Answer:
0 J
Explanation:
= Work done on the satellite in circular orbit about earth by earth
= Force on gravity on satellite by earth
= displacement of the satellite
= Angle between the force on gravity and displacement = 90
We know that, Work done is given as
-
Eddy Current Testing
Introduction
Basic Principles
History of ET
Present State of ET
The Physics
Properties of Electricity
Current Flow & Ohm's Law
Induction & Inductance
Self Inductance
Mutual Inductance
Circuits & Phase
Impedance
Depth & Current Density
Phase Lag
Instrumentation
Eddy Current Instruments
Resonant Circuits
Bridges
Impedance Plane
Display - Analog Meter
Probes (Coils)
Probes - Mode of Operation
Probes - Configuration
Probes - Shielding
Coil Design
Impedance Matching
Procedures Issues
Reference Standards
Signal Filtering
Applications
Surface Breaking Cracks
SBC using Sliding Probes
Tube Inspection
Conductivity
Heat Treat Verification
Thickness of Thin Mat'ls
Thickness of Coatings
Advanced Techniques
Scanning
Multi-Frequency Tech.
Swept Frequency Tech.
Pulsed ET Tech.
Background Pulsed ET
Remote Field Tech.
Quizzes
Formulae& Tables
EC Standards & Methods
EC Material Properties
-
Current Flow and Ohm's Law
Ohm's law is the most important, basic law of electricity. It defines the relationship between the three fundamental electrical quantities: current, voltage, and resistance. When a voltage is applied to a circuit containing only resistive elements (i.e. no coils), current flows according to Ohm's Law, which is shown below.
I = V / R 
Where:
I =
Electrical Current (Amperes)
V =
Voltage (Voltage)
R =
Resistance (Ohms)
Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. The formula can be reorganized so that the relationship can easily be seen for all of the three variables.
The Java applet below allows the user to vary each of these three parameters in Ohm's Law and see the effect on the other two parameters. Values may be input into the dialog boxes, or the resistance and voltage may also be varied by moving the arrows in the applet. Current and voltage are shown as they would be displayed on an oscilloscope with the X-axis being time and the Y-axis being the amplitude of the current or voltage. Ohm's Law is valid for both direct current (DC) and alternating current (AC). Note that in AC circuits consisting of purely resistive elements, the current and voltage are always in phase with each other.
Exercise: Use the interactive applet below to investigate the relationship of the variables in Ohm's law. Vary the voltage in the circuit by clicking and dragging the head of the arrow, which is marked with the V. The resistance in the circuit can be increased by dragging the arrow head under the variable resister, which is marked R. Please note that the vertical scale of the oscilloscope screen automatically adjusts to reflect the value of the current.
See what happens to the voltage and current as the resistance in the circuit is increased. What happens if there is not enough resistance in a circuit? If the resistance is increased, what must happen in order to maintain the same level of current flow?
The tangent looks good.
The curve is a bit crooked, at the 0.9 and 1.
But overall, cool graph.
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
