Std. XII Sci. Success Physics - II
14
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MAGNETIC EFFECT OF ELECTRIC CURRENT
Contents
14.0 Introduction
14.1 Ampere’s law and its applications
14.2 Moving coil Galvanometer
14.3 Ammeter
14.4 Voltmeter
14.5 Sensitivity and accuracy of M.C.G.
14.6 Cyclotron
Formulae
Solved Problems
Problems for practice
Multiple Choice Questions
Short Test
14.0 Introduction
1.
Ans.
HORIZON Publication
Define magnetic effect of electric
current. State the expression for force
acting on a current carrying conductor
placed in a uniform magnetic field.
i.
The effect in which an electric
current flowing through a conductor
produces magnetic field around it is
called magnetic effect of an electric
current.
ii. The expression for force acting on a
conductor of length l carrying
current I, kept in uniform magnetic
field of induction B is given by
F  Il  B
iii.
Ans.
Statement : The line integral of magnetic
field of induction ( B) around any closed
path in free space is equal to absolute
permeability of free space (0) times the
total current flowing through area
bounded by the path.
Mathematically,  B  dl   0 I
Explanation :
i.
Ampere’s law is generalization of
Biot-Savart’s law and is used to
determine magnetic field at any point
due to distribution of current.
ii. Consider a long straight conductor
XY, carrying current is placed in
vacuum. A steady current I flows
through it from the end Y to X as
shown in figure.
 F = I l B sin  (magnitude)
where  is the angle between the
length of the conductor and magnetic
induction.
If conductor is at right angle to the
magnetic field then  = 900
 F = I l B sin 900 = I l B
14.1 Ampere’s law and its applications
*2.
State and explain Ampere’s circuital
law.
[Oct. 11, Mar. 12, 14]
Magnetic Effects of Electric Current
iii.
Imagine a closed curve (amperian
loop) around the conductor having
radius r.
14.1
Std. XII Sci. Success Physics - II
www.horizonpublication.com
iv.
The loop is assumed to be made of
large number of small elements each
of length d l .
v.
Its direction is along the direction of
traced loop.
vi. Let B be the strength of magnetic
field around the conductor.
viii. All the scalar products of B and dl
give the product of 0 and I.
It is given by
iv.
 B  dl   B dl cos 
 B  d l   B dl

Ans.
iii.
 B  d l   B dl  B  dl
But  dl  2  r

where  = angle between B and dl
Derive an expression for magnetic
induction at a point near infinitely long
straight conductor carrying an electric
current on the basis of Ampere’s law.
[Mar. 09]
Expression for the magnetic induction
at a point due to a long straight current
carrying conductor.
i.
Consider a long straight conductor
XY carrying a current I. A point P is
at a distance r from the conductor.
We have to find the magnetic
induction at P due to this current
carrying conductor.
ii. Consider a circular Amperian loop of
radius r, drawn in a plane
perpendicular to
the straight
conductor, with the conductor
passing through the centre of the
circle. The direction of current is
outwards at right angles to the plane
of the circle.
[ cos 00 = 1]
[ B = constant]
 B . dl   B dl cos 
*3.
Hence, the line integral along the
closed loop is given by
v.
 B  dl  B (2r )
According to Ampere’s law
 B  dl  
vi.
….(1)
0
I
….(2)
Comparing equations (1) and (2)
we get,
B (2 r) = 0 I
 I

B 0
2 r
0 2 I

4 r
This is the required expression.

4.
Ans.
B
What is solenoid ? Obtain an expression
for magnetic induction along the axis of
long solenoid using Ampere’s law.
Solenoid : A coil of insulated wire wound
around an insulating hollow cylinder in
which diameter of coil is smaller than its
length is called solenoid.
Expression for magnetic induction :
i.
Consider a long straight solenoid as
shown in figure.
ii. Consider a rectangular closed path
PQRS with side PQ along the axis of
solenoid.
At every point of the loop, the
magnetic induction B is directed in
the tangential path. Therefore the
angle between B and the current
element d l is zero at all the points.
i.e.  = 00
Magnetic Effects of Electric Current
iii.
Let
PQ = RS = L = length of rectangular
path
14.2
Std. XII Sci. Success Physics - II
iv.
n = number of turns per unit length
of solenoid
nl = number of turns in length l of
solenoid i.e. number of turns in
PQRS of solenoid.
I = current through the solenoid
B = magnitude of magnetic induction
insider the solenoid
The line integral of magnetic
induction over closed path PQRS is
given by,
 B  dl 
v.
HORIZON Publication
Q
R
S
P
P
Q
R
S
According to Ampere’s circuital law,

 B  dl   0 n L I
x.
From equation (6) and (7) we have,
Bl = 0 (nL  I)
 B = 0n I
This is the required expression for
magnetic induction along the axis of
long solenoid.
Magnetic induction at a point near
end of solenoid is
1
B end  B axis
2
1

Bend   0 n I
2
xi.
 B  dl   B  dl   B  dl   B  dl
….(1)
The side PQ of loop is parallel to
magnetic induction, hence  = 00

ix.
Q
Q
5.
P
P
Ans.
 B  dl   B dl cos 
Q
  B dl cos 0
P
Q
= B  dl
What is toroid ? Derive expression for
magnetic induction due to toroid. [Mar. 14]
Toroid : A toroid is a ring shaped closed
solenoid.
Expression for magnetic induction :
i.
Consider a thin toroid having centre
O and radius r as shown in figure.
P
Q

 B  dl  BL
….(2)
P
vi.
The sides QR and SP are
perpendicular to magnetic field,
hence  = 900.

R
R
 B  dl   B dl cos 
Q
Q
R
  B dl cos 90
ii.
Q
R
 B  dl  0
….(3)
Q
P
Similarly  B  dl  0
….(4)
S
vii. The side RS is outside the solenoid,
hence, B = 0
S

 B  dl  0
….(5)
iii.
iv.
R
viii. From equation (1), (2), (3), (4) and
(5), we get


 B  dl  BL
 B  dl  BL
000
….(6)
Magnetic Effects of Electric Current
v.
Let,
N = Total number of turns per unit
length
r = mean
radius
of
toroid
(amperian loop)
I = current passing through the
toroid
B = magnitude
of
magnetic
induction inside the toroid.
Consider a closed path in the form of
a circle of radius r.
The
magnitude
of
magnetic
induction B is same everywhere on
closed path.
Due to the current in conductor,
magnetic field is produced around it.
The magnetic field is in the form of
concentric circles centred on the
conductor.
14.3
Std. XII Sci. Success Physics - II
vi.
www.horizonpublication.com
Consider a small element d l at a
point on the circle. The magnetic
induction is tangent to path. The
7.
angle between B and d l is 0.
Ans.
Distinguish between suspended coil type
and pivoted
coil type moving coil
galvanometer.
i.
vii. The line integral of B over this
closed path is given by,
ii.
 B  dl   B dl cos 
  B dl cos 0
  B d l ( cos 0 = 1)
 B  dl
o
iii. It
is
more
sensitive
as
compared
to
pivoted coil type
MCG.
iv. It
is
not
portable.
But,  d l  l

