AIM
To study the field at a point along the axis of a current carrying circular coil and hence to find moment of magnet
APPARATUS REQUIRED
Circular coil apparatus, compass box, rheostat, battery or power supply, ammeter, commutator, key and connecting wires.
FORMULA USED
Field at a point along the axis of the coil is,
nia2
F = ___________ ampere/metre
2(a2 + x2)3/2
n -Number of turns in the coil
i -current measued (A)
a -Radius of the coil (m)
x - distance between centre of coil and centre of compass box (m)
Moment of magnet is,
𝜋nia2𝜇0(d2-l2)2
M = __________________ weber.metre
d(a2 + x2)3/2
Substituting F and 𝜇0 , we can reframe the equation as,
8𝜋2F(d2-l2)2
M = ____________ x 10-7 weber.metre
d
Where,
𝜇0 -Permittivity of free space =4𝜋 x 10-7 Henry/metre
l -semilength of magnet (m)
d- distance between centre of compass box and centre of bar magnet (m)
CIRCUIT DIAGRAM
Figure 1: Circuit to find the field at points along the axis of the coil |
PROCEDURE
The preliminary adjustments are carried out as follows
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Figure 2: Circular coil apparatus with compass box |
The leveling screws are adjusted so that the circular coil is vertical.
The wooden bench is adjusted to be along the magnetic east-west direction
- A compass box is placed with its centre coinciding with the axis of the coil
- The compass box alone is rotated till the aluminium pointer reads 0° − 0°
Electrical connections are made as shown in the circuit diagram (figure 1). The compass box is placed along its axis, with its centre at a distance x from the centre of the coil on one side. A suitable current (say, 1A) is passed through the coil by adjusting rheostat so that the deflection of the aluminium pointer lies between 30 º and 60 º. Two readings θ1 and θ2 corresponding to two ends of the pointer are noted. Now a bar magnet is placed along the axis of coil at a distance d1 from the centre of compass box to make deflection 0-0 (as shown in figure 3).
Figure3: method to nulify the deflection using a magnet |
Number of turns in the coil n= 30 turns
Circumference of the coil (2πa) = 45 cm
Radius of the coil a =0.073m
semilength of the magnet=0.038m
Mean (M)= 1.7936 x10-6 wm
Result:
i)The field at points along the axis of a current carrying circular coil is studied.
ii)The value of moment of magnet is 1.7936 x10-6 wm
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