a) North – East (b) South – West (c) East – West (d) North – South
Answer» (d) North-South

Discussion

118.	Why docs a freely suspended magnet always rest in north- south direction ?
Answer» A freely suspended magnet always rest in north-south direction because the north-pole of the magnet lies in the geographic north direction and the south pole of the magnet lies in the geographic south direction. So it aligns itself in N-S direction. As unlike poles attract and like poles repel.

Discussion

119.	Fill in the blanks.1. A freely suspended magnet always comes to rest in the................ direction.2. Permanent magnets are generally made of steel, cobalt, nickel, or an alloy called...............3. The perpendicular bisector of the line joining the two poles of a magnet is called...................4. The north pole of the imaginary magnet of the earth is near the geographic .......... pole.5. Magnetic poles always exist in ...........
Answer» 1. A freely suspended magnet always comes to rest in the north- south direction. 2. Permanent magnets are generally made of steel, cobalt, nickel, or an alloy called ALNICO. 3. The perpendicular bisector of the line joining the two poles of a magnet is called magnetic equator. 4. The north pole of the imaginary magnet of the earth is near the geographic south pole. 5. Magnetic poles always exist in pairs.

119.

Fill in the blanks.1. A freely suspended magnet always comes to rest in the................ direction.2. Permanent magnets are generally made of steel, cobalt, nickel, or an alloy called...............3. The perpendicular bisector of the line joining the two poles of a magnet is called...................4. The north pole of the imaginary magnet of the earth is near the geographic .......... pole.5. Magnetic poles always exist in ...........

Answer»

1. A freely suspended magnet always comes to rest in the north- south direction.

2. Permanent magnets are generally made of steel, cobalt, nickel, or an alloy called ALNICO.

3. The perpendicular bisector of the line joining the two poles of a magnet is called magnetic equator.

4. The north pole of the imaginary magnet of the earth is near the geographic south pole.

5. Magnetic poles always exist in pairs.

Discussion

120.	The relation connecting magnetic susceptibility `chi_(m)` and relative permeability `mu_(p)`, isA. `x_(m)=mu_(r)`B. `x_(m)-1=mu_(r)`C. `mu_(r)=1+x_(m)`D. `mu_(r)=1-x_(m)`
Answer» Correct Answer - C

Discussion

121.	The range of magnetic susceptibility and relative magnetic permeability for diamagnetic subtance areA. ZeroB. Equal to unityC. less than unityD. Greater than unity
Answer» Correct Answer - C

Discussion

122.	A straight wire carring current I is turned into a circular loop. If the magnitude of magnetic moment associated with it in M.K.S. unit is M, the length of wire will beA. `(4pi)/(M)`B. `(Mpi)/(41)`C. `sqrt((4piM)/(1))`D. `sqrt((4piI)/(M))`
Answer» Correct Answer - C M=IA `M=Ipir^(2)` `therefore r=sqrt((M)/(piI))` length of wire=`l=2pir` `l=2pisqrt((M)/(pil))` `=sqrt((M4pi^(2)))/(piI)` `sqrt((4piM)/(I))`

Discussion

123.	A bar magnet made of steel has a magnetic moment of `2.5 A m^2` and mass of `6.6 xx 10^(-5) kg`. If the density of steel is `7.9 xx 10^3 kg m^(-3)`, find the intensity of magnetization of the magnet.A. `1.0 xx 10^(6) Am`B. `4.0 xx 10^(6) Am`C. `3.0 xx10^(6) Am`D. `8.0 xx 10^(6)Am`
Answer» Correct Answer - B `x=2.1xx10^(-5)` Increase in magnetic held =`B-B_(0)` `=B_(0)+B_(1)-B_(0)` `B_(1)` Now % increase in magnetic field =`(B_(1))/(B_(0))xx100` `=(mu_(0)I)/(mu_(0)H)xx100` `(I)/(H)xx100` `x xx 100` `2.1xx10^(-5)xx100` `=2.1xx10^(-3)`

Discussion

124.	A superconducting material isA. ferromagneticB. ferroelecticC. diamagneticD. paramagnetic
Answer» Correct Answer - C A superconducting material is diamagnetic .