 B  dl  B 2 r
….(1)
viii. According to Ampere’s circuital law,

ix.
x.
 B  dl   N  2 r  I ….(2)
0
From equation (1) and (2) we have,
B.2 r = 0 (N  2 r  I)
 B = 0N I
….(3)
This is the required expression for
magnetic induction due to toroid.
If n is the number of turns per unit
length of toroid then n = N/2r
 B = 0 n I
Suspended Coil
Type MCG
In
suspended
coil type MCG,
the rectangular
coil
is
suspended
in
radial magnetic
field by means
of thin phosphor
bronze wire.
The deflection
of the coil can
be measured by
lamp and scale
arrangement.
*8.
Ans.
Pivoted
Coil
Type MCG
In pivoted coil
type MCG, the
rectangular coil
is pivoted in
radial magnetic
field by means
of two spiral
springs.
The deflection
of the coil can
be measured by
a
long
aluminium
pointer on a
linear scale.
It
is
less
sensitive
as
compared
to
suspended coil
type MCG.
It is portable.
With the help of neat diagram explain
the construction of moving coil
galvanometer.
[Mar. 09]
Construction of MCG :
14.2 Moving coil Galvanometer
6.
Ans.
What is a galvanometer ? State
different types of galvanometer.
i.
Galvanometer is a device with the
help of which a very small electric
current can be detected.
ii. Moving Coil Galvanometers are of
two types :
a.
Suspended coil type
b.
Pivoted coil type
Magnetic Effects of Electric Current
14.4
Std. XII Sci. Success Physics - II
W : Phosphor bronze wire
M : Mirror
N-S : Concave pole pieces
H : Helical spring
I : Current
C : Soft iron core
PQRS : Rectangular coil
The MCG consists of the following four
main parts :
i.
Rectangular coil : A rectangular
coil (PQRS) of large number of turns
is kept suspended by means of a thin
phosphor bronze wire (W). The other
end of the coil is connected to a
helical spring (H) which helps the
coil to bring back to its original
position when current is cut off.
ii. Plane mirror : A small plane mirror
(M) is attached on phosphor bronze
wire suspension. This along with
lamp and scale arrangement is used
to measure the deflection.
iii. Magnet : A horseshoe magnet
having cylindrical concave pole
pieces is producing a strong
magnetic filed in the space between
the pole pieces.
iv. Soft iron core : A soft cylindrical
iron core (C) is fixed between the
pole pieces so that the coil is free to
turn in the gap between the core and
magnetic pole pieces.
*9.
Ans.
Explain the principle and working of
suspended coil type MCG.
[Sep. 08, Mar. 10]
OR
*Show that current through the coil of a
MCG is directly proportional to the
deflection of the coil.
Principle : When a current carrying coil
is suspended in a uniform magnetic field, a
torque is said to act on it. This torque
tends to rotate the coil about the axis of
suspension so that the magnetic flux
passing through the coil is maximum.
Magnetic Effects of Electric Current
HORIZON Publication
Working :
i.
ii.
iii.
iv.

Let PQRS represent a turn of
rectangular coil of wire of length l
and breadth b. Let A be the area of
the coil.
When an electric current I flows
through the coil PQRS, its vertical
conductors PQ and SR are acted
upon by forces each of magnitude
nBIl where n is the number of turns
in the coil, l is the length of the
conductor and B is the magnetic
induction of the magnetic filed.
Forces F1 and F2 acting on
conductors PQ and SR are equal in
magnitude but opposite direction
thus form a couple.
The horizontal sides PS and QR are
parallel to the magnetic field so do
not experience any force.
Now, moment of couple or torque :
 = (magnitude of one force) 
(perpendicular distance between two
parallel forces)
 = (nBIl) (b)
  = n BIA
( A = l b)
This torque deflects the coil and
hence is called deflecting torque.
Deflecting torque d = n BIA ….(1)
14.5
Std. XII Sci. Success Physics - II
v.
vi.
10.
Ans.
11.
Ans.
12.
Since the magnetic field is radial, the
deflecting torque in all positions of
the coil is the same and equal to
nBIA. As the coil is deflected, the
phosphor bronze wire is twisted.
This twist provides the restoring
torque. It is given by
R = k
….(2)
where k is twist constant and  is the
angle of twist.
For equilibrium of the coil,
Deflecting torque = Restoring torque
 D = R
 nI A B = k 
 k 
 
I  

 nB A 
This is the expression for current
flowing through the coil of MCG.
As k, n, A & B in equation (3) are
constant.
 I
Thus, the current flowing through the
MCG is directly proportional to
angle of deflection of the coil.
Draw a neat and labelled diagram of
suspended coil type M.C.G. [Mar. 15]
Ref. Q.8
State the main advantages and
disadvantages of MCG.
Advantages of MCG :
i.
It is not affected by strong magnetic
field.
ii. It has high torque/weight ratio.
iii. It is very accurate and reliable.
iv. As, I   therefore scales is uniform
Disadvantages of MCG :
i.
Changes in temperature affect
restoring torque.
ii. Restoring torque cannot be easily
changed.
iii. Possibility of damage of phosphor
bronze fibre suspension and hair
helical springs arising out of severe
stress.
iv. It can not be used for alternating
current (a.c.) measurement.
Why is it necessary to have radial
magnetic field for a MCG ? How is it
produced ?
Magnetic Effects of Electric Current
www.horizonpublication.com
Ans.
i.
ii.
If the field is radial whatever may be
the position of the coil, the field is
parallel to the plane of the coil and
deflecting torque acting on the coil is
of constant magnitude. Hence, the
current (I) through the coil is directly
proportional to the deflection () of the
coil. As a result, we have a linear scale.
To produce the radial magnetic field
cylindrical concave pole pieces are
used. They provide a strong radial
magnetic field in the space between
the poles.
14.3 Ammeter
*13.
Ans.
Explain, how a moving coil galvanometer
is converted into an ammeter. Derive
the necessary formula.
[Mar. 12]
i.
An ammeter is used to measure
current in a circuit.
ii. A moving coil galvanometer can be
converted into an ammeter by
connecting a low resistance called
shunt parallel to it.
iii. The value of shunt is so adjusted that
when current I flows, only the part Ig
of the current passes through the coil
of MCG and the remaining part
(I - Ig) = Is flows through shunt S.
iv.
v.
Let G be the resistance of the
galvanometer, S be the resistance of
the shunt, Ig be the current for which
the galvanometer shows a full scale
deflection, I be the current to be
measured.
As resistances S and G are parallel,
 P.D. across S = P.D. across G
 Is S = Ig G
 (I - Ig) S = Ig G
G Ig
 S
I  Ig
This is the required value of shunt
resistance.
14.6
Std. XII Sci. Success Physics - II
14.
Ans.
Derive the formula for current flowing
through a shunt in ammeter.
i.
Since, GIg = S (I - Ig)
GIg = SI - SIg
SI = GIg + SIg
SI = Ig (G + S)
 S 
 I
I g  
….(1)
G

S


Equation (1) gives the part of the
total current flowing through the
galvanometer coil.
ii. Let I be ‘n’ times Ig then
I = nIg
….(2)
 Ig 
G
Also S  
….(3)
I  I 
g 

