Experiment No……
ObjectiveTo plot a graph showing the variation of magnetic field with distance along the axis of a
Helmholtz galvanometer and calculate the reduction factor ‘k’, horizontal component of earth’s
magnetic field ‘H’ and the radius of coil ‘r’ from graph.
Apparatus requiredHelmholtz galvanometer, variable power supply, an ammeter, commutator and connecting wires
TheoryHelmholtz galvanometer-A Helmholtz galvanometer, like other galvanometers, is an instrument
which can be used to measure dc current in the circuit .It is an improved form of the tangent galvanometer.
It consists of two identical coils placed coaxially at a distance equal to the radius of the either coil.
The two coils are of the same radius on the same direction. The field at any point on the axis of
the coil is equal to the sum of the magnetic field due to the individual coil. It consists of two
identical circular coils, each having several hundred turns in it, mounted coaxially at a distance equal to
the radius.
In the side of each coil, four terminals are provided. First of which
is internally connected with the starting end of the coil, second is connected with the end of the
second turn, third with the end of the 50th turn and fourth with the 500th turns. With the help of these
terminals, equal number of turns of each coil can be connected in series to allow the same current
to flow in the same direction
For a coil consisting of n turns of wire and having a mean radius r, the magnetic field at a point
on the axis at a distance x from the center of the coil is given by
F=
2πnir2
3
(r2 +x2 ) ⁄2
Where I is the current passing through the coil.
The rate of variation is given by
dF
= -3x [(2πnir 2 )(r 2 + x 2 )
dx
d2 F
dx2
= -6πni𝑟 2 [(r 2 + x 2 )
−5⁄
2
d2 F
−5⁄
2]
- 5𝑥 2 (r 2 + x 2 )
−7⁄
2]
dF
From which x = ±r/2, if dx2 = 0 or dx = constant
If there are two identical coils having the same axis and caring the same current in the direction
with their centers r cm. apart, the rate of increase of field due to one coil at the midpoint between
the coils is equal to the rate of decrease of field due to the other at the same point. Therefore, if
one moves away along the axis from the midpoint, any diminution in the intensity of the field
due to one coil is compensated by the increase in the field due to the other so that the field
between the coils is practically uniform.
The field at the midpoint is given by
F=2*
2πnir2
3
[r2 +(r⁄2)2 ] ⁄2
=
32𝜋𝑛𝑖
𝑟 √125
If the coils are in magnetic meridian will be perpendicular to H. If the needle shows the
deflection θ,
32𝜋𝑛𝑖
𝑟√125
Or
I =
= H tan θ
r H √125 tan θ
32 𝜋 𝑛
= k tan θ , where k (expressed in emu) is a constant for the
galvanometer at a particular place called the reduction factor of the galvanometer.
Also k = I / tan θ
Where k and I are in amperes.
Procedure
1. Place the magnetometer compass box on the sliding bench so that its magnetic needle is
at the centre of the coil. By rotating the whole apparatus in the horizontal plane, set the
coil in the magnetic meridian roughly. In this case the coil, needle and its image all lie in
the same vertical plane. Rotate the compass box till the pointer ends read 0 – 0 on the
circular scale.
2. To set the coil exactly in the magnetic meridian set up the electrical connections as
shown in fig. Send the current in one direction with the help of commutator and note
down the deflection of the needle. Now reverse the direction of the current and again note
down the deflection. If the deflections are equal then the coil is in magnetic meridian.
Otherwise turn the apparatus a little,
3. Using rheostat Rh adjust the current such that the deflection of nearly 450 produced in the
compass needle placed at the centre of the two coils. Read both the ends of the pointer.
Reverse the direction of the current and again read both the ends of the mean deflection at
x =0.
4. The current is noted from the ammeter.
5. The magnetometer box is slowly moved over the bench and kept the distance
1,2,3,4,5,6,7,8,10,12,14,16,18& 20 cm. on east side from the center. The deflection for
both the sides of the pointer is noted at these points of direct as well as reverse currents.
The procedure is repeated for west arm.
6. Plot graph taking distances along X-axis and tan θ along Y- axis.
7. Mark the points of inflections on the curve. The distance between the two points will be
the radius of coil
ObservationsNumber of turns in the coil = ….
Current in ampere = ….amp
S.No. distance along
Deflection of needle
the axis from
East arm
West arm
center (x) in Direct Reverse
Direct Revers
cm.
e
Mean θ tan θ
θ1 θ2 θ3 θ4
θ1 θ2 θ3 θ4
1.
1 cm.
2.
2 cm.
3.
3 cm.
4.
4 cm.
5.
5 cm.
6.
6 cm.
7.
7 cm.
8.
8 cm.
9.
10 cm.
10.
12 cm.
11.
14 cm.
12.
16 cm.
13.
18 cm.
14.
20 cm.
Calculation-
Mean θ tan θ
K = I⁄tan 𝜃 amperes
H=
32𝜋𝑛𝑘
10 𝑟 √125
(k/10 is expressed in emu)
ResultRadius of the coil from the graph is ………………..cm.
The reduction factor of Helmholtz galvanometer for ……turns is ……..ampere.
Horizontal component of earth’s magnetic field in the lab ........( Standard value = 0.35 Gauss).
Precautions1. The coil should be carefully adjusted in the magnetic meridian.
2. All the magnetic materials and current carrying conductors should be at a considerable
distance from the apparatus.
3. The current passed in the coil should be of such a value as to produce a deflection of nearly
450.
4. Parallax should be removed while reading the position of the pointer. Both ends of the pointer
should be read.