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physics manual

Fig 1. Current carrying conductor
Fig 2. Circular coil carrying current
Determination of Magnetic Field along the axis of a coil
To study the variation of magnetic field with distance along the axis of a circular coil
carrying current.
Apparatus Required:
Circular coil, compass box, ammeter, rheostat, commutator, cell, key, connection wires,
etc. The purpose of the commutator is to allow the current to be reversed only in the coil,
while flowing in the same direction in the rest of the circuit.
Bx = B0 tan θ (T)
Bx – Magnetic field along the axis of the circular coil at a distance x (Tesla)
B0 – Horizontal component of Earth’s magnetic field (3.5x10-5 T)
Θ - Deflection at a distance x (degrees)
A current carrying wire generates a magnetic field. According to Biot-Savart’s law, the
magnetic field at a point O, due to an element PQ (as shown in fig 1) of a conductor carrying
current is,
1. Directly proportional to the strength of the current, i
2. Directly proportional to the length of the element, dl
3. Directly proportional to the Sine of the angle θ between the element and the line joining the
element to the point and
4. Inversely proportional to the square of the distance r between the element and the point.
Thus, the magnetic field at O is dB, such that,
is the proportionality constant and
is called the permeability of free space.
In vector form,
Fig 3. Compass Needle
Model graph
Consider a circular coil of radius r (as in fig 2), carrying a current I. Consider a point P,
which is at a distance x from the centre of the coil. We can consider that the loop is made up
of a large number of short elements, generating small magnetic fields. So the total field at P
will be the sum of the contributions from all these elements. At the centre of the coil, the field
will be uniform. As the location of the point increases from the centre of the coil, the field
decreases. By Biot- Savart’s law, the field dB due to a small element dl of the circle, centred
at A is given by,
This can be resolved into two components, one along the axis OP, and other PS, which is
perpendicular to OP. PS is exactly cancelled by the perpendicular component PS’ of the field
due to a current and centered at A’. So, the total magnetic field at a point which is at a
distance x away from the axis of a circular coil of radius r is given by,
If there are n turns in the coil, then
Since this field Bx from the coil is acting perpendicular to the horizontal intensity of earth’s
magnetic field, B0, and the compass needle align at an angle θ with the vector sum of these
two fields, we have from the fig 3. The horizontal component of the earth’s magnetic field
varies greatly over the surface of the earth. For the purpose of this simulation, we will assume
its magnitude to be B0 = 3.5x10-5 T. The variation of magnetic field along the axis of a
circular coil is shown here.
Circular coil apparatus:
The apparatus consists of a circular coil C of 5 to 50 turns, having diameter about 10 cm.
There is a brass frame in which the coil is wound. The frame is fitted to a stand, with its plane
vertical. It can be moved along a rectangular wooden board. B is the deflection
magnetometer. There is a scale fitted to the wooden board, from which the distance of the
centre of the magnetic needle from the centre of the coil can be measured. Procedure:
To find the magnetic field:
The connections are made as shown in the diagram and the initial adjustments of the
apparatus are made as follows:
First, the coil is fixed at the middle of the platform and the compass box is placed at the
centre of the coil.
The compass box is rotated till the 90-90 line becomes parallel to the plane of the coil.
Then the apparatus as a whole is rotated till the aluminium pointer reads 0-0.
Close the circuit.
To plot the graph between distance and magnetic field intensity:
B0= 3.5x10-5 T.
Current, I = ……… A
No: of turns of the coil, n = …..
Radius of the circular coil, r = ….. cm.
from the
centre, x
Deflection in
compass box on
direct reversed
Θ1 Θ2
Deflection in
compass box on
direct reversed
Θ1 Θ2
Mean θ
tan θ
Bx = B0 tan θ
Adjust the rheostat until the deflection lies between 30° and 60° . Note down the deflection of
the compass needle and the current.
Then current through the coil is reversed using the commutator and again the deflection and
current are noted.
Average the magnitude of the two deflections and calculate the magnetic field at the centre of
the coil from the equation.
Without changing the current or the number of turns, place the compass box at a
particular distance from the centre of the coil. Note the deflection. Again reverse the current
and average the magnitudes of the two deflections. Note the average, and the distance.
The same procedure is repeated with the compass box at the same distance on the other side
of the arm, keeping number of turns and current constant.
Take the average of the two values of θ measured on opposite sides of the coil.
Then calculate the magnetic field Bx from the coil using equation (3).