Substituting equation (2) in (3),
 Ig

G
S
 nI  I 
g
g


 G 

S  
 n  1
G  S
….(4)
n

 S 
Since IS = I - Ig
 G 
 I [From eqn (4)]
I S  

S  G 

iii.
15.
Ans.
16.
Ans.
State the advantage of using a shunt to
convert galvanometer into ammeter.
i.
Shunt is a low resistance connected
in parallel to any electric circuit.
ii. It reduces the effective resistance of
an ammeter.
iii. It increases the range of an ammeter.
iv. It provides an alternate path for
excess current to pass, which
protects the galvanometer from the
damage.
Explain why an ammeter should have
low resistance.
i.
To measure current in a circuit, an
ammeter is connected in series with
the circuit. As the galvanometer coil
has some resistance, its introduction
in the circuit increases the total
resistance of the circuit. This in turn
reduces the current in the circuit.
Magnetic Effects of Electric Current
HORIZON Publication
ii.
iii.
As a result, the current measured is
less than the actual current through
the conductor.
In order to reduce the inaccuracy in
measuring current, an ammeter must
have low resistance. Lower the
resistance, more accurate is the
current measured.
14.4 Voltmeter
*17.
Ans.
Explain
how
a
moving
coil
galvanometer
is
converted
into
voltmeter.
Derive
the
necessary
formula.
[Mar. 08]
i.
A voltmeter is an instrument which
is used to measure the potential
difference in volt between any two
points in a circuit.
ii. A moving coil galvanometer can be
converted into a voltmeter by
connecting a high resistance in series
with the galvanometer.
iii.
iv.
Let G be the resistance of the
galvanometer, Ig be the current
required for the full scale deflection,
V be the maximum potential
difference to be measured, R be the
high resistance connected in series.
In the series combination, the
potential difference V gets divided
across the galvanometer (G) and
resistance (R)
 V = G Ig + R Ig
 V = Ig (G + R)

V
GR
Ig

R 
V
G
Ig
This is the required
resistance in series.
value of
14.7
Std. XII Sci. Success Physics - II
18.
Ans.
19.
Ans.
20.
www.horizonpublication.com
Why a voltmeter must have very high
resistance ?
i.
A voltmeter is used to measure
potential difference between two
points in an electrical circuit. In a
circuit, a voltmeter is always
connected in parallel.
ii. So when a voltmeter is introduced in
a circuit to measure voltage, it
provides a parallel path to the current
and decreases the current through the
conductor.
iii. As a result, the voltage measured is
less than the actual voltage across the
conductor. In order to reduce the
inaccuracy in measuring potential
difference, a voltmeter must have a
high resistance.
iv. Higher the resistance, more accurate
is the voltage measured.
v.
Since large value resistance with
high precision is not easily available,
voltmeter of high voltage range
cannot be obtained.
14.5 Sensitivity and accuracy of M.C.G.
*21.
Ans.
What is the function of high resistance
in volt meter ?
i.
To increase the effective resistance
of the voltmeter.
ii. To protect the galvanometer from
damage due to large current.
iii. To increase the range of voltmeter.
Distinguish
voltmeter.
between
ammeter
and
22.
Ans.
i.
ii.
iii.
iv.
Ammeter
It
measures
current in the
circuit.
It is connected
in series.
It
has
low
effective
resistance.
Resistance of
ideal ammeter
is zero.
Voltmeter
It
measure
potential
difference.
It is connected
in parallel.
It has high
effective
resistance.
Resistance of
ideal voltmeter
is 
Magnetic Effects of Electric Current
Define sensitivity of a moving coil
galvanometer. Derive expression for
current sensitivity of a moving coil
galvanometer.
[Mar. 13]
i.
Sensitivity of M.C.G. : Sensitivity of
M.C.G. is defined as the ratio of
change in deflection (d) to the
change in current (dI) passing
through it.
Mathematically
d
Si 
dI
ii. For M.C.G. we have,
C
…..(1)
I

nBA
Where,
C = restoring torque per unit twist
n = number of turns of coil
B = magnetic induction
A = area of coil
iii. Differentiating eqn (1) w.r.t. ,
we get,
C
dI
d  C 


 
nBA
d d  nBA 
d nBA


dI
C
d
 Si  sensitivit y
iv. But
dI
nBA
 Si 
C
This is the expression for sensitivity
of a M.C.G.
Ans.
State the factors on which current
sensitivity of MCG depends.
OR
*State formula for sensitivity of MCG.
How can sensitivity be increased.
[Oct. 10]
i.
Current sensitivity of MCG is given by,
nBA
Si 
….(1)
C
where,
n = number of turns
B = magnetic induction
A = area of coil
C = restoring torque per unit twist
14.8
Std. XII Sci. Success Physics - II
ii.
23.
Ans.
24.
Ans.
From equation (1) it is clear that
sensitivity of MCG can be increase by
a.
increasing the number of turns
(n),
b. increasing
the
magnetic
induction (B) ,
c.
increasing the area of coil (A),
d.
decreasing restoring torque per
unit twist (C).
Why sensitivity of MCG cannot be
increased beyond a certain limit.
i.
If a strong magnet is taken, the
instrument become bulky.
ii. Similarly, if no. of turns and area of
coil increases coil become heavy.
iii. Then suspension fibre may not be
able to support it.
iv. If we take thick suspension fibre
twist constant (C) would increase
and sensitivity decreases. To
minimise twist constant, a flat strip
of phosphor bronze wire is taken.
v.
Thus, there is a limit beyond which
sensitivity of M.C.G. can not be
increased.
What do you mean by accuracy of
MCG? How can it be increased in MCG?
Accuracy of MCG :
i.
Moving coil galvanometer is said to
be more accurate if the relative error
in current is less and vice versa.
ii. Current flowing through coil of
MCG :
 k 
….(1)
I

 nBA 
iii. Differentiate with
respect
to
variables :
 k 
dI  
 d ….(2)
 nBA 
iv. Dividing equation (2) by (1), we get
dI d

….(3)
I

This is the condition for accuracy of
MCG.
v.
For an accurate galvanometer, the
fractional error must be least.
Thus, to increase the accuracy
a.
increase the deflection  and
b.
decrease d
Magnetic Effects of Electric Current
HORIZON Publication
14.6 Cyclotron
25.
Draw a labeled diagram of cyclotron
showing accelerated charged particle by
dees.
Ans.
26.
Ans.
What is a cyclotron ? State its principle.
Cyclotron is a cyclic accelerator used to
accelerate charged particles to acquire
enough energy to carry out nuclear
reactions.
Principle : When a positively charged
particle moves with a periodic time again
and again perpendicular to uniform
magnetic field and is accelerated
continuously by high frequency electric
field, it traces a spiral path of increasing
radius. The positively charged particle is
accelerated to high speed and gains
sufficient large amount of energy.
27.
Explain the construction and working
of a cyclotron.
[Oct. 10]
Construction :
Ans.
14.9
Std. XII Sci. Success Physics - II
It consists of two D-shaped hollow
semicircular metal boxes called
Dees. There is a small gap between
the two dees (D1 and D2).
ii. The dees are externally connected to
a high alternating potential of the
order of 10000 volts and frequency
10 MHz.
iii. These dees are mounted between two
powerful pieces of electromagnet
producing vertical magnetic field. A
source of positive ion is kept at the
centre of the dees gap.
Working :
i.
The positive ions are produced at the
centre of the gap between D1 and D2.
Since the dees D1 and D2 are
connected to high frequency
alternating potential, its polarity
changes. If D1 is at positive potential,
then dee D2 is at negative potential
and vice versa.
ii. Now when D1 is at positive potential
and D2 at negative potential the
positive ion produced at P moves
away from D1 and gets attracted to
D2 and enters in dee D2.
iii. When it enters into the dee, there is
no electric field inside the dee. The
ions are only in the magnetic field
produced by the electromagnet.
iv. Thus, the positive ion moves in
strong magnetic field perpendicular
to the plane of dees, hence it
experiences a force. This force
provides the necessary centripetal
force due to which it revolves in a
semicircular path in the dee D2
before arriving at the gap.
v.
Soon after the completion of
semicircle in D2, positive ions arrive
at the gap between the dees. At that
instant, the electric field reverses its
direction so that dee D1 becomes
negative and the dee D2 becomes
positive.
vi. Now the charged particle is
accelerated towards
D1
with
increasing speed and hence in dee
D1, the charged particle moves along
the semicircle with greater radius.
www.horizonpublication.com
i.
Magnetic Effects of Electric Current
vii. This process repeats itself again and
again. Ultimately highly energetic
positive ions escape from the
window (W) in either of the dees and
strike the target.
28.
Ans.
Show that time taken by a charged
particle to move along a semicircular
path is independent of the speed and
radius of semicircular path.
i.
The time taken by the particle to
complete the semicircular path inside
the dee is given by
Distance
Time 
Speed
r
v
Where r is the circumference of
semicircular path.



