# Chapter 5 – Magnetism And Matter Questions and Answers: NCERT Solutions for Class 12 Physics

Class 12 Physics NCERT book solutions for Chapter 5 - Magnetism And Matter Questions and Answers.

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Class 12 Physics NCERT book solutions for Chapter 5 - Magnetism And Matter Questions and Answers.

a) What are the three independent quantities which are conventionally used to describe the magnetic field on the Earth’s surface specifically?

I. Magnetic declination

II. The horizontal component of Earth’s magnetic field and

III. Magnetic inclination or angle of dip.

The dip angle is maximum at the poles and minimum at the equator. The angle of dip would be greater at the location in Britain as it would be closer to the North Pole (and for a fact, it is about 70∘70∘) as compared to that at a location in southern India because this location is closer to the equator.

Hence, the magnetic field lines emanate from close to the south pole of the Earth and terminate at the North Pole. Now, in a map depicting Earth’s magnetic field lines, the field lines at Melbourne, Australia, which is close to the Earth’s South Pole, would come out of the surface of the Earth.

Hence, a compass on the geomagnetic North Pole or the South Pole that too aligned in the vertical plane would point upwards and downwards at the geomagnetic south and North Pole, respectively.

If it were kept horizontally, the needle would be perpendicular to the field lines of the Earth, and hence would be free to move in every direction.

B=μ0M4πr3B=μ0M4πr3

Here,

MM

is the magnetic moment, and is given as

MM

= 8×1022J/T8×1022J/T,

rr

is the radius of Earth,

r=6.4×106mr=6.4×106m

μ0μ0

is the permeability of the free space =

4π×10−7Tm/A4π×10−7Tm/A

Hence,

B=4π×10−7×8×10224π(6.4×106)3B=4π×10−7×8×10224π(6.4×106)3

⇒B=0.3G=3G×(10−1)⇒B=0.3G=3G×(10−1)

The order of magnitude of the magnetic field is −1−1.

a) Does the Earth’s magnetic field vary in time as it varies from point to point in space. If it does, then on what time scale does it change significantly?

The change in the Earth’s magnetic field with time should be neglected if we consider multiple decades’ time scales.

Some of these rocks have had ferrous metals, and the magnetic fields got weakly recorded in them. Hence the geologists get clues about the geomagnetic history from the analysis of such rocks.

10−1210−12

T.

Note: The above exercise is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown and could be a great field of research. For details, one should consult a good text on geomagnetism.

A constant field would make a moving charge go in circles. Since the interstellar field is very weak, the deviation in the path of the charged particle may not be noticeable.

For instance, if a charged particle is moving with a constant speed and is affected by the interstellar magnetic field perpendicular to its path, then it will experience a force that will make it go in a circle of a radius so large it can barely be noticed in small length scales.

However, with reference to the gigantic interstellar space, the deflection can affect the passage of charged particles since the distance scales under consideration are huge.

30∘30∘

with a uniform external magnetic field of

0.25T0.25T

experiences a torque of magnitude equal to

4.5×10−2J4.5×10−2J

. What is the magnitude of magnetic moment of the magnet?

Magnetic field strength

B=0.25TB=0.25T

Torque on the bar magnet,

T=4.5×10−2JT=4.5×10−2J

Angle between the given bar magnet and the external magnetic field,

θ=30∘θ=30∘

Torque is related to magnetic moment (M) as:

T=MBsin(θ)T=MBsin(θ)

⇒M=4.5×10−20.25×sin30∘=0.36J/T⇒M=4.5×10−20.25×sin30∘=0.36J/T

Clearly, the magnetic moment of the magnet is

0.36J/T0.36J/T

M=0.32J/TM=0.32J/T

is placed in a uniform magnetic field of

0.15T0.15T

. If the bar is free to rotate in the plane of the field, which orientation and would correspond to its

a) Stable?

M=0.32J/TM=0.32J/T

External magnetic field,

B=0.15TB=0.15T

It is considered as being in stable equilibrium, when the bar magnet is aligned along the magnetic field. Therefore, the angle

θθ

, between the bar magnet and the magnetic field is

0∘0∘

.