Repeat for various distances.
Draw graph of Bx on the vertical axis vs. distance x on the horizontal axis.
Then plot the graph.
The variation of magnetic field with distance along the axis of a circular coil carrying current
is studied.
State Biot-Savart’s law.
Define permeability.
Why θ should always lie between 30° and 60°?
What is the purpose of commutator in this circuit?
Explain the variation of magnetic field with respect to distance from the coil?
Mass of the solid metal
Initial Temperature of the
hot metal (oC)
Specific heat of calorimeter
Mass of calorimeter and
stirrer (kg)
Mass of calorimeter, stirrer,
and water (kg)
Mass of cold water
Specific heat of cold water
Initial Temperature of cold
water (oC)
Final equilibrium
Metal 1:
Metal 2:
900 J/kgCo
900 J/kgCo
4186 J/kgCo
4186 J/kgCo
Determination of specific heat of solids-Calorimeter
Aim: To determine the specific heat of various solids by the method of heat
Apparatus Required:
Calorimeter with stirrer, weighing scale, Thermometer, boiler (beaker and hotplate)
Small pieces of ice, two metal solids (made of different materials) and
Paper towels
Specific heat of given solid, c =
m2 c2 (T3 −T2 )+ m1 c1 (T3 −T2 )
m(T3 – T1)
J/kg oC or J/kg K
m = Mass of the solid (Kg)
m1= Mass of calorimeter with stirrer (Kg)
m2= Mass of cold water (Kg)
c1= Specific heat of calorimeter( Aluminum) = 900 J/kg oC
c2= Specific heat of water = 4186 J/kg oC
T1= Initial temperature of (hot) solid ( oC)
T2= Initial temperature of cold water ( oC)
T3= Final equilibrium temperature of mixture ( oC)
Heat capacity of a body is the quantity of heat required to raise the temperature
of the body by 1oC. The specific heat of a substance is the heat capacity per unit
Thus, heat capacity = mass x specific heat. The specific heat is essentially a
measure of how thermally insensitive a substance is to the addition of energy.
The water equivalent of a body is the mass of water, which would require the
same amount of heat as the body in order to raise the temperature through one
degree Celsius.
The method of mixtures makes use of the principles that when two bodies at
different temperatures exchange heat, the quantity of heat lost by the warmer
body is equal to the heat gained by the cooler body, and some intermediate
equilibrium temperature is finally reached. This is true provided no heat is lost or
gained from/to the surroundings. The purpose of the calorimeter is to prevent
heat lose to the surroundings. There are three methods of heat transfer:
conduction, convection and radiation.
In this experiment, a heated solid of known mass and temperature is dropped
in the calorimeter containing known mass of cold water. The equilibrium
temperature is then measured. The magnitude of the heat lost by the solid
must be equal to the magnitude of the heat gained by the water, and
calorimeter with stirrer.
Mass of the solid (metal) in kg
Specific heat of the metal
Initial temperature of (hot) solid
Mass of calorimeter with stirrer
Mass of cold water
Initial temperature of cold water
Final equilibrium temperature of mixture
Specific heat of calorimeter( Aluminum)
Specific heat of water
Heat lost = -Heat gained
= T3
= c1 = 900 J/kgCo
= c2 = 4186 J/kgCo
Heat lost by the solid
= ( m )( c )[(T2 - T1)]
Heat gained by water + (calorimeter & stirrer) = (m2)(c2)(T3 – T2) + (m1)(c1)(T3 –
Equating heat lost to heat gained:
-(m )(c)(T3 – T1) = m2 c2 (T3 – T2) + m1c1(T3 – T2)
c =
m2 c2 (T3 −T2 )+ m1 c1 (T3 −T2 )
m(T3 – T1)
This equation is used to compute the unknown specific heat, c, of the metal.
1. Fill the beaker about half way with water and start heating it.
2. Record the name of the metal being used in the data table.
3. Weigh the solid (m) metal, and then lower it into the beaker of hot
water by means of a thread, to avoid burns.
4. While the solid is being heat go to step 5.
5. Weigh the inner chamber of calorimeter and the stirrer, together.
6. Fill the inner chamber of the calorimeter about half way with cool
water and add one or two small pieces of crushed ice to the water.
7. Weigh the inner chamber of the calorimeter, stirrer and cold water (m1 + m2).
8. Place the inner chamber of the calorimeter into the outer calorimeter jacket and
9. place the lid on, then record the temperature of the cold water (T2).
