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Current carrying conductor in a magnetic field.
N S
+ -
XN S
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.
Current carrying conductor in a magnetic field.
N S
+ -
XN S
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.
Right Hand Wire Rule
A .............................. field is generated around any conductor when an electric current flows through I t.
Magnetic EffectMagnetic Effect
Electric current
Wire with current coming ...................... you Wire with current
going ............... from you
x
Magnetic EffectMagnetic 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
SOLENOIDSOLENOID A coil generates a very concentrated (strong)
magnetic field in its center.Electric current
x
N
S
x
x
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 ..............!!!
Right Hand Rule - Solenoid
Solenoid Field
• Note field same as bar magnet • Inside S N!!!!!
Web Applet Demo>>
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
C
THE ELECTRIC MOTOR
N
current
S
CurrentCurrent
Magnetic Magnetic FieldField
ThrustThrust
A
B
D
+ -
Rotation
ThrustThrust
THE MOTOR EFFECT
N S
+ -
XN S
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.
………………..
………………....………………..
ForefingerSecond
Thumb Magnetic fields ……………………..
THE MOTOR EFFECT
N S
+ -
XN S
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.
CurrentCurrent
FieldFieldThrustThrust
ForefingerSecond
Thumb Magnetic fields strengthened!
THE ELECTRIC MOTOR
• Supply the missing labels and indicate the polarity of the battery.• Show how the direction of rotation supports your decision.
THE ELECTRIC MOTOR
- +-- ++
Electromagnetic Induction
N S
+-An electrical conductor that is accelerated through a magnetic field will experience an …………….. .electrical current according to Fleming's .................Hand Rule.
………..
..................................
..................................
........................
Forefinger
SecondThumb
Faraday’s Law
The size of the induced ……………….is directly proportional to the …………….of change of the magnetic …….. ……………….
.............. current
Electromagnetic Induction
N S
+-An electrical conductor that is accelerated through a magnetic field will experience an INDUCED electrical current according to Fleming's Right Hand Rule.
Thrust
CurrentCurrent
FieldField
ThrustThrust
Forefinger
SecondThumb
Faraday’s Law
The size of the induced current is directly proportional to the rate of change of the magnetic flux linkage.
Alternating Current• Voltage & current …………. constantly.
Alternating Current• Voltage & current vary constantly.
Coil vertical – 90o to the magnetic field
A maximum current would be produced at the same time as the maximum voltage.
V = vmax sin 2ftI = Imax sin 2ft
0o 90o 180o 270o 360o
0o
360o
90o
180o
270o
Alternating and Direct Current • What would be the equivalent constant/direct current /voltage to
give the same effect as an alternating current?
??
X
~
X
Power Transmission
Reducing the current ensures ......................................in transmission.
25 000V0,6A
0.6 A
Cable resistance 200 km 10 Ω
Voltage dropV =
Power lost =
250 000V0,06A
0.6 A
Cable resistance 200 km 10 Ω
Voltage dropV =
Power lost =
P = ...........................................
P = ..............................................
Power Transmission
Reducing the current ensures less voltage/power lost in transmission.
25 000V0,6A
0.6 A
Cable resistance 200 km 10 Ω
Voltage dropV = IR = (0,6)(10) = 6.0 V
Power lost = VI = (6)(0.6) = 3,6 W
250 000V0,06A
0.06 A
Cable resistance 200 km 10 Ω
Voltage dropV = 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
RECTIFIED CURRENT