MPS/PHY-XII-2011/S8
PHYSICS
ASSIGNMENT-8
SOLUTIONS
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1.
A magnetic dipole is placed in the position of stable equilibrium in a uniform electric
field B (i) How much is the potential energy of the magnet? (ii) If it is rotated through
1800, then what will be the amount of work done?
2.
(i)
In stable equilibrium, U = -MBcos00= -MB
(ii)
If dipole is rotated through 1800, work done is W=MB(cos1800-cos00)=-2MB
What is the basic difference between electric and magnetic lines of force?
Magnetic field lines always form closed loops while electric field lines do not.
3.
Name the parameters needed to completely specify the earth’s magnetic field at a point
on the earth’s surface?
The 3 components of Earths magnetism are (i) declination (θ)(ii) Inclination or Dip(δ) (iii)
Horizontal component of Earth’s Magnetic Field(BH)
4.
Horizontal component of Earth’s magnetic field at a place is
3 times the vertical
component. What is the value of angle of dip at this place?
BH=√3BV =>BV/BH = 1/√3 =>tan δ =1/√3 =>δ=tan-1(1/√3)= 300
5.
Suppose you are facing the north pole of the earth. An electron flying horizontally away
from you in the meridian is deflected to the left. Are you in the northern hemisphere or
southern hemisphere?
r
Using the rt. Hand thumb rule, magnetic field B on the electron must have a component acting
vertically upwards. This can happen only in the southern hemisphere.
6.
Does the magnetization of a paramagnetic salt depend on temperature? Give reason for
your answer?
Yes, Magnetization of a paramagnetic salt decreases with temperature. This is because of the
tendency to disrupt the alignment of atomic dipoles arising from the thermal random motion
increases with the increase in temperature.
7.
Why should the material used for making permanent magnets have high coercivity?
High coercivity means a strong opposite demagnetizing field will only be able to destroy the
residual magnetism which is required of permanent magnets.
8.
Two identical looking iron bars A and B are given, one of which is definitely known to be
magnetized. (We do not which one). How would one ascertain whether or not both are
magnetized? If only one is magnetized, how does one ascertain which one? [Use nothing
else but the two bars A and B]
Repulsion is the surer test of magnetism. If on bringing different ends of two bars closer to one
another, repulsions occur in anyone situation then both the iron rods are magnetized. If the force is
always attractive, then one of them is magnetized. To check whether A of B is magnetized, place the
bar B on a table. Hold the bar A in hand and lower its end on the middle of bar B. If there is an
attraction, then bar A is magnetized other wise B is magnetized.
9.
A coil of n turns and radius R carries a current I. It is unwound and rewound to make
another coil of radius R/2, current remaining the same. Calculate the ratio of the
magnetic moments of the new coil and the original coil.
If L = length of the wire, then L = n2
n2πR = n’2πR/2
No. of turns in new coil n’=2n
Original magnetic moment M=niA = ni
niπR2
New magnetic moment M’=n’iA=2ni(
M’=n’iA=2ni(πR/2)2 = niπR2/2
M’/M = 1:2
10.
The susceptibility of a magnetic material is -0.085.
0.085. Identify the magnetic type of the
substance. A specimen of this material is kept in a uniform magnet
magnetic
ic field. Draw the
modified field pattern.
-ve
ve susceptibility means the substance is diamagnetic. Therefore the
field lines will prefer not to pass through the material.
11.
Two identical magnetic dipoles of magnetic moments 1.0 Am2 each are placed at a
separation of 2m with their axes perpendicular to each other. What is the resultant
magnetic field at a point mid
mid-way between the poles?
r
BA = Baxial =
µ0 2M
along axis of A and
4π (d / 2)3
r
BB= Bequatorial =
µ0 M
parallel to the axis
4π (d / 2)3
The resultant field at point P is B =
2
2
Baxial
+ Bequatorial
.
Resultant field makes an angle θ with BA => tanθ=BB/BA
Substitute the values and get the answer.
12.
An electron in an atom revolves around the nucleus in an orbit of radius 0.5Ǻ. Calculate
the equivalent magnetic moment if the frequency of revolution of the electron is 1010
MHz.
Electron revolving around the nucleus in circular orbit is equivalent to a current loop. Its magnetic
moment M=iA=efπR2=1.257X10-23Amp-m2
13.
A current of 7 A is flowing in a plane circular coil of radius 1 cm having 100 turns. The
coil is placed in a uniform magnetic field of 0.2 Wb m-2. If the coil is free to rotate, what
orientations would correspond to its (i) stable equilibrium and (ii) uns
unstable
table equilibrium?
Calculate the potential energy of the coil in these cases?
N=100, i=7Amp;r=1X10-2m; B=0.2T
M=NiA=Niπr2=0.22Amp=m2
14.
(i)
Stable equilibrium: M parallel to B=> Umin=-MBcos00 = -0.044J
(ii)
Unstable equilibrium: M antiparallel to B => Umax = -MBcos1800 = +0.044J
r
r
µs and µ l associated with the intrinsic spin angular
r
r
momentum S and orbital angular momentum l , respectively, of an electron are
The magnetic moment vectors
predicted by quantum theory
(and verified experimentally to a high accuracy) to be
r
 -e  r r
 e  r Which of these relations is in accordance with the result
given by: µ
s = -
 S,µ l = - 
l
m
 2m 
expected ‘classically’? Outline the deriv
derivation of the classical result.
In contrast to µl/l, the magnitude of µ s/S is e/m i.e., twice the classically expected value. This latter
result(verified experimentally) is an outstanding success of modern quantum theory and cannot be
derived from classical physics.
15.
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its
plane normal to an external field of magnitude 5.0 X 10-2T. The coil is free to turn about
an axis in its plane perpendicular to the field direction. When tthe
he coil is turned slightly
and released, it oscillates about its stable equilibrium with a frequency 2.0 s-1. What is
the moment of inertia of the coil about its axis of rotation?
N=16, r=10cm=0.10m ; i=0.75A, B=5X10-2T, f=2s-1
M=NiA=Niπr2 ; f =
I=
16.
1
2π
MB
; I=moment of inertia.
I
MB
Niπ r 2 B
=
=1.2 X 10-4kgm-2
4π 2 f 2 4π 2 f 2
If δ1 and δ2 be the angles of dip observed in two vertical planes at right angles to each
other and δ is the true angle of dip, prove that cot2 δ1 + cot2 δ2 = cot2 δ.
r
Let BH and BV be the horizontal and vertical components of earth
earth’ss magnetic field B . δ = true dip;
tan δ = BV/BH => cot δ = BH/BV
Let Plane 1 and 2 are two mutually perpendicular planes and making angle θ and 90-θ with
magnetic meridian. The vertical components of earth
earth’ss magnetic field remain same in the two
planes but the effective horizontal co
components in the two
planes is given by B1 = BHcosθ
and B2 = BHsinθ.
Then tanδ1 = BV/B1 = BV/BHcosθ
θ =>cotδ1 = BHcosθ/BV
tanδ1 = BV/B1 = BV/BHsinθ =>cot
=>cotδ2 = BHsinθ/BV
BH2
BH2
2
2
Therefore cot δ1 + cot δ 2 = 2 (cos θ + sin θ ) = 2
BV
BV
2
2
cot 2 δ1 + cot 2 δ 2 = cot 2 δ
17.
A sample of paramagnetic salt contains 2.0 X 1024 atomic dipoles each of dipole moment
1.5 X 10-23 JT-1. The sample is placed under homogeneous magnetic field of 0.84 T and
cooled to the temperature of 4.2 K. The degree of magnetic saturation achieved is equal
to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and
a temperature of
2.8 K(assume Curie’s Law)?
Diple moment of each atomic dipole M= 1.5 X 10-23J/T
Total no. of dipoles = N =2 X1024 =>Mtotal=1.5 X10-23 X 2 X 1024=30J/T
Initial total magnetic moment at temp T1= 4.2 K =M=15% of Mtotal= 4.5 J/T
By Curie’s Law, I = CH/T =>
I 2 H 2 T1 µ H 2 T1 B2 T1
=
=
=
I1 H1 T2 µ H1 T2 B1 T2
B1=0.84T ; T1 = 4.2 K; B2=0.98 T; T2=2.8K
∴ I 2 = I1
18.
B2 T1
= 7.9T
B1 T2
At a certain location in Africa, a compass points 120 west of the geographic north. The
north tip of the magnetic needle of a dip circle placed in the plane of the magnetic
meridian points 600 above the horizontal. The horizontal component of the earth’s field
is measured to be 0.16G. Specify the direction and magnitude of earth’s field at the
location.
BH = 0.16G; δ = 600 ∴ B =
19.
BH
0.16
=
= 0.32G
0
cos 60
cos 600
A solenoid has a core of a material with relative permeability 400. The windings of the
solenoid are insulated from the core and carry a current of 2A. If the number of turns in
1000 per meter, calculate (a) H (b) M (c) B and (d) the magnetizing intensity IM .
n = 1000 turns/m ; i=2A; µr = 400
(a) H=ni=2 X 103A/m
(b) B = µH =µr µ0H = 1 T
(c) I = χmH = (µr-1)H≈8X105 Amp/m
(d) im= I/n = 8 X 103 A
20.
What is the physical significance of the area under the B~H (Hysteresis) curve?
The area within a B-H loop represents the energy dissipated per unit volume in the material when it
is carried through a cycle of magnetization
[Area]=[BH]=[B][B/µ]=[B2]/[ µ]=[B2]/ µoµr= Dimensions of Energy/Volume