10. Record the temperature of the hot solid when the temperature
becomes steady (T1). This should occur after the water boils. Hint:
The metal should be the same temperature as the hot water.
11. Now quickly transfer the solid from the hot water to the calorimeter
without splashing any water.
12. Place the lid onto the calorimeter and stir the water very gently and
record the final equilibrium temperature (T3).
11. Repeat the previous steps for a different metal.
Specific heat of various solids by the method of heat transfer =
Define specific heat.
Define the unit of specific heat.
How heat capacity is related to universal gas constant?
What is the specific heat of salt water?
Which metal has highest specific heat?
J/kg oC
Circuit Diagram
Model graph
Determination of capacitance using LCR bridge
Aim: To determine the capacitance using LCR bridge.
Apparatus Required: A PC oscilloscope such as the 2205 with PicoScope Education
software, A capacitor of unknown value (approximately 1 µF for the values used here), A
battery (a 9 V PP3 was used but any type or value can be used), Suitable leads and
A storage oscilloscope enables the voltage/time graph for a capacitor charging through a
resistor to be displayed and, from the print-out, a value of the time constant for the circuit to
be calculated. This provides all the information required for calculating the value of the
Capacitance C = t/R (Farad)
Where, R-resistance included in the circuit (ohm)
t- The time constant of the circuit, it is the time taken for the value of Vc to reach 63.2% of
the final value which can be taken as Vs (second)
Vc - the potential difference across the parallel plate capacitor (voltage)
Vs – voltage of the DC supply (voltage)
When an uncharged capacitor is connected to a DC supply through a resistor, the
potential difference across its plates (Vc) increases rapidly at first but then slows down until
the value is equal to that of the DC supply (Vs). When Vc = Vs then there is no current
flowing through the resistor and the capacitor is fully charged. The time constant of the
circuit, t, is defined as the time taken for the value of Vc to reach 63.2% of the final value
which can be taken as Vs. We also have that t = RC where R is the value of the resistor in
ohms and C is the capacitor value in farads. For a full treatment of the mathematics behind
this definition see the Appendix at the end of this note.
The PicoScope 2205 oscilloscope can give a time axis of almost any value and so the values
of R and C can be any convenient value. It should be noted however that the input impedance
of the PicoScope 2205 is approximately 1 MΩ and will be connected in parallel with the
capacitor. If this input impedance is not to significantly change the maximum value of Vc
then the charging resistor should have a value < 1 MΩ by a factor of about 100.
Experiment setup
Arrange the components to give a series RC circuit as shown. Some means of discharging the
capacitor between measurements is required and a flying lead will suffice. In our experiment
a 1 µF capacitor was used with a 3900 Ω resistor and a 9 V battery.
The PC oscilloscope settings are typically: Timebase 2 ms/div, Channel A input +10 V DC,
Trigger Single shot, rising to capture a pulse when capture is started. Use a sensitivity of 150
mV or so at a pre-trigger value of about 5%.
Circuit Diagram
Discharge the capacitor by shorting across its terminals with the flying lead. Press the space
bar on the PC and close the switch. A trace similar to that shown should be obtained. Adjust
the timebase settings as required and repeat the run. To repeat a run, open the switch,
discharge the capacitor, press the space bar on the PC and close the switch again. Save the
screen image.
Measurements can either be made from the computer printout or the screen cursors available
in PicoScope can be used to determine the maximum value of Vc. Once the time constant
value of Vc has been determined the cursors can be repositioned so that the time can be read
from the display. Measure the maximum value of Vc reached on the printout. Call this value
Vs. Calculate 63.2% of Vs and draw a dotted line across the graph parallel to the x–axis.
From the point where the 63.2% line crosses the trace, drop a line down to the x-axis (see
diagram above). Calculate the value of t from the graph.
If t = RC then C = t/R.
From the example trace Vs = 9 V and 63.2% is 5.7 V giving a t value of 4.1 ms
Since R = 3900 we get C = 4.1 x 10–3 / 3900.
Which gives C = 1.05 µF
The value of the given unknown capacitance =
1. What is the use of a capacitor in a circuit?
2. Why the capacitor blocks d.c?
3. For which medium capacitance is high? Air, Mica, Water or Metal
4. How can we change the value of capacitance in a variable capacitor?
5. Which type of capacitor is preferred when there is high frequency in the circuit?
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