Discussion

125.	A steel bar is magnetized by keeping it inside a long coil of insulated copper wire and passing a current through the coil. Discuss what happen to the pole strength of the magnet as the time for which the electric current passes through the coil increases.
Answer» When current passes though the coil, the end at which current enters in an anti-clockwise direction will become the north pole and the other end becomes the south pole. As we are increasing the current more number of molecular magnets are arranged in the form of long straight chains, when all the molecular magnets are arranged in the form of long straight chains, the substance is said to the saturated with magnetism. once a given magnetic substances is saturated, it cannot be magnetized further.

Discussion

126.	Which of the following quantities: (I) magnetic declination (II) dip is used to determine the strength of earths magnetic field at a point on the earths surfaceA. Both `I&II`B. Neither `I` nor `II`C. `I` OnlyD. `II` only
Answer» Correct Answer - A

Discussion

127.	Draw a neat labelled diagram for angle of dip.Write a short note on Earth’s magnetic field. Mention the extreme values of magnetic field at magnetic poles and magnetic equator.
Answer» i. Magnetic force experienced per unit pole strength is magnetic field \(\overset\rightarrow{B}\) at that place. ii. This field can be resolved in components along the horizontal (\(\overset\rightarrow{B}_H\)) and along vertical (\(\overset\rightarrow{B}_v\)). iii. The two components are related with the angle of dip (ø) as, B_H = B cos ø, B_v= B sin ø \(\frac{B_v}{B_H}\) = tan ø B² = \(B^2_v+B^2_H\) ∴ B = \(\sqrt{B^2_V+B^2_H}\) iv. At the magnetic North pole: \(\overset\rightarrow{B}=\overset\rightarrow{B}_V\) directed upward, \(\overset\rightarrow{B}_H\) = 0 and ø = 90°. v. At the magnetic south pole: \(\overset\rightarrow{B}=\overset\rightarrow{B}_V\) directed downward, \(\overset\rightarrow{B}_H\) = 0 and ø = 270°. vi. Anywhere on the magnetic equator (magnetic great circle): B = B_H along South to North, \(\overset\rightarrow{B}_V\) = 0 and ø = 0

127.

Draw a neat labelled diagram for angle of dip.Write a short note on Earth’s magnetic field. Mention the extreme values of magnetic field at magnetic poles and magnetic equator.

Answer»

i. Magnetic force experienced per unit pole strength is magnetic field \(\overset\rightarrow{B}\) at that place.

ii. This field can be resolved in components along the horizontal (\(\overset\rightarrow{B}_H\)) and along vertical (\(\overset\rightarrow{B}_v\)).

iii. The two components are related with the angle of dip (ø) as, B_H = B cos ø, B_v= B sin ø

\(\frac{B_v}{B_H}\) = tan ø

B² = \(B^2_v+B^2_H\)

∴ B = \(\sqrt{B^2_V+B^2_H}\)

iv. At the magnetic North pole: \(\overset\rightarrow{B}=\overset\rightarrow{B}_V\) directed upward, \(\overset\rightarrow{B}_H\) = 0 and ø = 90°.

v. At the magnetic south pole: \(\overset\rightarrow{B}=\overset\rightarrow{B}_V\) directed downward, \(\overset\rightarrow{B}_H\) = 0 and ø = 270°.

vi. Anywhere on the magnetic equator (magnetic great circle): B = B_H along South to North, \(\overset\rightarrow{B}_V\) = 0 and ø = 0

Discussion

128.	State the SI units of pole strength and magnetic dipole moment.
Answer» 1. SI unit of pole strength (q_m) is Am. 2. SI unit of magnetic dipole moment (m) is Am².

128.

State the SI units of pole strength and magnetic dipole moment.

