1. Current carrying conductor in a magnetic field. NS + - X NS A current carrying conductor in a magnetic field experiences a..................... The direction is given by............................................... Motor rule. Force Fields in............................... direction they cancel out/become weakened - resulting in a downward force.

2. Current carrying conductor in a magnetic field. NS + - X NS A current carrying conductor in a magnetic field experiences a force. The direction is given by Flemmings Left Hand Motor rule. Force Fields in opposite direction they cancel out/become weakened - resulting in a downward force.

3. Right Hand Wire Rule

4. A.............................. field is generated around any conductor when an electric current flows through I t. Magnetic Effect Electric current Wire with current coming...................... you Wire with current going............... from you x

5. Magnetic Effect A magnetic field is generated around any conductor when an electric current flows through it. Electric current Wire with current coming towards you Wire with current going away from you x

6. SOLENOID A coil generates a very concentrated (strong) magnetic field in its center. Electric current x N S xx Increasing the number of coils strengthens the magnetic field. The.............. Hand Rule can also be applied to a solenoid! INSIDE THE COIL THE FIELD GOES FROM...............TO..............!!!

7. Right Hand Rule - Solenoid

8. Solenoid Field Note field same as bar magnet Inside S  N!!!!! Web Applet Demo>>

9. THE ELECTRIC MOTOR Indicate the direction of rotation with an arrow. Show the application of the rule used to decide which way the coil will turn. N current S + - C A B D

10. C THE ELECTRIC MOTOR N current SCurrentMagneticField Thrust A B D + - Rotation Thrust

11. THE MOTOR EFFECT NS + - X NS A current carrying conductor in a magnetic field experiences a ……………. The direction is given by ……………………………….. rule. Fields in opposite direction they ………………………… ………………… - resulting in a …………………….. force.……….……….. ………. Forefinger Second Thumb Magnetic fields ……………………..

12. THE MOTOR EFFECT NS + - X NS A current carrying conductor in a magnetic field experiences a force. The direction is given by Fleming's Left Hand Motor rule. Force X Fields in opposite direction they cancel out/become weakened - resulting in a downward force.CurrentField Thrust Forefinger Second Thumb Magnetic fields strengthened!

13. THE ELECTRIC MOTOR Supply the missing labels and indicate the polarity of the battery. Show how the direction of rotation supports your decision.

14. THE ELECTRIC MOTOR - +

16. Electromagnetic Induction NS + - An electrical conductor that is accelerated through a magnetic field will experience an ……………...electrical current according to Fleming's.................Hand Rule. ………................................................ Forefinger Second Thumb Faraday’s Law The size of the induced ……………….is directly proportional to the …………….of change of the magnetic …….. ………………............... current

17. Electromagnetic Induction NS + - An electrical conductor that is accelerated through a magnetic field will experience an INDUCED electrical current according to Fleming's Right Hand Rule. Thrust Current Field Thrust Forefinger Second Thumb Faraday’s Law The size of the induced current is directly proportional to the rate of change of the magnetic flux linkage.

18. Alternating Current Voltage & current …………. constantly.

19. Alternating Current Voltage & current vary constantly. Coil vertical – 90 o to the magnetic field A maximum current would be produced at the same time as the maximum voltage. V = v max sin 2  ft I = I max sin 2  ft 0 o 90 o 180 o 270 o 360 o 0 o 360 o 90 o 180 o 270 o

20. Alternating and Direct Current What would be the equivalent constant/direct current /voltage to give the same effect as an alternating current? ?? X ~ X

21. Power Transmission Reducing the current ensures......................................in transmission. 25 000V 0,6A 0.6 A Cable resistance 200 km 10 Ω Voltage drop V = Power lost = 250 000V 0,06A 0.6 A Cable resistance 200 km 10 Ω Voltage drop V = Power lost = P =........................................... P =..............................................

22. Power Transmission Reducing the current ensures less voltage/power lost in transmission. 25 000V 0,6A 0.6 A Cable resistance 200 km 10 Ω Voltage drop V = IR = (0,6)(10) = 6.0 V Power lost = VI = (6)(0.6) = 3,6 W 250 000V 0,06A 0.06 A Cable resistance 200 km 10 Ω Voltage drop V = IR = (0,06)(10) = 0.6 V Power lost = VI = (0.6)(0.06) = 0.036 W P = VI = (25000)(0.6) = 15 000 W P = VI = (250000)(0.06) = 15 000 W

23. RECTIFIED CURRENT