iii.
29.
Ans.
t
mv


 Bq 
m
t 
Bq
t

v

mv
 r 

B q 

From the above expression, we can
see that time taken by a charged
particle to move along a semicircular
path is independent of the speed and
radius of semicircular path.
Show that cyclotron frequency for a
qB
charged particle is
2 r
i.
Let q = magnitude of charge on
charged particle
v = velocity of charged particle
t = time required to traverse
semicircle
r = radius of semicircle
ii. Time taken by charged particle to
traverse semicircular path in the
cyclotron is given by

t
 r  mv
 
v
v qB
t
m
qB

mv 
 r 

qB 

Time required to traverse a
circular path is
14.10
Std. XII Sci. Success Physics - II
HORIZON Publication
2m
qB
Since magnetic resonance frequency
is the reciprocal of its time period
(T).
1
1
f  

T 2m
qB
qB

f 
2 m
t  2t 
iii.
30.
Ans.
31.
Ans.
Show that velocity of charged particle in
a cyclotron is directly proportional to
the radius of circular path.
i.
When a positively charged particle
(q) moves at right angles to the
magnetic field (B) a force due to
magnetic field acts on the particle
and is given by
F = Bqv sin 
 F = Bqv
(  = 90, sin 90 = 1)
ii. This force produces the necessary
centripetal force to the charged
particle to move in a circular path of
radius r.
 Force due to magnetic field =
centripetal force
m v2

Bq v 
r
mv

Bq 
r
Bq r
v

m
 v  r ( q, B, m are constant)
Thus, velocity of charged particle in
a cyclotron is directly proportional to
the radius of circular path.
Derive an expression for kinetic energy
of positively charged particle.
Expression for kinetic energy of
charged particle :
i.
Force due to magnetic field is given
by,
FB = Bqv
Centripetal force required to move
the charged particle in circular path
is given by,
Magnetic Effects of Electric Current
ii.
iii.
m v 2max
FCP 
r
In equilibrium,
FCP = FB
m v 2max

 Bq v
r
Bq r

v max 
m
Kinetic energy of a charged particle
is given by
K.E. 
1
1  Bq r 
m v 2max  m 

2
2  m 
2
B2 q 2 r 2
2m
This is the required expression for
kinetic energy of charged particle in
the cyclotron.
K.E. 
32.
Ans.
State the limitations of cyclotron.
Limitations :
i.
It cannot accelerate uncharged
particles like neutrons.
ii. It cannot accelerate electrons
because it has a small mass, hence
moves with very high speed.
Therefore, it cannot remain in phase
with the field.
iii. The charged particles cannot move
with speed beyond certain limit in
the cyclotron.
iv. It is not possible to design a machine
capable of producing highly
energetic particles having energy of
the order of 500 MeV.
Formulae
1.
Force on a current carrying conductor :
F = I/B sin 
2.
Torque acting on the coil :
 = InAB cos 
3.
Magnetic induction due to a long
straight conductor :
 2I
B 0
4 r
14.11
Std. XII Sci. Success Physics - II
4.
5.
6.
Magnetic induction due to long solenoid
i.
Induction at a point inside the
solenoid
B = 0nI
ii. Induction at a point near the end of
solenoid
1
B end   0 Ni
2
Magnetic induction due to toroid :
B = 0nI
Moving coil galvanometer (M.C.G.) :
i.
Deflecting torque acting on coil :
d = nIAB cos 
ii. Restoring torque : r = k
7.
Current flowing through galvanometer:
 C 
I

 nAB 
8.
Deflection in moving coil galvanometer :
nIAB

C
9.
Current sensitivity of M.C.G. :
 nAB
Si  
I
C
10.
Ammeter :
I  Is  Ig
i.
ii.
iii.
iv.
11.
G
S
n 1
Voltmeter :
V
R
 G  G ( n  1)
i.
Ig
ii.
12.
IsS = IgG
GI g
S
I  Ig
Ig 
V
RG
Radius of circular path in a cyclotron :
mv
r
qB
Magnetic Effects of Electric Current
www.horizonpublication.com
13.
Time period for charge particle to
complete circular path :
2 m
T
qB
14.
Frequency of charged particle :
1
qB
f  
T 2 m
15.
q 2 B2 r 2
K.E. of charged particle K.E. =
2m
Solved Problems
*1.
A rectangular coil in a moving coil
galvanometer has 50 turns each of
length 5 cm and breadth 3 cm, which is
suspended in a radial magnetic filed of
0.050 Wb/m2. The twist constant of
suspension is 1.5  10-9 Nm/deg.
Calculate the current through the coil
which will deflect it through 300.
Solution : n = 50, C = 1.5  10-9 Nm/deg,
B = 0.05 Wb/m2,  = 300
A = l  b = 5  3 = 15 cm2
= 15  10-4 m2
I=?
C
I
nAB
1.5  10 9  30
I

50  15  10  4  0.05
 I = 1.2  10-5 A
*2.
A galvanometer with a coil of resistance
40 ohm gives a full scale deflection for a
current of 5 mA. How will you convert
it into an ammeter of range 0 – 5 A ?
Solution : G = 40 , Ig = 5 mA = 5  10-3 A,
I = 5 A, S = ?
 Ig 
G
S
I  I 
g





 5  10 3 
  40  0.04 
S  
3 
5

5

10


S = 0.04 
A shunt of 0.04  is connected in
parallel with galvanometer.
14.12
Std. XII Sci. Success Physics - II
Calculate the value of resistance needed
to convert a moving coil galvanometer
of 60  into an ammeter of range 5 A
which gives full scale deflection for a
current of 50 mA and into voltmeter of
range 0 – 50 V.
Solution : G = 60 , V = 50 V, I = 5 A,
Ig= 50 mA = 50  10-3 A
= 5  10-2 A
S = ?, Rs = ?
G Ig
S
I  Ig
HORIZON Publication
*3.

S
60  0.05
3

5  0.05
4.95

S = 0.6061 
V
G
Also, R s 
Ig
From the figure,
Ig
S

I
(G  X )  S
Magnetic Effects of Electric Current
10
 0.04
2  10 2
(16 + X + 0.04) = 20
X = 20 – 16.04
X = 3.96 
(16  X  0.04 ) 
A resistance of 3 ohm is connected in
parallel to a galvanometer of resistance
297 ohm. Find the fraction of current
passing through galvanometer.
Solution : S = 3 , G = 297 
Ig
I
*4.


*5.
Rs = 1000 – 60
Rs = 940 
A resistance of 0.6061  in parallel
and 940  in series is connected to
M.C.G.
A galvanometer has a resistance of 16 
and gives a full scale deflection when a
current of 20 mA is passed through it.
The only shunt resistance available is
0.04  which is not appropriate to
convert galvanometer into an ammeter.
How much resistance should be
connected in series with coil of
galvanometer so that the range of
ammeter is 10 A ?
Solution : Let ‘X; be the resistance connected
in series with galvanometer.
Since S is not sufficient for I = 10 A
G = 16 ,
(G  X )  S 



50
Rs 
 60
5  10  2



I
S
Ig

?

Ig G = (I - Ig) S
G
I
1
S
Ig

GS
I

S
Ig

Ig

Ig

Ig
I
I

S
3

G  S 297  3

3
 3


 100 
300  300

I
=1%
*6.
The
combined
resistance
of
galvanometer of resistance 1000 ohm
and its shunt is 25 ohm. Calculate the
value of shunt.
Solution : G = 1000 , Req. = 25 
S=?
1
1 1


R eq G S

1
1
1 G  R eq .



S R eq . G
R eq .  G

S

S

S = 25.64 
R eq G
G  R eq .
25  1000 25  1000

1000  25
975
14.13
Std. XII Sci. Success Physics - II
*7.
A rectangular coil of a moving coil
galvanometer contains 50 turns each
having area 12 cm2. It is suspended in
radial magnetic field of induction
0.025 Wb/m2 by a fibre of twist
constant 15  10-10 Nm/deg. Calculate
the sensitivity of a moving coil
galvanometer.
Solution : n = 50, A = 12 cm2 = 12  10-4 m2,
B = 0.025 Wb/m2,
C = 15  10-10 Nm/deg
Si = ?
nAB
Si 
C
50  12  10 4  0.025
 Si 
15  10 10
 Si = 106 div/A
8.
Find the magnetic induction at a point
at a distance of 2.5 m from a long
straight conductor carrying a current of
50 A.
Solution : r = 2.5 m, I = 50 A
B=?
 2I
B 0
4 r
2  50
100
 10 7 
 10 7 
2 .5
2 .5
-7
= 10  40
 B = 4  10-6 T
A solenoid is 2 m long and 3 cm in
diameter, it has 5 layers of winding of
1000 turns each and carries a current of
5 A. What is the magnetic filed at its
centre ?
Solution : l = 2 m, d = 3 cm
 r = 1.5 cm
5  1000
N
 2500
2
I=5A
The magnetic induction is,
B = 0 N I
= 4  10-7  2500  5
= 4  25  5  10-5
= 100  5  10-5
= 5  10-3
= 5  3.14  10-3
 B = 15.7  10-3 T
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10.
A toroid has a core (non-ferromagnetic)
of inner radius 20 cm and outer radius
25 cm around which 1500 turns of a
wire are wound. If current in the wire is
2A. Calculate the magnetic field,
i.
inside the toroid
ii. outside the toroid
Solution :
20  25
i.
Mean radius of toroid (r) 
2
45

 22 .5 cm
2
 r = 22.5  10-2 m
The length of the toroid (l) = 2r
= 2  22.5  10-2 m
1500

N
2  22.5  10  2
1500

2  225  10  3
1
15


 10 5
2 225
1
N
 105
2  15
B = o N I
 4 10  7 
4
 10  2  0.267  10  2
15
 B = 2.67  10-3 T
Magnetic field outside the toroid is
always zero.

9.
Magnetic Effects of Electric Current
1
 105  2
2 15
ii.
11.
A plane coil of M.C.G. of 50 turns each
of area 10 cm2 is suspended freely in a
radial magnetic field of induction B. If
the torsional constant of suspension
fibre is 3  10-8 Nm per radian. The coil
deflects through an angle of 0.2 radian
when a current of 3 A is passed
through it. Find B.
Solution : n = 50
A = 10 cm2 = 10  10-4 m2
C = 3  10-8 Nm/rad
 = 0.2 rad
I = 3 A = 3  10-6 A
B=?
14.14
Std. XII Sci. Success Physics - II
HORIZON Publication
We have,
B

C
In A
 3

15 
 I2 = 40  10-6 A
 I2 = 40 A
Now, /15 = 12o

12
 Si  2 
I2
40
 Si = 0.3 degree/A
= 4  10-4  102
B = 4  10-2 Wb/m2
A coil of a moving coil galvanometer is
40  10-3 m long, 20  10-3 m wide. It has
50 turns the field is 10-2 Wb/m2 and
suspension has a torsional constant of
10-8 Nm/degree. Calculate the current
sensitivity of moving coil galvanometer.
Solution : A = l  b
= 40  10-3  20  10-3
= 800  10-6 m2
N = 50
B = 10-2 Wb/m2
C = 10-8 Nm/degree
Si = ?
We have,
nBA
Si 
C
50  10 2  800  10 6

10 8
= 50  800
 Si = 40,000 degree/A
 Si = 0.04 degree/A
12.
A current of 200 A deflects the coil of
an M.C.G. through 600. What should be
the current to cause the rotation
through /15 radian ? What is the
sensitivity of the meter ?
Solution : I1 = 200 A = 200  10-6 A

1 = 600 = 60 
180
 1 = /3
2 = /15
I2 = ?
Si = ?
We have,
C
C
I1 
1 , I 2 
2
nAB
nAB
I2 2


I1
1
14.
A galvanometer needs 50 mV for a full
scale deflection of 100 division. Find the
voltage sensitivity. What must be its
resistance, if its current sensitivity is 1
div/20 A ?
Solution : Vg = 50 mV = 50  10-3 V
 = 100 division
Sv = ?
Si = 1 div/20 A
= 1 div/20  10-6 A
1

 10  6 div / A
20
= 0.05  106 div/A
 Si = 5  104 div/A
The voltage sensitivity is,

Sv 
Vg
100
 2  103 div/volt
3
50  10
The resistance of the galvanometer is,
S
G i
Sv

13.
Magnetic Effects of Electric Current
2
1
 200  10  6 
3  10 8  0.2

3  10  6  50  10  10  4

I 2  I1 


5  10 4 5
  10
2  10 3 2
G = 25 
15.
A galvanometer of resistance 100 ohm,
gives a full scale deflection for 1 mA
current. How will you use it to measure
the current upto 1 A.
Solution : G = 100 Ohm
Ig = 1 mA = 1  10-3 A
I=1A
S=?
The shunt resistance which is connected in
parallel with galvanometer is,
14.15
Std. XII Sci. Success Physics - II
 I 
S  G g 
I  I 
g 

 1  10 3 
 10 3


 100 

100
3 
 1  1  10 
 0.999


www.horizonpublication.com
10
10
 4
10
 1 10  1
3
10
10
10


10000  1 9999




 S = 0.001 
To convert the galvanometer into
voltmeter the high resistance connected in
series is,
V
R
G
Ig
100  10 3 100

999  10  3 999
S = 0.1 ohm
16.
The
combine
resistance
of
a
galvanometer of a 500 ohm and its
shunt is 25 ohm. Calculate the value of
shunt.
Solution : G = 500 ohm
X = 25 ohm
S=?
Let X is the combine resistance,
1
1
1
 
X S G
1
1
1
500  S

 

25 S 500
500 S
500 S

25 
500  S
 25 (500 + S) = 500 S
 12500 + 25 S = 500 S
 12500 = 475 S
12500
 S
475
 S = 26.32 ohm
17.
A galvanometer having a resistance of
10  gives a full scale deflection for a
current of 1 mA. How will you convert
it into (i) an ammeter of range 10 A and
(ii) a voltmeter of range 100 V ?
Solution : G = 10 
Ig = 1 mA = 1  10-3 A
I = 10 A
V = 100 V
S=?
R=?
The shunt resistance is,
Ig
S
G
I  Ig

S
G
I
1
Ig
Magnetic Effects of Electric Current
100
 10
10  3
= 105 – 10
= 1,00,000 – 10
R = 99,990 


18.
A voltmeter has range of 25 volts and
resistance of 500 . How will you
convert it into a voltmeter capable of
reading upto 250 volts.
Solution : Vg= 25 volts
G = 500 
V = 250 volts
R=?
We have,
Vg = Ig G
25 = Ig 500
25