Answer»

1. SI unit of pole strength (q_m) is Am.

2. SI unit of magnetic dipole moment (m) is Am².

Discussion

129.	Explain the pole strength and magnetic dipole moment of a bar magnet.
Answer» i. The bar magnet said to have pole strength +q_m and -q_m near the north and south poles respectively. ii. As bar magnet has two poles with equal and opposite pole strength, it is called as a magnetic dipole. iii. The two poles are separated by a distance equal to 2l. iv. The product of pole strength and the magnetic length is called as magnetic dipole moment. ∴ \(\overset\rightarrow{m}\) = q_m \((2\overset\rightarrow{l})\) where, \(2\overset\rightarrow{l}\) is a vector from south pole to north pole.

129.

Explain the pole strength and magnetic dipole moment of a bar magnet.

Answer»

i. The bar magnet said to have pole strength +q_m and -q_m near the north and south poles respectively.

ii. As bar magnet has two poles with equal and opposite pole strength, it is called as a magnetic dipole.

iii. The two poles are separated by a distance equal to 2l.

iv. The product of pole strength and the magnetic length is called as magnetic dipole moment.

∴ \(\overset\rightarrow{m}\) = q_m \((2\overset\rightarrow{l})\)

where, \(2\overset\rightarrow{l}\) is a vector from south pole to north pole.

Discussion

130.	The perpendicular drawn to magnetic axis and passing through centre of magnetic dipole is called asA. Magnetic lengthB. Mafnetic equatorC. Magnetic dipole momentD. None of these
Answer» Correct Answer - B

Discussion

131.	The ability of interaction of the either pole of the dipole isA. Pole stregthB. Magnetic momentC. magnetic dipoleD. None of these
Answer» Correct Answer - A

Discussion

132.	The coupl acting on a bar magnet of pole strength `2 Am` when kept in a magnetic field of intensity `10 A//m`, such that axis of the magnet makes an angle `30^(0)` with the direction of the field is `80xx10^(-7)Nm`. The distance between the poles of the magnet isA. `(2)/(pi)m`B. `(pi)/(2)m`C. `63.36m`D. `(1)/(2pi)m`
Answer» Correct Answer - A `C=MB sin theta`

Discussion

133.	A bar magnet of pole strnegth `2A-m` is kept in a magnetic field of induction `4xx10^(-5)wbm^(-2)` such that the axis of the magnet makes an angle `30^(@)` with the directon of the field. The couple acting on the magnet is found `80xx10^(-7)N-m`. Then the distance between the poles of the magnet isA. `20m`B. `2m`C. `3cm`D. `20cm`
Answer» Correct Answer - D

Discussion

134.	The magnetic induction and the intensity of magnetic field inside an iron core of an electromagnet are `1 Wbm^(-2)` and `150Am^(-1)` respectively. The relative permeability of iron is : `(mu_(0)=4pixx10^(-7)"henry"//m)`A. `(10^(6))/(4pi)`B. `(10^(6))/(6pi)`C. `(10^(5))/(4pi)`D. `(10^(5))/(6pi)`
Answer» Correct Answer - D `mu_(r)=(B)/(mu_(0)H)`

Discussion

135.	………… suggested that earth behaves as a giant bar magnet
Answer» Gilbert suggested that earth behaves as a giant bar magnet.

Discussion

136.	A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current 1 = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the center of the spiral is(a) 5 μT (b) 7 μT (c) 8 μT (d) 10 μT
Answer» Correct answer is (b) 7 μT

Discussion

137.	Let I1 and I2 be the steady currents passing through a long horizontal wire XY and PQ respectively. The wire PQ is fixed in horizontal plane and the wire XY be is allowed to move freely in a vertical plane. Let the wire XY is in equilibrium at a height d over the parallel wire PQ as shown in figure.
Answer» If the wire XY is slightly displaced and released, it executes simple uarmonic motion due to force of repulsion produced between the current carrying wire. Acceleration of the wire, a = -ω²y Time period of oscillation of the wire, T = 2π\(\sqrt{\frac{d}{g}}\)

137.