Potential energy of the system

=−MBcos(θ)=−MBcos(θ)

⇒−MBcos(θ)=−0.32×0.15×cos(0)=−4.8×10−2J⇒−MBcos(θ)=−0.32×0.15×cos(0)=−4.8×10−2J

Hence the potential energy is

=−4.8×10−2J=−4.8×10−2J

M=0.32J/TM=0.32J/T

External magnetic field,

B=0.15TB=0.15T

When the bar magnet is aligned opposite to the magnetic field, it is considered as being in unstable equilibrium,

θ=180∘θ=180∘

Potential energy of the system is hence

=−MBcos(θ)=−MBcos(θ)

⇒−MBcos(θ)=−0.32×0.15×cos(180∘)=4.8×10−2J⇒−MBcos(θ)=−0.32×0.15×cos(180∘)=4.8×10−2J

Hence the potential energy is

=4.8×10−2J=4.8×10−2J

800800

turns and area of cross section

2.5×10−4m22.5×10−4m2

carries a current of

3.0A3.0A

. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

n=800n=800

Area of cross-section,

A=2.5×10−4m2A=2.5×10−4m2

Current in the solenoid,

I=3.0AI=3.0A

A current-carrying solenoid is analogous to a bar magnet because a magnetic field develops along its axis, i.e., along its length joining the north and south poles.

The magnetic moment due to the given current-carrying solenoid is calculated as:

M=nIA=800×3×2.5×10−4=0.6J/TM=nIA=800×3×2.5×10−4=0.6J/T

Thus, the associated magnetic moment

=0.6J/T=0.6J/T

0.25T0.25T

is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of

30∘30∘

with the direction of applied field?

B=0.25TB=0.25T

Magnetic moment,

M=0.6/TM=0.6/T

The angle,

θθ

between the axis of the turns of the solenoid and the direction of the external applied field is

30∘30∘

Hence, the torque acting on the solenoid is given as:

τ=MBsin(θ)τ=MBsin(θ)

⇒τ=0.6×0.25sin(30∘)⇒τ=0.6×0.25sin(30∘)

⇒τ=7.5×10−2J⇒τ=7.5×10−2J

Hence the magnitude of torque is

=7.5×10−2J=7.5×10−2J

1.5J/T1.5J/T

lies aligned with the direction of a uniform magnetic field of

0.22T0.22T

a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

Magnetic moment,

M=1.5J/TM=1.5J/T

Magnetic field strength,

B=0.22TB=0.22T

θ1=0∘θ1=0∘

Final angle between the magnetic field and the axis is,

θ2=90∘θ2=90∘

The work that would be required to make the magnetic moment perpendicular to the direction of magnetic field would be:

W=−MB(cosθ2−cosθ1)W=−MB(cosθ2−cosθ1)

⇒W=−1.5×0.22(cos90∘−cos0∘)⇒W=−1.5×0.22(cos90∘−cos0∘)

⇒W=−0.33(0−1)⇒W=−0.33(0−1)

⇒W=0.33J⇒W=0.33J

θ1=0∘θ1=0∘

Final angle between the magnetic field and the axis,

θ2=180∘θ2=180∘

The work that would be required to make the magnetic moment opposite (180 degrees) to the direction of magnetic field is given as:

W=−MB(cosθ2−cosθ1)W=−MB(cosθ2−cosθ1)

⇒W=−1.5×0.22(cos180∘−cos0∘)⇒W=−1.5×0.22(cos180∘−cos0∘)

⇒W=−0.33(−1−1)⇒W=−0.33(−1−1)

⇒W=0.66J⇒W=0.66J

θ=θ1=90∘θ=θ1=90∘

Hence the Torque,

τ⃗ =M⃗ ×B⃗ τ→=M→×B→

And its magnitude is:

τ=MBsin(θ)τ=MBsin(θ)

⇒τ=1.5×0.22sin(90∘)⇒τ=1.5×0.22sin(90∘)

⇒τ=0.33Nm⇒τ=0.33Nm

Hence the torque involved is

=0.33Nm=0.33Nm

For the second-(ii) case:

θ=θ1=180∘θ=θ1=180∘

And its magnitude of the torque is:

τ=MBsin(θ)τ=MBsin(θ)

⇒τ=1.5×0.22sin(180∘)⇒τ=1.5×0.22sin(180∘)

⇒τ=0Nm⇒τ=0Nm

Hence the torque is zero.

