Download question bank tabulated UNIT 3

KENDRIYA VIDYALAYA SANGATHAN, AHMEDABAD REGION
CLUSTER LEVEL WORKSHOP
AHMEDABAD-GANDHINAGAR CLUSTER
DATE:12/12/2015
TOPIC: MAGNETIC EFFECTS OF CURRENT AND MAGNETISM
One mark question
Q1
What is the work done by the magnetic force on a charged particle moving perpendicular to the
magnetic field?
A
zero
Q2
A
Why does the Kinetic Energy of the charged particle does not change while moving through a
perpendicular uniform magnetic field?
The speed of the particle moving through a perpendicular magnetic field is v= qBr/m.
This speed does not change wit time, therefore, kinetic energy remain constant.
TWO MARKS QUESTION
Q3
A
Q4
The vertical component of earth’s magnetic field at a place is √3 times the horizontal
component. What is the value of angle of dip at this place?
Bv=√3BH
BV/BH =√3
tan δ= √3
therefore, the angle of dip is
δ= 600
Define one ampere.
Definition : One ampere is the value of steady current which when maintained in each of the
two very long, straight, parallel conductors of negligible cross section and placed one metre
apart in vacuum, would produce on each of these conductors a force equal of 2 x 10-7 N/m of
its length.
Q5
Ans
Q6
ans
Q7.
Explain why steel is preferred for making permanent magnets while soft iron is preferred for
making electromagnets.
Steel has high retentivity and High coercivity therefore it is preferred for making permanent
magnets. Whereas soft iron has low retentivity and low coercivity, therefore it is preferred for
making electromagnets.
How will the magnetic filed intensity at the centre of a circular coil carrying current change if
the current through the coil is doubled and the radius of the coil is halved?
B = μ0n x 2I / 2 x (R/2) = 4B
Write the factors on which current sensitivity and voltage sensitivity of MCG depends
Factors affecting the sensitivity:
1. Number of turns N in its coil, I s  N
2. Magnetic field B
Is  B
3. Area A of the coil, I s  A
4. Torsion constant k of the spring, I s 
Q8
1
k
Draw the variation of magnetisation (M or I)with the magnetising field (H) for a)diamagnetic
substances b) paramagnetic substances c)ferromagnetic substances
Three marks questions
Q9
Draw the variation of magnetic susceptibility (χ) with temperature(T) for
a)diamagnetic substances b) paramagnetic substances c)ferromagnetic substances
Q10.
What are the elements of Earth’s magnetic field? Define these with the help of a labelled
diagram?
Five marks questions
Q11
Differentiate between diamagnetic, paramagnetic and ferromagnetic substances with two
examples of each.
S.No. DIA
1
Diamagnetic substances are
PARA
FERRO
those substances which are
feebly repelled by a magnet.
Paramagnetic substances are
those substances which are
feebly attracted by a magnet.
Eg. Antimony, Bismuth,
Copper, Gold, Silver,
Eg. Aluminium, Chromium,
Platinum, Oxygen
Ferromagnetic
substances are those
substances which are
strongly attracted by a
magnet.
Eg. Iron, Cobalt, Nickel,
Gadolinium,
Dysprosium,
2
When placed in magnetic field,
the lines of force tend to avoid
the substance.
The lines of force prefer to
pass through the substance
rather than air.
The lines of force tend
to crowd into the
specimen.
3.
Induced Dipole Moment (M) is
a small – ve value.
Induced Dipole Moment (M) is
a small + ve value.
Induced Dipole
Moment (M) is a large
+ ve value.
4
Magnetic permeability μ is
always less than unity.
Magnetic permeability μ is
more than unity.
Magnetic permeability
μ is large i.e. much
more than unity.
5
Magnetic susceptibility cm has a
small – ve value.
Magnetic susceptibility cm has
a small + ve value.
Magnetic susceptibility
cm has a large + ve
value.
6.
Q12
They do not obey Curie’s Law.
i.e. their properties do not
change with temperature.
They obey Curie’s Law. They
lose their magnetic properties
with rise in temperature.
They obey Curie’s Law.
At a certain
temperature called
Curie Point, they lose
ferromagnetic
properties and behave
like paramagnetic
substances.
Describe the principle, construction and working of a moving coil galvanometer. Hence define
current sensitivity and voltage sensitivity of moving coil galvanometer.
Principle and working : A current carrying coil, placed in a uniform magnetic field, (can)
experience a torque
Consider a rectangular coil for which no. of turns = N,
Area of cross- section = l×b=A,
Intensity of the uniform magnetic field=B,
Current through the coil=I
Deflecting torque = 𝐵𝐼𝐿 × 𝑏 = 𝐵𝐼𝐴
𝐹𝑜𝑟 𝑁 𝑡𝑢𝑟𝑛𝑠 𝜏 =NBIA
Restoring torque in the spring = k𝜃
(k=restoring torque per unit twist)
∴ 𝑁𝐵𝐼𝐴 = 𝑘𝜃
𝑘
∴𝐼=(
)𝜃
𝑁𝐵𝐴
∴𝐼 ∝ 𝜃
The deflection of the coil, is, therefore, proportional to the current flowing through it.
Current sensitivity: It is the deflection produced in the galvanometer when a unit
current flows through it.
 NBA
Current sensitivity, I s  
I
k
Voltage sensitivity: It is the deflection produced in the galvanometer when a unit
potential difference is applied across its ends.
 NBA
Voltage sensitivity, Vs  
V k RG
Q13.
Write an expression for force experienced by a charged particle moving in a uniform
magnetic field? With the help of labelled diagram, explain principle and working of a
cyclotron. Show that cyclotron frequency does not depend upon the speed of particles. Write
its two limitations.
Principle: A charged particle can be accelerated to a high speed by passing it through a
moderate electric field number of times in presence of an uniform perpendicular magnetic field
which through the charged particle in circular motion.
Construction: The cyclotron is made up of two hollow semi-circular disc like metal containers,
D1 and D2, called dees.
It uses crossed electric and magnetic fields. The electric field is provided by an oscillator of
adjustable frequency.
Working: In a cyclotron, the frequency of the applied alternating field is adjusted to be equal to
the frequency of revolution of the charged particles in the magnetic field. This ensures that the
particles get accelerated every time they cross the space between the two dees. The radius of
their path increases with increase in energy and they are finally made to leave the system via
an exit slit.
When a particle of mass ‘m’ and charge ‘q’ , moves with a velocity 𝛎, in a uniform magnetic
field 𝑩 , it experiences a force 𝑭 where
F= qv B sinθ = q v B sin90o= q v B
This provides the necessary centripetal
∴ Centripetal force
∴𝑟=
𝑚𝑣
𝑞𝐵
𝑣
𝑚𝑣 2
𝑟
=𝑞𝑣𝐵
𝑞𝐵
frequency =
=
2𝜋𝑟 2𝜋𝑚
∴ frequency is independent of the velocity or the energy of the particle.
Limitations :
1. Neutral particles like neutrons cannot be accelerated .
2. Due to relativity mass of the particle increases with increase in the speed of the particle.
As a result frequency and time period of the particle change.
Due to this light particles such as electrons cannot be accelerated using this.
Q14.
State Ampere Circuital law. Using it derive the expression for the magnetic field intensity due a
current carrying torroid at any point lying in its interior and exterior.
(a) Statement of law:
The line integral of the magnetic field around a closed loop is µo times of the total current
threading through the loop.
Expression of the law in integral form:
⃗⃗⃗ = 𝜇0 𝑖
⃗ . 𝑑𝑙
∮𝐵
(b) In the interior of the torroid
⃗⃗⃗ = 𝜇0 𝑁𝑖
⃗ . 𝑑𝑙
∮𝐵
𝐵𝑥 2𝜋𝑟 = 𝜇0 𝑁𝑖
𝐵 = 𝜇0 𝑁𝑖/2𝜋𝑟 = 𝜇0 𝑛𝑖
(c) In the exterior of the torroid
here n = N/2𝜋𝑟
⃗⃗⃗ = 𝜇 𝑁(0)
⃗ . 𝑑𝑙
∮𝐵
0
⃗⃗⃗ = 0
⃗ . 𝑑𝑙
∮𝐵
𝐵=0
Q15.
Using Biot - Savart’s law deduce the expression for the magnetic field intensity due to a
circular current carrying loop at any point on it axial line.
Consider a circular loop of wire of radius a and carrying current I, as shown in figure.
Let the plane of the loop be perpendicular to the plane of paper. We wish to find field
P at a distance r from the centre C.