Let I1 and I2 be the steady currents passing through a long horizontal wire XY and PQ respectively. The wire PQ is fixed in horizontal plane and the wire XY be is allowed to move freely in a vertical plane. Let the wire XY is in equilibrium at a height d over the parallel wire PQ as shown in figure.

Answer»

If the wire XY is slightly displaced and released, it executes simple uarmonic motion due to force of repulsion produced between the current carrying wire.

Acceleration of the wire, a = -ω²y

Time period of oscillation of the wire,

T = 2π\(\sqrt{\frac{d}{g}}\)

Discussion

138.	A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is σ. The disc rotates about an a × is perpendicular to its plane passing through the center with angular velocity ω. Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field whose strength is B which is directed perpendicular to the a × is of rotation(a)\(\frac{1}{4}\)σωπ BR(b)\(\frac{1}{4}\) σωπ BR2 (c) \(\frac{1}{4}\) σωπ BR3(d) \(\frac{1}{4}\)σωπ BR4
Answer» Correct answer is (d) \(\frac{1}{4}\)σωπ BR⁴

Discussion

139.	A circular coil of radius 5 cm and 50 turns carries a current of 3 ampere. The magnetic dipole moment of the coil is-(a) 1.0 amp – m2 (b) 1.2 amp – m2 (c) 0.5 amp – m2 (d) 0.8 amp – m2
Answer» Correct answer is (b) 1.2 amp – m²

Discussion

140.	The magnetic moment produced in a substance of `1 gm is 6xx10^(-7) "ampere","metre"^(2)`. If its density is `5 gm//cm^(3)`, then the intensity of magnetisation in `A//m` will beA. `8.3 xx10^(6)`B. `3.0`C. `1.2xx10^(-7)`D. `13xx10^(-6)`
Answer» Correct Answer - B `I=(M)/(V)` `=(M)/((m)/(rho))=(Mrho)/(m)=3.0`

Discussion

141.	The magnetic moment produced in a substance of `1 gm is 6xx10^(-7) "ampere","metre"^(2)`. If its density is `5 gm//cm^(3)`, then the intensity of magnetisation in `A//m` will beA. `8.3xx10^(6)`B. `3.0`C. `1.2xx10^(-7)`D. `3 xx 10^(-6)`
Answer» Correct Answer - B Intensity of magnetisation , `l=(M)/V=-(M)/("mass/density")` Given , mass =1 g =`10^(-3)kg ` and density ` =5g//cm^(3)=(5 xx 10^(-3)kg)/((10^(-2))^(3))=5 xx 10^(3) kg m^(-3)` Hence , `l=( 6xx 10^(-7) xx 5 xx 10^(3))/( 10^(-3))=3Am^(-1)`

Discussion

142.	Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond curie temperature, then it will showA. paramagnetismB. anti-ferromagnetismC. no magnetic propertyD. diamagnetism
Answer» Correct Answer - A If the temperature of a ferromagnetic material is raised above a certain critical value , called the Curie temperature , the exchange coupling ceases to be effective . Most of such materials become simply paramagnetic , i.e. the dipoles still tend to align with an external field but much more weakly and thermal agitation can now more easily disrupt the alignment .

Discussion

143.	The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain?
Answer» Britain is close to magnetic north pole. Therefore, the angle of dip is greater in Britain than in India. It is about 70°.

Discussion

144.	A magnetising field of `2xx10^(3)Am^(-1)` produces a magnetic flux density of `8piT` in an iron rod. The relative permeability of the rod will beA. `10^(2)`B. `10^(0)`C. `10^(3)`D. `10^(4)`
Answer» Correct Answer - D `mu_(r)=(mu)/(mu_(0))=(B)/(mu_(0)H)` `=(8pi)/((4pixx10^(-7))(2xx10^(3)))` `mu_(r)=10^(4)`

Discussion

145.	A magnetising field of `5000 A//m` produces a magnetic flux of `5xx10^(-5)` weber in an iron rod. If the area of cross section of the rod is `0.5 cm^(2)`, then the permeability of the rod will beA. `1xx10^(-3)`B. `2xx10^(-4)`C. `3xx10^(-5)`D. `4xx10^(-6)`
Answer» Correct Answer - B `phi_(B)=bar(B).bar(A)`

