Electron and Proton Atomic Energy - Assignment Example

• The Relationship between B and I for a wire-carrying current is electromagnetic interaction due to associating elementary particles. Direct Relationship
• Relationship between B and R for a wire-carrying current Inverse Relationship
• Ohm’s law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. I=V/R
• In Ohm’s law, I and Rare inversely proportional to each other = True
• In Ohm’s law, V and Rare directly proportional to each other= True
• A typical galvanometer deflects full scale (the maximum current that you can pass through it) to a draft of 0.010A, and it has a resistance of 50ohms. How can it be converted into a:
  • 5A ammeter?

V = 0.010*50 = 0.5 V

Rs =   0.5 /4.99 = 0.1002 Ω.  By connecting a shunt resistor of 0.1002 Ω in parallel.

• 20V voltmeter?

Rt = 20 / 0.010 = 2000 Ω. By connecting a resistor of 1950 Ω in series.

• 50V voltmeter?

Rt = 50 / 0.01 = 5000 Ω. By connecting a resistor of 4950 Ω in series.

1. Exit Questions
   • What is the unit of a magnetic field? B Teslas ( T)
   • What is the company of the magnetic flux? ƟB Webers (Wb)
   • As given by Ampere’s law, the magnetic field increases the distance of the point from the wire.
   • Is there a magnetic field outside of a solenoid? No, outside the solenoid, the area is also fragile due to this cancellation effect. For a solenoid which is long compared to its diameter, the site is very close to zero.
   • What is the force's direction (up or down) on an electron moving along the positive x-axis and entering a magnetic field perpendicular to the chalkboard (going inside)? Up
   • A galvanometer is used to measure current. True
   • A voltmeter has a high internal resistance. True
   • An ammeter has a low internal resistance. True
   • An ammeter is used in parallel in an electrical circuit. False it will cause a short circuit
   • A voltmeter is used in series in an electrical circuit. True

2. Practice Home Work
   • A typical galvanometer has resistance of 60ohms and it can detect full scale for a voltage of 120mV across its terminals. How can it be made into a 4A ammeter?

L = 0.12 / 60 = 0.002 A

Rs     = 0.12 / 3.998 = 0.030015 Ω.  By connecting a shunt resistor of 0.030015 Ω in parallel

• A typical galvanometer deflects full scale of (the maximum current that you can pass through it) to a draft of 0.015A, and it has a resistance of 40Ohms. How can it be converted into a: