# NCERT : Magnetism & Matter

Q1. Answer the following questions regarding earth’s magnetism:

(a) A vector needs three quantities for its specification . Name the three independent quantities conventionally used to specify the earth’s magnetic field.

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

(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground ?

(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole ?

(e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 1022 J T–1 located at its centre . Check the order of magnitude of this number in some way.

(f ) Geologists claim that besides the main magnetic N-S poles , there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all ?

Sol. (a) the three quantities are : (i) Magnetic declination (ii) Magnetic Inclination or angle of dip (iii)Horizontal component of Earth magnetic field .

(b) Since Britain is closer to the magnetic north pole hence angle of dip would be greater .

(c) Since Australia is closer to magnetic south pole hence magnetic field lines would come out of the ground .

(d) At geometric poles the direction of magnetic field would be in the vertical direction hence a magnetic needle moving in the horizontal plane would point in any direction .

(e) Here, M= 8×1022 JT-1

Let us calculate magnetic field intensity at magnetic equator of earth, i.e., at a point on equatorial line of short dipole for which, d = R = radius of earth = 6400 km =6.4×106 m.

$\displaystyle B = \frac{\mu_0}{4\pi} \frac{M}{d^3}$

$\displaystyle B = 10^{-7} \times \frac{8 \times 10^{22}}{(6.4 \times 10^6)^3}$

B = 0.31×10-4 T = 0.31 gauss

The value is in good approximation with the observed values of earth’s magnetic field.

(f) The earth’s magnetic field is only approximately a dipole field. Therefore, local N – S poles may exist oriented in different direction. This is possible due to deposits of magnetised minerals.

(a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably ?

(b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why ?

(c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents ?

(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past ?

(e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion ?

(f ) Interstellar space has an extremely weak magnetic field of the order of 10–12 T . Can such a weak field be of any significant consequence ? Explain.

Sol. (a) Yes , due to the motion of its plates and core the earth’s magnetic field does shift . Noticeable changes occur over a large period of time (few hundred years ) and variations are there even in small durations of time (few years ) which cannot be ignored .

(b) The core contains molten iron (magnetic domains are destroyed in molten form ) which is not ferromagnetic hence it can not contribute to magnetism of earth .

(c) No  one really knows .

(d) By study rock magnetism i.e. the solidification of rocks leads to trapping of materials in their magnetism states . Analysing these we can get clues to the history of geomagnetism .

(e) At large distances external factors such as solar winds , ionospheric particles etc. influences the earths magnetic field , hence it distorted .

(f) From cyclotron relation r = mv/qB , We can say that small field can affect the path of moving charged particle whose radius is large .

Q3. A short bar magnet placed with its axis at 30º with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10–2 J. What is the magnitude of magnetic moment of the magnet ?

Sol. B = 0.25T , τ = 4.5 × 10–2 J , θ = 30º

By using formula , τ = MBsinθ , we get M= 0.36 Am2

Q:4. A short bar magnet of moment 0.32 JT-1 is placed in a uniform external magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientations would correspond correspond to its,

(i) stable and

(ii) unstable equilibrium ? what is the potential energy of the magnet in each case ?

Sol.

Here, M = 0.32 JT-1 ,B = 0.15 T

In stable equilibrium, the bar magnet is aligned along the magnetic field, i.e., θ = 0°

Potential Energy=-MB cos⁡0° =-0.32×0.15×1 = -4.8×10-2 J

In unstable equilibrium, the magnet is so oriented that magnetic moment is at 180 to the magnetic field i.e. θ = 180°

Potential energy = -MB cos⁡180° =- 0.32 × 0.15(-1) = 4.8 × 10-2 J.

Q:5.A closely wound solenoid of 800 turns and area of cross section 2.5 × 10-4 m2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment ?

Sol.
Here, n = 800, A = 2.5 × 10-4 m2 ; I=3.0 A

A magnetic field develops along the axis of the solenoid. Therefore, the current carrying solenoid behaves like a bar magnet.

M = n I A

M = 800 × 3.0 × 2.5 × 10-4

= 0.6 JT-1,along the axis of solenoid.

Q:6.If the solenoid in the above question is free to turn about the vertical direction, and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of the torque on the solenoid when its axis makes on angle of 30° with the direction of the applied field?

Sol.
Here, M = 0.6 JT-1 (from Ques.5)

B = 0.25 T ;  τ = ?  ; θ = 30°

As τ = MB sin⁡θ

= 0.6 × 0.25 sin⁡30°

= 0.075 N m.

Q: 7. A bar magnet of magnetic moment 1.5 JT-1 lies aligned with the direction of a uniform magnetic field of 0.22 T.

(a) What is the amount of work done to turn the magnet so as to align its magnetic moment

(i) Normal to the field direction, (ii) opposite to the field direction

(b) What is the torque on the magnet in case (i) and (ii) ?

Sol.
Here, M = 1.5 JT-1 , B = 0.22 T, W = ?

(i) Here, θ1 = 0° (along the field) , θ2 = 90°(⊥ to the field)

As W = -MB(cos⁡θ2-cos⁡θ1 )

W =-1.5×0.22 (cos⁡90° – cos⁡0° )

=-0.33(0-1) =0.33 J

(ii) Here,θ1 = 0° ,θ1 = 180°

W=-1.5×0.22(cos⁡180° – cos⁡0° )

W =-0.33(-1-1) = 0.66 J

Torque τ = M B sin θ

(i) Here, θ = 90°,

τ =1.5×0.22 sin⁡90 =0.33 Nm

(ii) Here, θ = 180° ,

τ =1.5×0.22 sin⁡180 =0

Q:8. A closely wound solenoid of 2000 turns and area of cross section 1.6×10^(-4) m^2, carrying a current of 4 amp. Is suspended through its centre allowing it to turn a horizontal plane :

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

(b)What are the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5×10-2 T is set up at an angle of 30° with the axis of the solenoid

Sol.
N=2000, A=1.6×10-4 m2, I =4 amp.M = ?, B=7.5 ×10-2 T.

As, M = N IA

M = 2000×4×1.6×10-4

=1.28 JT-1

Net force on the solenoid = 0

Torque, τ = M B sinθ

=1.28×7.5×10-2 sin⁡30

=1.28×7.5×102×(1/2)

=4.8×10-2 Nm.

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