Discussion

146.	A magnetic field of `1600Am^-1` produces a magnetic flux of `24xx10^-5` weber in a bar of iron of cross section `02cm^2`. Calculate permeability and susceptibility of the bar.
Answer» Magnetic induction, `B=(phi)/(A)=(2.4xx10^(-5))/(0.2xx10^(-4))=1.2Wb//m^(2)` i) Permeability, `mu=(B)/(H)=(1.2)/(1600)=7.5xx10^(-4)TA^(-1)m` ii) As `mu=mu_(0)(1+chi)` then Susceptibility, `chi=(mu)/(mu_(0))-1=(7.5xx10^(-4))/(4pixx10^(-7))-1=596.1`

Discussion

147.	How do we determine strength of magnetic field at a given point due to a magnet? Write down units of magnetic field in SI and CGS system and their interconversion.
Answer» i. Density of lines of force i.e., the number of lines of force per unit area around a particular point determines the strength of the magnetic field at that point. ii. The magnitude of magnetic field strength B at a point in a magnetic field is given by, Magnetic Field = \(\frac{magnetic\,flux}{area}\) i.e., B = \(\frac{ø}{A}\) iii. SI unit of magnetic field (B) is expressed as weber/m² or Tesla. iv. 1 Tesla = 10⁴ Gauss

147.

How do we determine strength of magnetic field at a given point due to a magnet? Write down units of magnetic field in SI and CGS system and their interconversion.

Answer»

i. Density of lines of force i.e., the number of lines of force per unit area around a particular point determines the strength of the magnetic field at that point.

ii. The magnitude of magnetic field strength B at a point in a magnetic field is given by,

Magnetic Field = \(\frac{magnetic\,flux}{area}\)

i.e., B = \(\frac{ø}{A}\)

iii. SI unit of magnetic field (B) is expressed as weber/m² or Tesla.

iv. 1 Tesla = 10⁴ Gauss

Discussion

148.	If a magnet of length `10cm` and pole strength `40 A-m` is placed at an angle of `45^(@)` in an uniform induction field of intensity `2xx10^(-4)T`, the couple acting on it isA. `0.5656xx10^(-4)N-m`B. `0.5656xx10^(-3)N-m`C. `0.656xx10^(-4)N-m`D. `0.656xx10^(-5)N-m`
Answer» Correct Answer - B

Discussion

149.	Relative permitivity and permeability of a material `epsilon_(r) and mu_(r)`, respectively . Which of the following values of these quantities are allowed for a diamagnetic material?A. `epsilon_(r)=0.5,mu_(r)=1.5`B. `epsilon_(r)=1.5,mu_(r)=0.5`C. `epsilon_(r)=0.5,mu_(r)=0.5`D. `epsilon_(r)=1.5,mu_(r)=1.5`
Answer» Correct Answer - C for diamangnetic `in_(r)gt1` `mu_(r)lt1`

Discussion

150.	Relative permitivity and permeability of a material `epsilon_(r) and mu_(r)`, respectively . Which of the following values of these quantities are allowed for a diamagnetic material?A. `epsilon_(r)=1.5,mu_(r)=0.5`B. `epsilon_(r)0.5, mu_(r)=0.5`C. `epsilon_(r)1.5,mu_(r)=1.5,mu_(r)=1.5`D.
Answer» Correct Answer - A

Discussion

Explore topic-wise InterviewSolutions in .

Distinguish between natural and artificial magnets.

Match the following :I.Natural magnets(a)Levitating propellerII.Artificial magnets(b)Wooden spoonIII.Magnetic substance(c)HeatingIv.Non magnetic substance(d)Bar magnetv.Demagnetization(e)Pin holdersvI.Demagnetizaiton(f)Magnetite

Why are the artificial magnets preferred over the natural magnets ?