Ig 
500
 Ig = 0.05 A
Now,
V
R
G
Ig
250
250  500  0.05
 500 
0.05
0.05
250  25
250


0.05
0.05
R = 4500 


19.
A galvanometer has a current sensitivity
of 5 div/mA and voltage sensitivity
1 div/mV. If the galvanometer has
60 div, how will you modify it to read
(i) 5 A (ii) 50 V ?
Solution : Si = 5 div/mA = 5 div/10-3 A
= 5  103 div/A
Sv = 1 div/mV
= 1 div/10-3 V = 1  103 div/V
14.16
Std. XII Sci. Success Physics - II
HORIZON Publication
 = 60 division
I =5A
V = 50 V
S = ?, R = ?
The resistance of the galvanometer,
S
5  10 3
G i 
Sv 1  10 3
 G =5

The current sensitivity is, Si 
Ig

Ig 
5

50
50  10 3

5

5
12  10  3
12
50 ,000  60 49 ,940

12
12
R = 4161 
In a cyclotron, magnetic field of 14000
gauss is used to accelerate protons. How
rapidly should the electric field between
the Dees be reversed ? (Given : qP =
1.6  10-19 C, mP = 1.67  10-27 kg)
Solution : qP = 1.6  10-19 C
mP = 1.67  10-27 kg
B = 14000 gauss = 1.4 T
t=?

27
 m P 3.14  1.67  10

qP B
1.6  10 19  1.4
t = 2.34  10-8 s
Magnetic Effects of Electric Current
1  102  2.56  1038  1010

3.34  10 27
K.E. = 0.766  10-23 = 7.66  10-22 J
In a cyclotron, magnetic field of 3.5
Wb/m2 is used to accelerate protons.
What should be the time interval in
which the electric field between the Dees
be reversed ?
[Mar. 13]
[Mass of proton = 1.67 × 10–27 kg,
charge on proton = 1.6 × 10–19 C]
Solution : B = 3.5 Wb/m2
m = 1.67 × 10–27 kg
q = 1.6 × 10–19 C
T=?
T m
t 
2 qB
3.14  1.67 1027

r
1.6  1019  3.5
 t = 9.36 × 10–9 sec.
23.
20.
t
(0.1) 2  (1.6  10 19 ) 2 (10 5 ) 2
2  1.67  10  27
22.




5
5
5  103

1
1
12  10  3
12
5
60


5000  12 4988
12
 S = 0.012 
The high resistance is,
V
R
G
Ig

What is the kinetic energy of proton in
the cyclotron. If the radius of circular
path is 0.1 m and the magnetic field
1  10-5 T.
Solution : r = 0.1 m
B = 1  10-5 T
K.E. = ?
1 r 2 q 2 B2
K.E. 
2 mp

60

Si 5  10 3
 Ig = 12  10-3 A
The shunt resistance is,
Ig  G
G
S

I
I  Ig
1
Ig

21.
A circular coil of 250 turns and
diameter 18 cm carries a current of
12A. What is the magnitude of magnetic
moment associated with the coil ?
[Mar. 13]
Solution : N = 250, D = 18, i = 12 A
D 18
r

 9 cm  0.09 m
2
2
M = NiA
= 250 × 12 ×  × (0.09)2
= 76.30 A-m2
24.
A solenoid /2 meter long and 5.0 cm in
radius has ‘two’ layers of windings 500
turns each and carries a current of 5A.
Find the magnetic induction at its
centre along the axis.
[Oct. 12]
14.17
Std. XII Sci. Success Physics - II
[Given : 0 = 4 × 10–7 Wb/Am]
Solution : B = 0 n i
4107 1000  5  2


–3
 B = 4 × 10 T
Higher Order Problems
A circular coil of wire consisting of 100
turns, each of radius 8.0 cm carries a
current of 0.40 A. What is the
magnitude of the field B at the centre
of the coil?
Solution : I = 0.40 A, a = 8.0 cm = 0.08 m
n = 100
 2nI
B= 0 
4
a
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4.
A horizontal overhead power line
carries a current of 90 A in the east to
west direction. What is the magnitude
and direction of magnetic field due to
the current at a point 1.5 m below the
line?
Solution : I = 90 A, a = 1.5 m

2I
B= 0 
4 a
1.
= 10–7 ×
2   100  0.40
0.08
= 3.14 × 10–4 T
2.
A long straight wire carries a current of
35 A. What is the magnitude of the field
= 10–7 
Direction of magnetic field at a point
below the line will be towards south.
5.
A closely wound solenoid 80 cm long
has 5 layers of windings of 400 turns
each. The diameter of the solenoid is 1.8
cm. If the current carried is 8.0 A,
estimate the magnitude of B inside the
solenoid near its centre.
Solution : I = 8.0 A; L = 80 cm = 0.8 m
Total number of turns of the solenoid,
N = 400 × 5 = 2,000
Therefore, number of turns per unit length
of the solenoid,
n=
B at a point 20 cm from the wire.
Solution :
I = 35 A, a = 20 cm = 0.2 m
B=?
 2I
B= 0 
4 a
= 10–7 
3.
2  35
= 3.5 × 10–5 T
0.2
2  50
= 10 
= 4.0 × 10–6 T
2.5
6.
Ans.
–7
Right hand thumb rule tells that for a
current in north to south direction, the
magnetic field at a point due east of the
wire will be in outward direction.
Magnetic Effects of Electric Current
N 2, 000
= 2,500 m–1

L
0.8
Now, B = 0 n I
= 4  × 10–7 × 2,500 × 8.0
 B = 2.51 × 10–2 T
A long straight wire in the horizontal
plane carries a current of 50 A in north
to south direction. Give the magnitude
and direction of B at a point 2.5 m east
of the wire.
Solution : I = 50 A, a = 2.5 m
 2I
B= 0 
4 a
2  90
= 1.2 × 10–5 T
1.5

Two moving coil galvanometers M1 and
M2 have the following particulars:
R1 = 10 ; N1 = 30; A1 = 3.6 × 10–3 m2;
B1 = 0.25 T; R2 = 14 ; N2 = 42;
A2 = 1.8 × 10–3 m2; B2 = 0.50 T
The spring constants are identical for
the two Meters. Determine the ratio of
i.
current sensitivity and
ii. voltage sensitivity of M2 and M1.
Let k be the spring constant of both the
meters.
i.
Now, current sensitivity of a meter,
d N B A

dI
k
current sensitivity of meter M 2
current sensitivity of meter M 1

N 2 B2 A 2
k2

k1
N1 B1 A1
14.18
Std. XII Sci. Success Physics - II
=
N 2 B2 A 2
N1 B1 A1
( k1  k 2 )
42  0.50  1.8  10 3
= 1.4
30  0.25  3.6  10 3
Voltage sensitivity of a meter,
d d N B A


dV IR
kR
voltage sensitivity of meter M 2
voltage sensitivity of meter M1
HORIZON Publication
7.

ii.