20002000

turns and area of cross-section

1.6×10−4m21.6×10−4m2

, carrying a current of

4.0A4.0A

, is suspended through its center allowing it to turn in a horizontal plane.

a) What is the magnetic moment associated with the solenoid?

n=2000n=2000

Area of cross-section of the solenoid,

A=1.6×10−4m2A=1.6×10−4m2

Current in the solenoid,

I=4AI=4A

The magnetic moment inside the solenoid at the axis is calculated as:

M=nAI=2000×1.6×10−4×4=1.28Am2M=nAI=2000×1.6×10−4×4=1.28Am2

7.5×10−2T7.5×10−2T

is set up at an angle of

30∘30∘

with the axis of the solenoid?

Magnetic field,

B=7.5×10−2TB=7.5×10−2T

Angle between the axis and the magnetic field of the solenoid,

θ=30∘θ=30∘

Torque,

τ=MBsin(θ)τ=MBsin(θ)

⇒τ=1.28×7.5×10−2sin(30∘)⇒τ=1.28×7.5×10−2sin(30∘)

⇒τ=4.8×10−2Nm⇒τ=4.8×10−2Nm

Given the magnetic field is uniform, and the force on the solenoid is zero. The torque on the solenoid is

4.8×10−2Nm4.8×10−2Nm

1616

turns and radius

10cm10cm

carrying a current of

0.750.75

A rests with its plane normal to an external field of magnitude

5.0×10−2T5.0×10−2T

. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of

2.0/s2.0/s

. What is the moment of inertia of the coil about its axis of rotation?

The number of turns in the given circular coil solenoid,

N=16N=16

Radius of the coil,

r=10cm=0.1mr=10cm=0.1m

Cross-section of the coil,

A=πr2=π×(0.1)2m2A=πr2=π×(0.1)2m2

Current in the coil,

I=0.75AI=0.75A

Magnetic field strength,

B=5.0×10−2TB=5.0×10−2T

Frequency of oscillations of the coil,

υ=2.0/sυ=2.0/s

Therefore, magnetic moment,

M=NAI=NIπr2M=NAI=NIπr2

⇒M=16×7.5×π×0.12⇒M=16×7.5×π×0.12

⇒M=0.3777J/T⇒M=0.3777J/T

Frequency is given by the relation:

ν=12πMBI−−−−√ν=12πMBI

where,

I=I=

Moment of inertia of the coil

⇒I=MB4π2ν2⇒I=MB4π2ν2

⇒I=0.377×5×10−24π2×22⇒I=0.377×5×10−24π2×22

⇒I=1.19×10−4kgm2⇒I=1.19×10−4kgm2

Clearly, the moment of inertia of the coil about its axis of rotation

1.19×10−4kgm21.19×10−4kgm2

22∘22∘

with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be

0.35G0.35G

. Determine the magnitude of the earth’s magnetic field at the place.

The horizontal component of earth’s magnetic field,

BH=0.35GBH=0.35G

The angle made by the needle with the horizontal plane (angle of dip)

=δ=22∘=δ=22∘

Earth’s magnetic field strength is

BB

We can relate

BB

and

BHBH

as:

BH=BcosδBH=Bcosδ

⇒B=BHcosδ⇒B=BHcosδ

⇒B=0.35cos22∘=0.377G⇒B=0.35cos22∘=0.377G

Clearly, the strength of earth’s magnetic field at the given location is

0.377G0.377G

12∘12∘

west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points

60∘60∘

above the horizontal. The horizontal component of the earth’s field is measured to

0.16G0.16G

. Specify the direction and magnitude of the earth’s field at the location.