A magnet is placed in iron powder and then taken out, then maximum iron powder is atA. Some away from north poleB. Some away from south poleC. The middle of the magnetD. The end of the magnet

A magnet is placed in iron poweder and the taken out , them maximum iron powder is atA. some distance away from north poleB. some distance away from south poleC. the middle of the magnetD. the ends of the magnet

The electron in hydrogen atom moves with a speed of `2.2xx10^(6)m//s` in an orbit of radius `5.3xx10^(-11)cm`. Find the magnetic moment of the orbiting electron.A. `8.3 xx 10^(-23)A-m^(2)`B. `9.3 xx 10^(-24)A-m^(2)`C. `7.2 xx 10^(-24)A-m^(2)`D. `6 xx 10^(-24)A-m^(2)`

The electron in hydrogen atom moves with a speed of `2.2xx10^(6)m//s` in an orbit of radius `5.3xx10^(-11)cm`. Find the magnetic moment of the orbiting electron.A. `8.27 xx 10^(-26) Am^(2)`B. `9.27 xx 10^(-27) Am^(2)`C. `9.3 xx 10^(-26)Am^(2)`D. `8.8 xx 10^(-27)Am^(2)`

A closlly wound coil of 1000 turns and cross sectional area `2xx10^(-4) m^(2)` carries a current of 1.0 A.The magnetic moment of the coil isA. `0.1Am^(2)`B. `0.2 AM^(2)`C. `0.4 Am^(2)`D. `0.6 Am^(2)`

A magnetic needle is made to vibrate in uniform field `H`, then its time period is `T`. If it vibrates in the field of intensity `4H`, its time period will beA. 2TB. `T//2`C. `2//T`D. T

A straight solenoid of length `50cm` has `1000` turns and a mean cross sectional area of `2xx10^-4m^2`. It is placed with its axis at `30^@` with a uniform magnetic field of `0*31T`. Find the torque acting on the solenoid when a current of `2A` is passed through it.

The magnetic moment per unit volume of the material is termed as ……

In diamagnetic materials the net magnetic moment of atoms is ………

A freely suspended magnet always comes to rest in the (a) North – East (b) South – West (c) East – West (d) North – South

Why docs a freely suspended magnet always rest in north- south direction ?

The relation connecting magnetic susceptibility `chi_(m)` and relative permeability `mu_(p)`, isA. `x_(m)=mu_(r)`B. `x_(m)-1=mu_(r)`C. `mu_(r)=1+x_(m)`D. `mu_(r)=1-x_(m)`

The range of magnetic susceptibility and relative magnetic permeability for diamagnetic subtance areA. ZeroB. Equal to unityC. less than unityD. Greater than unity

A straight wire carring current I is turned into a circular loop. If the magnitude of magnetic moment associated with it in M.K.S. unit is M, the length of wire will beA. `(4pi)/(M)`B. `(Mpi)/(41)`C. `sqrt((4piM)/(1))`D. `sqrt((4piI)/(M))`

A bar magnet made of steel has a magnetic moment of `2.5 A m^2` and mass of `6.6 xx 10^(-5) kg`. If the density of steel is `7.9 xx 10^3 kg m^(-3)`, find the intensity of magnetization of the magnet.A. `1.0 xx 10^(6) Am`B. `4.0 xx 10^(6) Am`C. `3.0 xx10^(6) Am`D. `8.0 xx 10^(6)Am`

A superconducting material isA. ferromagneticB. ferroelecticC. diamagneticD. paramagnetic

A steel bar is magnetized by keeping it inside a long coil of insulated copper wire and passing a current through the coil. Discuss what happen to the pole strength of the magnet as the time for which the electric current passes through the coil increases.

Which of the following quantities: (I) magnetic declination (II) dip is used to determine the strength of earths magnetic field at a point on the earths surfaceA. Both `I&II`B. Neither `I` nor `II`C. `I` OnlyD. `II` only

Draw a neat labelled diagram for angle of dip.Write a short note on Earth’s magnetic field. Mention the extreme values of magnetic field at magnetic poles and magnetic equator.

State the SI units of pole strength and magnetic dipole moment.

Explain the pole strength and magnetic dipole moment of a bar magnet.

The perpendicular drawn to magnetic axis and passing through centre of magnetic dipole is called asA. Magnetic lengthB. Mafnetic equatorC. Magnetic dipole momentD. None of these

The ability of interaction of the either pole of the dipole isA. Pole stregthB. Magnetic momentC. magnetic dipoleD. None of these

The magnetic induction and the intensity of magnetic field inside an iron core of an electromagnet are `1 Wbm^(-2)` and `150Am^(-1)` respectively. The relative permeability of iron is : `(mu_(0)=4pixx10^(-7)"henry"//m)`A. `(10^(6))/(4pi)`B. `(10^(6))/(6pi)`C. `(10^(5))/(4pi)`D. `(10^(5))/(6pi)`

………… suggested that earth behaves as a giant bar magnet

A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current 1 = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the center of the spiral is(a) 5 μT (b) 7 μT (c) 8 μT (d) 10 μT

A circular coil of radius 5 cm and 50 turns carries a current of 3 ampere. The magnetic dipole moment of the coil is-(a) 1.0 amp – m2 (b) 1.2 amp – m2 (c) 0.5 amp – m2 (d) 0.8 amp – m2

The magnetic moment produced in a substance of `1 gm is 6xx10^(-7) "ampere","metre"^(2)`. If its density is `5 gm//cm^(3)`, then the intensity of magnetisation in `A//m` will beA. `8.3 xx10^(6)`B. `3.0`C. `1.2xx10^(-7)`D. `13xx10^(-6)`

The magnetic moment produced in a substance of `1 gm is 6xx10^(-7) "ampere","metre"^(2)`. If its density is `5 gm//cm^(3)`, then the intensity of magnetisation in `A//m` will beA. `8.3xx10^(6)`B. `3.0`C. `1.2xx10^(-7)`D. `3 xx 10^(-6)`

Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond curie temperature, then it will showA. paramagnetismB. anti-ferromagnetismC. no magnetic propertyD. diamagnetism

The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain?

A magnetising field of `2xx10^(3)Am^(-1)` produces a magnetic flux density of `8piT` in an iron rod. The relative permeability of the rod will beA. `10^(2)`B. `10^(0)`C. `10^(3)`D. `10^(4)`

A magnetising field of `5000 A//m` produces a magnetic flux of `5xx10^(-5)` weber in an iron rod. If the area of cross section of the rod is `0.5 cm^(2)`, then the permeability of the rod will beA. `1xx10^(-3)`B. `2xx10^(-4)`C. `3xx10^(-5)`D. `4xx10^(-6)`

A magnetic field of `1600Am^-1` produces a magnetic flux of `2*4xx10^-5` weber in a bar of iron of cross section `0*2cm^2`. Calculate permeability and susceptibility of the bar.

How do we determine strength of magnetic field at a given point due to a magnet? Write down units of magnetic field in SI and CGS system and their interconversion.

If a magnet of length `10cm` and pole strength `40 A-m` is placed at an angle of `45^(@)` in an uniform induction field of intensity `2xx10^(-4)T`, the couple acting on it isA. `0.5656xx10^(-4)N-m`B. `0.5656xx10^(-3)N-m`C. `0.656xx10^(-4)N-m`D. `0.656xx10^(-5)N-m`

Relative permitivity and permeability of a material `epsilon_(r) and mu_(r)`, respectively . Which of the following values of these quantities are allowed for a diamagnetic material?A. `epsilon_(r)=1.5,mu_(r)=0.5`B. `epsilon_(r)0.5, mu_(r)=0.5`C. `epsilon_(r)1.5,mu_(r)=1.5,mu_(r)=1.5`D.

A magnetic field of `1600Am^-1` produces a magnetic flux of `24xx10^-5` weber in a bar of iron of cross section `02cm^2`. Calculate permeability and susceptibility of the bar.