N 2 B2A 2

N 2 B2A 2R 1

k 2R 2

8.
9.
k1R 1
N1B1A1
N1B1A1R 2
10.
42  0.50  1.8  10 3  10
1
30  0.25  3.6  10 3  14
Problems for practice
11.
1.
2.
3.
4.
5.
6.
A moving coil galvanometer of resistance
100 ohm gives a full scale deflection of 50
divisions for a current of 25 milliampere.
How will you convert it into an ammeter
to read 1 ampere for 10 divisions ?
A moving coil galvanometer of resistance
200  gives full scale deflection of 100
divisions for a current of 50 mA. How will
you convert it into an ammeter to read 2 A
for 20 divisions ?
A galvanometer has a capacity to carry a
maximum current of 25 mA. How can it
be used as ammeter to read the current
upto 0.1 A ?
A galvanometer of resistance 100  gives
a full scale deflection for a current of 2
mA. How will you use it to measure
i.
current upto 2 A ?
ii. voltage upto 5 V ?
Calculate the value of the shunt which
when connected across a galvanometer of
resistance 38 ohm will allow 1/20th of the
current to pass through the galvanometer.
A rectangular coil of effective area 0.05
m2 is suspended freely in a radial field of
0.01 Wb/m2. If the torsional constant of
the suspension fibre is 5  10-9 Nm per
degree, find the angle through which the
coil rotates when a current of 300 A is
passed through it.
Magnetic Effects of Electric Current
A milliammeter reading from 0 to 50 mA
has a resistance of 396 . How will you
change it so that it can be used as an
ammeter to measure a maximum current
of 5 A ?
A galvanometer has current range
0.25 mA and voltage range 0 – 500 mV.
How will you adopt it to real (i) the
current upto 10 A, (ii) the potential
difference upto 10 V ?
A galvanometer of resistance 100  gives
a full-scale deflection for 2 mA. Explain
how will you use it (a) to measure currents
upto 2A and (b) to measure voltage upto
5V.
The magnetic flux density applied in a
cyclotron is 3.5 T. What will be the
frequency of electric field that must be
applied between the dees in order to
accelerate protons ?
Mass of proton = 1.67  10-27 kg.
A cyclotron having dees of radii 50 cm is
used to accelerate protons. If the frequency
of the oscillator used is 10 MHz, what
should be the value of magnetic induction
used ? Calculate the maximum velocity
and maximum energy of the protons
produced. Mass of proton = 1.67  10-27
kg, charge of proton = 1.6  10-19 C.
Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
0.503  should be connected in parallel
with the galvanometer.
A shunt of resistance 1.005  should be
connected in parallel with galvanometer.
Connecting a shunt of resistance G/3 ohm
in parallel with the galvanometer.
i.
0.1001  should be connected in
parallel.
ii. 2400  should be connected in
series.
2
30o
4
0.05012 , 380 
100
, 2400 
999
5.35  107 Hz
3.14  107 m/s, 5.14 MeV
14.19
Std. XII Sci. Success Physics - II
www.horizonpublication.com
8.
Multiple Choice Question
1.
The line integral of magnetic filed ( B )
around any closed path through which
current I is flowing is given by  B dl 
c
0
I
I
(c) 0 I
(d)
0
Ampere’s circuital law is used to find out
(a) magnetic induction
(b) magnetic potential
(c) electric intensity
(d) electric potential
The magnetic induction at any point due to a
long straight wire carrying a current is
(a) proportional to the distance from the wire
(b) inversely proportional to the distance
from wire
(c) inversely proportional to the square of
the distance from the wire
(d) does not depend on distance
If the value of current flowing in a
conducting wire is doubled, then the value
of magnetic filed produced by it at a point
will
(a) remain same
(b) be half
(c) be doubled
(d) be four times
The strength of the magnetic field at a point
r near a long straight current carrying wire is
r
B. The field at a distance
will be
2
B
B
(a)
(b)
2
4
(c) 2B
(d) 4B
If the strength of the magnetic field
produced 10 cm away from a infinitely long
straight conductor is 10-5 weber/m2, the
value of the current flowing in the conductor
will be
(a) 5 ampere
(b) 10 ampere
(c) 500 ampere
(d) 1000 ampere
The value of magnetic induction inside a
solenoid along the radius
(a) is zero
(b) decreases with distance from the axis
(c) is uniform
(d) increases with distance from the axis
(a)
2.
3.
4.
5.
6.
7.
0 I2
(b)
Magnetic Effects of Electric Current
9.
10.
11.
12.
13.
14.
A solenoid is 1.0 metre long and it has 4250
turns. If a current of 5.0 ampere is flowing
through it, what is the magnetic field at its
centre ? [0 = 4  10-7 weber/amp-m]
(a) 5.4  10-2 weber/m2
(b) 2.7  10-2 weber/m2
(c) 1.35  10-2 weber/m2
(d) 0.675  10-2 weber/m2
A solenoid 1.5 m long and 0.4 cm in
diameter possesses 10 turns per cm length.
A current of 5 A flows through it. The
magnetic field at the axis inside the solenoid
is
(a) 2  10-3 T
(b) 2  10-5 T
(c) 4  10-2 T
(d) 4  10-3 T
A toroid has N number of turns per unit
length, current I, then the magnetic field is
(a) 0NI
(b) 0N2I
(c) 0I/N
(d) 0N2I2
Toroid is a solenoid of
(a) infinite length
(b) infinite length of non-uniform radius
(c) finite length bent into a circle
(d) infinite length bent into a circle
A toroid has a core (non-ferromagnetic) of
inner radius 25 cm and outer radius 26 cm,
around which 3,500 turns of a wire are
wound. If the current in the wire is 11 A, the
magnetic field inside the core of the toroid
will be
(a) 30.2  10-2 T
(b) 30.2  10-3 T
(c) 40.2  10-2 T
(d) 40.2  10-3 T
A charge of 1 C is moving in a magnetic
field of 0.5 tesla perpendicular to the filed. If
the force experienced by the charge is 5 N,
then the velocity of charge is
(a) 10 m/s
(b) 20 m/s
(c) 30 m/s
(d) 30.5 m/s
In cyclotron, the applied magnetic field
(a) increases the speed of particle only
(b) changes the direction of particle only
(c) changes the direction of particle and
increase speed
(d) neither changes the direction nor
increases the speed
14.20
Std. XII Sci. Success Physics - II
15.
16.
17.
18.
19.
20.
21.
22.
In a cuclotron, a charged particle is
accelerated to high velocity to gain kinetic
energy from
(a) electric and magnetic fields
(b) only the magnetic filed
(c) only the static electric field
(d) only the alternating electric field.
Cyclotron is used to accelerate
(a) Electrons
(b) Neutrons
(c) Positive ions
(d) Negative ions
The path followed by a charge in a cyclotron
is
(a) linear
(b) circular
(c) hyperbolic
(d) spiral
In a cyclotron, period of the path of the
projectile is independent of
(a) mass of the particle.
(b) charge of the particle
(c) velocity of the particle
(d) magnetic induction of the field inside
the dees
The cyclotron frequency of an electron
gyrating in a magnetic field of 1 T is
approximately
(a) 28 MHz
(b) 280 MHz
(c) 2.8 GHz
(d) 28 GHz
The oscillating frequency of a cyclotron is
10 MHz. If the radius of its Dees is 0.5 m,
the kinetic energy of a proton, which is
accelerated by the cyclotron is
(a) 10.2 MeV
(b) 2.55 MeV
(c) 20.4 MeV
(d) 5.1 MeV
A cyclotron is used to accelerate protons
with a K.E. of 5 MeV. If the strength of
magnetic filed in the cyclotron is 2 Wb/m2,
the frequency needed for the applied voltage
and radius of cyclotron respectively are
(a) 3.104  107 Hz; 0.161 m
(b) 3.104  108 Hz; 0.161 m
(c) 3.06  107 Hz; 0.161 m
(d) 3.06  108 Hz; 0.171 m
A current loop of area A, number of turns N
is placed in a uniform magnetic induction B.
The angle between the plane of the loop and
B is . The torque acting on the loop will be
(a) NiAB
(b) NiAB sin 
(c) NiAB cos 
(d) NiAB tan 
Magnetic Effects of Electric Current
HORIZON Publication
23.
24.
25.
26.
27.
28.
29.
A coil carrying electric current is placed in a
uniform magnetic field, then
(a) Torque is formed
(b) E.M.F. is induced
(c) Both ‘a’ and ‘b’ are correct
(d) None of these
To make the field radial in a moving coil
galvanometer
(a) the number of turns in the coil is
increased.
(b) magnet is taken in the form of horseshoe.
(c) poles are cylindrically cut
(d) coil is wounded on an aluminium
frame.
A small cylindrical soft iron piece is kept in
a galvanometer so that
(a) a radial uniform magnetic field is
produced
(b) a uniform magnetic field is produced
(c) there is a steady deflection of the coil
(d) all of these
In a moving coil galvanometer, the
deflection of the coil  is related to the
electrical current I by the relation
(a) I  tan 
(b) I  
(c) I  2
(d) I  
The pole pieces of the magnet used in a
pivoted coil galvanometer are
(a) plane surfaces of a bar magnet
(b) plane surfaces of a horse-shoe magnet
(c) cylindrical surfaces of a bar magnet
(d) cylindrical surfaces of a horse-shoe
magnet
When a galvanometer is shunted, the current
flowing through it
(a) decreases
(b) increases
(c) remains the same
(d) increases to a certain value
The deflection in a moving coil
galvanometer falls from 50 divisions to 10
divisions, when a shunt of 12  is connected
across it. The resistance of the galvanometer
is
(a) 24 
(b) 36 
(c) 48 
(d) 60 
14.21
Std. XII Sci. Success Physics - II
30.
31.
32.
33.
34.
35.
A rectangular coil of 500 turns with an
average area per turn 4.00 cm2 carries a
current of 0.2 A. The coil is placed at an
angle of 600 with the direction of magnetic
induction of 10-3 Wb/m2. The torque acting
on the coil is
(a) 10-5 N-m
(b) 0.5  10-5 N-m
(c) 2  10-5 N-m
(d) 4  10-5 N-m
The sensitivity of MCG is maximum when it
is suspended by
(a) thick wire
(b) fine gauge wire
(c) thin wire
(d) very thick wire
the sensitiveness of a moving coil
galvanometer can be increased by
decreasing
(a) the number of turns in the coil
(b) the area of the coil
(c) the magnetic field
(d) the couple per unit twist of the
suspension
The current sensitivity of a moving coil
galvanometer can be increased by
(a) decreasing the magnetic field of the
permanent magnet.
(b) decreasing the area of the deflecting
coil
(c) increasing the number of turns in the coil
(d) increasing the restoring couple of the
coil
The coil of a galvanometer consists of 100
turns each of area 15 cm2. The torsional
constant of the suspension fibre is 1.5  10-8
N-m/radian. It is suspended in a radial
magnetic field of induction 3  10-2 Wb/m2.
The current sensitivity of this galvanometer
will be
(a) 1.5  104 rad/ampere
(b) 4.5  10-6 rad/ampere
(c) 6  10-7 rad/ampere
(d) 3  105 rad/ampere
A moving coil galvanometer of resistance
5  has voltage sensitivity 2 div per mV. Its
current sensitivity is
(a) 10 div/mA
(b) 10  10-3 div/mA
(c) 0.4 div/mA
(d) 0.4  10-3 div/mA
Magnetic Effects of Electric Current
www.horizonpublication.com
36.
37.
The parallel combination of galvanometer
and shunt is called
(a) Voltmeter
(b) Ammeter
(c) Ohmmeter
(d) Speedometer
When a galvanometer of resistance G is
converted to an ammeter of range I A, then
the current passing through S is
 G 
 I
(a) IS  
 G  S
 S 
 I
IS  
S