Angle of declination,

θ=12∘θ=12∘

Angle of dip,

δ=60∘δ=60∘

Horizontal component of earth’s magnetic field,

BH=0.16GBH=0.16G

The magnetic field of the Earth at the given location

=B=B

We can relate

BB

and

BHBH

as:

B=BHcosδB=BHcosδ

⇒B=0.16cos60∘=0.32G⇒B=0.16cos60∘=0.32G

Magnetic field of Earth lies in the vertical plane,

θ=12∘θ=12∘

west of the geographic meridian, making an angle of

δ=60∘δ=60∘

(upward) with the horizontal direction. Its magnitude is

0.32G0.32G

0.48J/T0.48J/T

. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of

10cm10cm

from the center of the magnet on

a) the axis,

MM

is

0.48J/T0.48J/T

Given distance,

d=10cm=0.1md=10cm=0.1m

The magnetic field at d-distance, from the centre of the magnet on the axis is given by the relation:

B=μ04π2Md3B=μ04π2Md3

here,

μ0=μ0=

Permeability of free space

=4π×10−7Tm/A=4π×10−7Tm/A

Substituting these values,

BB

becomes as follows:

⇒B=4π×10−74π2×0.480.13⇒B=4π×10−74π2×0.480.13

⇒B=0.96×10−4T=0.96G⇒B=0.96×10−4T=0.96G

The magnetic field is

0.96G0.96G

along the South-North direction.

d=10cm=0.1md=10cm=0.1m

away on the equatorial of the magnet is given as:

B=μ04πMd3B=μ04πMd3

⇒B=4π×10−74π0.480.13⇒B=4π×10−74π0.480.13

⇒B=0.48×10−4T=0.48G⇒B=0.48×10−4T=0.48G

The magnetic field is

0.48G0.48G

along the North-South direction.

14cm14cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36G0.36G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e.,14cm14cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field).

The magnetic field of Earth at the given place,

H=0.36GH=0.36G

The magnetic field at a

dd

-distance, on the axis of the magnet is given as:

B1=μ04π2Md3=HB1=μ04π2Md3=H

Here,

μ0=μ0=

Permeability of free space

=4π×10−7Tm/A=4π×10−7Tm/A

M=M=

The magnetic moment

The magnetic field at the same distance

dd

, on the equatorial line of the magnet is given as:

B2=μ04πMd3B2=μ04πMd3

⇒B2=H/2⇒B2=H/2

(comparing with

B1B1

)

Therefore, the total magnetic field,

B=B1+B2B=B1+B2

⇒B=H+H/2⇒B=H+H/2

⇒B=0.36+0.18=0.54⇒B=0.36+0.18=0.54

Clearly, the magnetic field is 0.54 G along the direction of earth’s magnetic field.

d1=14cmd1=14cm

, can be written as:

B1=μ04π2Md13=HB1=μ04π2Md13=H

here,

MM

is the magnetic moment

μ0μ0

is the permeability of free space

HH

is the horizontal component of the given magnetic field at d1d1

When the bar magnet is turned through

180∘180∘

, then the neutral point will lie on the equatorial line.

Also, the magnetic field at a distance

d2d2

on the equatorial line of the magnet can be written as:

B2=μ04πMd23=HB2=μ04πMd23=H

Equating

B1B1

and

B2B2

we get:

2d13=1d232d13=1d23

⇒d2=d1(12)1/3⇒d2=d1(12)1/3

⇒14×0.794=11.1cm⇒14×0.794=11.1cm

Thus, the new null point will be located

11.1cm11.1cm

on the normal bisector.

. Ignore the length of the magnet in comparison to the distances involved. At what distance from the center of the magnet, the resultant field is inclined at 45∘45∘ with earth’s field on

a) its normal bisector and

The magnetic moment of the bar magnet,

M=5.25×10−2J/TM=5.25×10−2J/T

The magnitude of the Earth’s magnetic field at a place,

G=0.42G=0.42×10−4TG=0.42G=0.42×10−4T

The magnetic field at the distance of R from the centre of the magnet on the normal bisector is given by the relation:

B=μ04πMR3B=μ04πMR3

Here,

MM

is the above-mentioned magnetic moment

μ0μ0

is the permeability of free space

When the resultant field is inclined at

45∘45∘

with earth’s field,

B=HB=H

μ04πMR3=H=0.42×10−4μ04πMR3=H=0.42×10−4

⇒R3=4π×10−74π5.25×10−20.42×10−4=12.5×10−5⇒R3=4π×10−74π5.25×10−20.42×10−4=12.5×10−5

⇒R=5×10−2m=5cm⇒R=5×10−2m=5cm

Clearly, at a distance of 5cm from the center of the magnet, the resultant field is inclined at

45∘45∘

with earth’s field on its normal bisector.