B
at an axial point

dl at the top of the loop. It has an outward coming current.


s be the position vector of point P relative to the element dl , then from Biot-Savart law, the field at
Consider a current element
If
point P due to the current element is
dB 
Since
 0 I dl sin 
4
r2


dl  s , i.e.,
dB 
 0 Idl
4 s 2
θ = 90 , therefore


The field dB lies in the plane of paper and is perpendicular to s , as shown by PQ. Let
between OP and CP. Then dB can be resolved into two rectangular components.

 be the angle

1. dB sin
along the axis,
2. dB cos
perpendicular to the axis.
For any two diametrically opposite elements of the loop, the components perpendicular to the axis of the
loop will be equal and opposite and will cancel out. Their axial components will be in the same direction,
i.e., along CP and get added up.
Therefore, total magnetic field at the point P in the direction CP is
B=
 dB sin 
a
s and
 0 Idl a
 4 . s 2 . s
sin  
But
B=
Since
B =
dB 
 0 Idl
4 s 2
 0 and I are constant, and s and a are same for all points on the circular loop, we have
 0 Ia
 0 Ia
0 I a 2
dI

.
2

a

4 s 3 
4 s 3
2 s3
As,
s  (r 2  a 2 )
1
2
[  dI  Circumference  2 a]
 B
0 I a 2
2 (r 2  a 2 )
3
2
B
0 N I a 2
2 (r  a )
2
2
3
.
2
If the coil consists on N turns, then
Magnetic field at the centre: For the field at the centre of the loop, r = 0. Therefore
B
0 N I
2a