G


 SI  G 
(c) IS  

 S 
 SI  G 
(d) IS  

 G 
The series combination of galvanometer and
high resistance is called as
(a) ammeter
(b) potentiometer
(c) voltmeter
(d) ohmmeter
To increase the range of voltmeter the series
resistance should be
(a) increased
(b) decreased
(c) constant
(d) low
A milliammeter reading from 0 to 50 mA
has a resistance of 396 . It is to be used to
measure a maximum current of 5 A. This
requires
(a) a series resistance of large value
(b) a series resistance of small value
(c) a shunt of resistance 4 
(d) a shunt of resistance 2 
(b)
38.
39.
40.
Answers
1.
6.
11.
16.
21.
26.
31.
36.
c
a
c
c
c
b
c
b
2.
7.
12.
17.
22.
27.
32.
37.
a
c
a
d
c
d
d
a
3.
8.
13.
18.
23.
28.
33.
38.
b
b
a
c
a
a
c
c
4.
9.
14.
19.
24.
29.
34.
39.
c
a
b
d
c
c
d
a
5.
10.
15.
20.
25.
30.
35.
40.
c
a
d
d
d
c
a
c
14.22
Std. XII Sci. Success Physics - II
HORIZON Publication
SHORT TEST
Q.4. Attempt any one question.
[4]
i.
Explain the principle, construction and
working
of
a
moving
coil
galvanometer.
ii. A galvanometer has a current
sensitivity of 5 div/mA and voltage
sensitivity 1 div/mV. If the
galvanometer has 60 div, how will you
modify it to read (i) 5 A (ii) 50 V ?
Time : 1 hr.
Marks : 20
Q.1. Select and write most appropriate option
from given alternatives.
[4]
i.
The magnetic field is made radial in a
M.C.G. to
(a) make the field stronger
(b) make field weaker
(c) make scale linear
(d) reduce its resistance
ii. The resistance of an ideal ammeter is
(a) zero
(b) low
(c) high
(d) infinite
iii. An
ammeter
consists
of
a
galvanometer having resistance 50 
and external resistance of 5 . The
resistance of ammeter will be
(a) 5 
(b) 50 
(c) 45
(d) 60 
iv. The SI unit of current sensitivity of
MCG is
(a) /degree
(b) amp/degree
(c) rad/A
(d) volt/degree
Q.2. Answer any three questions.
[6]
i.
Distinguish between ammeter and
voltmeter.
ii. How can you convert MCG into
voltmeter ?
iii. Find the magnetic induction at a point
at a distance of 2.5 m from a long
straight conductor carrying a current of
50 A.
iv. Derive an expression for current
sensitivity of MCG.
Answer Key
Q.1. i.
iii.
(c)
(c)
ii.
iv.
(a)
(c)
Q.2. i.
ii.
iii.
iv.
Ref. Q. 20 (Theory Question)
Ref. Q. 17 (Theory Question)
Ref. Q. 8 (Solved Problems)
Ref. Q. 21 (Theory Question)
Q.3. i.
ii.
iii.
Ref. Q. 2, Q. 4 (Theory Question)
Ref. Q. 16 (Solved Problems)
Ref. Q. 18 (Solved Problems)
Q.4. i.
ii.
Ref. Q. 8, Q. 9 (Theory Question)
Ref. Q. 19 (Solved Problems)
Q.3. Attempt any two questions.
[6]
i.
State Ampere’s law. Derive an
expression for magnetic induction at a
point due to a solenoid.
ii. The combine resistance of a
galvanometer of a 500 ohm and its
shunt is 25 ohm. Calculate the value of
shunt.
iii. A voltmeter has range of 25 volts and
resistance of 500 . How will you
convert it into a voltmeter capable of
reading upto 250 volts.
Magnetic Effects of Electric Current
14.23