The magnetic moment of the bar magnet,

M=5.25×10−2J/TM=5.25×10−2J/T

The magnitude of the Earth’s magnetic field at a place,

G=0.42G=0.42×10−4TG=0.42G=0.42×10−4T

The given magnetic field at

RR

distanced from the center of the magnet on a point on its axis is given as:

B′=μ04π2MR3B′=μ04π2MR3

The resultant field is inclined at

45∘45∘

with earth’s field

B′=HB′=H

⇒μ04π2M(R′)3=H⇒μ04π2M(R′)3=H

⇒(R′)3=4π×10−74π2×5.25×10−20.42×10−4=2.5×10−4⇒(R′)3=4π×10−74π2×5.25×10−20.42×10−4=2.5×10−4

⇒R=6.3×10−2m=6.3cm⇒R=6.3×10−2m=6.3cm

Clearly, at a distance of 6.3cm from the center of the magnet, the resultant field is inclined at

45∘45∘

with earth’s field on its axis.

a) Why does a paramagnetic sample display greater magnetization (for the same magnetizing field) when cooled?

On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetization when its temperature is lowered i.e., it is cooled.

Hence, the change in temperature that leads to a change in the internal motion of the atoms does not affect the diamagnetism of a material.

Hence the total field generated by the toroid will be slightly less than the empty-core-toroid.

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

It can be seen from the above-given curve that magnetization-B persists even when the external field-H is removed. This shows the irreversibility of a ferromagnet, i.e., the magnetization will not drop by reducing the magnetization field just the same way it was increased by increasing the magnetization field.

The value of magnetization is memory or record of hysteresis loop cycles of magnetization.

Ceramic is usually used for coating magnetic tapes in memory storage devices like cassette players and also for building memory stores in today’s computers.

2.5A2.5A

in the direction

10∘10∘

south of west to

10∘10∘

north of east. The magnetic meridian of the place happens to be

10∘10∘

west of the geographic meridian. The earth’s magnetic field at the location is

0.33G0.33G

, 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.)

Current in the wire,

I=2.5AI=2.5A

The angle of dip at the location,

δ=0∘δ=0∘

The Earth’s magnetic field,

H=0.33G=0.33×10−4TH=0.33G=0.33×10−4T

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

HH=Hcosδ=0.33×10−4×cos0∘=0.33×10−4THH=Hcosδ=0.33×10−4×cos0∘=0.33×10−4T

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

HH=μ02πIRHH=μ02πIR

here,

μ0μ0

= Permeability of free space

=4π×10−7Tm/A=4π×10−7Tm/A

⇒R=μ02πIHH⇒R=μ02πIHH

⇒R=4π×10−72π2.50.33×10−4=15.15×10−3m=1.51cm⇒R=4π×10−72π2.50.33×10−4=15.15×10−3m=1.51cm

Clearly, a set of neutral points lie on a straight line parallel to the cable at a perpendicular distance of

1.51cm1.51cm

above the plane of the paper.

1.0A1.0A

in the same direction east to west. The earth’s magnetic field at the place is

0.39G0.39G

, and the angle of dip is

35∘35∘

. The magnetic declination is nearly zero. What are the resultant magnetic fields at points

4.0cm4.0cm

below and above the cable?

The number of horizontal wires in the telephone cable,

n=4n=4

Current in each wire,

I=1.0AI=1.0A

Earth’s magnetic field at any location,

H=0.39G=0.39×10−4TH=0.39G=0.39×10−4T

The angle of dip at the location,

δ=35∘δ=35∘

and the angle of declination,

θ∼0∘θ∼0∘

For a point that is 4cm below the cable:

Distance,

r=4cm=0.04mr=4cm=0.04m

The horizontal component (parallel to Earth’s Surface) of Earth’s magnetic field is:

HH=H.cosδ−BHH=H.cosδ−B

Here,

BB

is the magnetic field at 4 cm due to current I in the four wires and

B=4μ02πIrB=4μ02πIr

Here,

μ0μ0

is the permeability of free space

=4π×10−7Tm/A=4π×10−7Tm/A

⇒B=44π×10−72π10.01⇒B=44π×10−72π10.01

⇒B=0.2×10−4T=0.2G⇒B=0.2×10−4T=0.2G

∴HH=0.39×cos35∘−0.2≈0.12G∴HH=0.39×cos35∘−0.2≈0.12G

The vertical component (perpendicular to Earth’s surface) of earth’s magnetic field is given as:

Hv=H.sinδHv=H.sinδ

⇒Hv=0.39×sin35∘=0.22G⇒Hv=0.39×sin35∘=0.22G

The angle between the field with its horizontal component is given as:

θ=tan−1HvHHθ=tan−1HvHH

⇒θ=tan−10.220.12=61.39⇒θ=tan−10.220.12=61.39

The resultant field at the point is obtained as:

H1=HH2+Hv2−−−−−−−−−√H1=HH2+Hv2

⇒H1=0.222+0.122−−−−−−−−−−−√=0.25G⇒H1=0.222+0.122=0.25G

For a point that is 4 cm above the cable,

Horizontal component of earth’s magnetic field:

HH=H.cosδ−BHH=H.cosδ−B

⇒HH=0.39cos35∘+0.2=0.52⇒HH=0.39cos35∘+0.2=0.52

Vertical component of earth’s magnetic field:

Hv=H.sinδHv=H.sinδ

⇒Hv=0.39sin35∘=0.22⇒Hv=0.39sin35∘=0.22

The angle

θ=tan−1HvHHθ=tan−1HvHH

⇒θ=tan−10.220.52=22.9∘⇒θ=tan−10.220.52=22.9∘

And the resultant field is:

H1=HH2+Hv2−−−−−−−−−√H1=HH2+Hv2

⇒H1=0.222+0.522−−−−−−−−−−−√=0.56G⇒H1=0.222+0.522=0.56G

Clearly, the resultant magnetic field below the cable is

0.25G0.25G

and above the cable is

0.56G0.56G

45∘45∘ with the magnetic meridian. When the current in the coil is 0.35A0.35A, the needle points west to east.

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

The number of turns in the given circular coil,

N=30N=30

The radius of the given circular coil,

r=12cm=0.12mr=12cm=0.12m

Current in the coil,

I=0.35AI=0.35A

Angle of dip,

δ=45∘δ=45∘

The magnetic field due to the current I , at a distance r , is given as:

B=4μ02πIrB=4μ02πIr

here,

μ0μ0

= Permeability of free space

=4π×10−7Tm/A=4π×10−7Tm/A

⇒B=4π×10−74π2π×30×0.350.12⇒B=4π×10−74π2π×30×0.350.12

⇒B=54.9×10−4T=0.549G⇒B=54.9×10−4T=0.549G

The compass needle points West to East. Hence, the horizontal component of earth’s magnetic field is given as:

BH=B.sinδBH=B.sinδ

⇒BH=0.549sin45∘=0.388G⇒BH=0.549sin45∘=0.388G

Magnitude of one of the magnetic fields,

B1=1.2×10−2TB1=1.2×10−2T

Magnitude of the other magnetic field is

B2B2

Angle between the above-mentioned two fields,

θ=60∘θ=60∘

At the state of stable equilibrium, the angle between the dipole and field

B1B1

is

θ1=15∘θ1=15∘

Angle between the dipole and field

B2B2

is

θ2=θ−θ1=60∘−15∘=45∘θ2=θ−θ1=60∘−15∘=45∘

At a rotational equilibrium, the torques experienced by the dipole, due to both the fields must balance each other.

Therefore, torque due to field

B1=B1=

Torque due to field

B2B2

MB1sinθ1=MB2sinθ2MB1sinθ1=MB2sinθ2

Where,

M=M=

Magnetic moment of the dipole

⇒B2=B1sinθ1sinθ2⇒B2=B1sinθ1sinθ2

⇒B2=4.39×10−3T⇒B2=4.39×10−3T

Clearly, the magnitude of the other magnetic field is

4.39×10−3T4.39×10−3T

. 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.

Energy of an electron beam,

E=18keV=18×103eVE=18keV=18×103eV

Charge on an electron,

e=1.6×10−19Ce=1.6×10−19C

∴E=18×103×1.6×10−19V=2.88×10−15V∴E=18×103×1.6×10−19V=2.88×10−15V

The magnetic field,

B=0.04GB=0.04G

The mass of an electron,

me=9.11×10−31kgme=9.11×10−31kg

Distance till where the electron beam travels,

d=30cm=0.3md=30cm=0.3m

We can write the kinetic energy carried by the electron beam as:

E=12mv2E=12mv2

⇒v=2Em−−−√⇒v=2Em

⇒v=2×2.88×10−159.11×10−31−−−−−−−−−−−−−−√=0.795×108m/s⇒v=2×2.88×10−159.11×10−31=0.795×108m/s

The electron beam deflects and gets in a circular path of radius,rr

The force experienced due to the magnetic field balances the centripetal force of the path.

BeV=mv2rBeV=mv2r

⇒r=mvBe⇒r=mvBe

⇒r=9.11×10−31×0.795×1080.4×10−4×1.6×10−19=11.3m

Let the up-down deflection of the beam be

x=r(1−cosθ)x=r(1−cosθ)

Where,θθ

is the angle of declination given by,

θ=sin−1(d/r)=1.521∘θ=sin−1(d/r)=1.521∘

And

x=11.3(1−cos1.521∘)=0.0039mx=11.3(1−cos1.521∘)=0.0039m

Clearly, the up and down deflection of the bean

=3.9mm=3.9mm

The number of atomic dipoles,

n=2.0×1024n=2.0×1024

Dipole moment for each atomic dipole,

M=1.5×10−23J/TM=1.5×10−23J/T

The given magnetic field,

B1=0.64TB1=0.64T

The sample is then cooled to a temperature,

T1=4.2KT1=4.2K

Total dipole moment of the atomic dipole,

Mtot=n×M=2×1024×1.5×10−23=30J/TMtot=n×M=2×1024×1.5×10−23=30J/T

Magnetic saturation is achieved at

1515

Hence, effective dipole moment,

M1=1510030=4.5J/TM1=1510030=4.5J/T

Now when the magnetic field is

B2=0.98TB2=0.98T

Temperature,

T2=2.8KT2=2.8K

Its total dipole moment

=M2=M2

According to Curie’s law, the ratio of the two magnetic dipoles at different temperatures is:

M2M1=B2B1T1T2M2M1=B2B1T1T2

⇒M2=B2T1M1B1T2⇒M2=B2T1M1B1T2

⇒M2=10.336J/T⇒M2=10.336J/T

Clearly, it can be seen that,

10.336J/T10.336J/T

is the total dipole moment of the sample for a magnetic field of

0.98T0.98T

when its temperature is

2.8K2.8K

in the core for a magnetizing current of 1.2A1.2A?

r=15cm=0.15mr=15cm=0.15m

And the number of turns on a ferromagnetic core,

N=3500N=3500

Relative permeability of the core material,

μr=800μr=800

Given magnetizing current is,

I=1.2AI=1.2A

The magnetic field generated by this current is given by the relation:

B=μrμ02πIrNB=μrμ02πIrN

where,

μ0μ0

is the permeability of free space

=4π×10−7Tm/A=4π×10−7Tm/A

⇒B=800×4π×10−72π1.20.153500=4.48T⇒B=800×4π×10−72π1.20.153500=4.48T

Clearly, the magnetic field in the core is obtained as

4.48T4.48T

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