MAGNETISM
MAGNETIC FIELD
TORQUE ON A COIL
FORCE BETWEEN 2 PARALLEL CURRENT CARRYING CONDUCTOR
FORCE ON A CURRENT CARRYING CONDUCTOR IN A UNIFORM MAGNETIC FIELD
APPLICATION OF MOTION OF CHARGED PARTICLE
MAGNETIC FIELD DIAGRAMS
FORCE ON MOVING CHARGED PARTICLE IN AN UNIFORM MAGNETIC FIELD
DEFINE
a region around a magnet where a magnetic force can be experienced
vector quantity
2 poles : North, South
N-N or S-S ( repel)
S-N or N-S (attract)
LINES
1) Magnetic field lines leaves the north pole and enter south pole of a magnet
2) Can be represented by straight lines/ curves
3) Do not intersect one another
SOURCES
4) Direction of magnetic field vector B is tangent to field line
5) Number of line per unit area is proportional to the strength of the magnetic field
Earth
Bar Magnet
A moving charge/ current carrying conductor creates a magnetic field in the surrounding space
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INTO
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OUT
Long Straight Wire
Circular Coil
Solenoid
B = μI/2πr
B = μNI/2r
Centre : B = μnI
End : B = 1/2 μnI
n = N/L
Torque (C) on the coil = BANI sinθ
Maximum torque (Co) = BANI
F = qvB sinθ
If the moving charge is (-), the direction of the force is opposite to that indicated by RIGHT HAND PALM RULE
Circular motion of a charge in a uniform magnetic field
1) a magnetic force, FB will be exerted on the charged particle.
2) The FB always perpendicular to the velocity and directed towards the centre of circular path
3) The Fg causes particle to move on circular path
4) Since Fg=Fc
qvB sin 90 = mv^2 / r
5) r = mv/qB
6) T = 2pir/v = 2pi/w = 2pi*m/qB
Two parallel conductors will exert a force on each other when current is passing through both of them
The force depends on the direction of current. Same direction : attractive. Opposite direction : repulsive
Each conductor is exposed to the other's wire magnetic field, hence experiences magnetic force
F/L = μI1I2/2nd
1) By the Right Hand Rule, force on the B field on the charged point upwards
2) If the electric force (Fe), is just balanced by the (FB), it will follows a straight line path.
3) As Fe = FB , qE = qvB sin 90, v = E/B
4) Force on the E field on the charge point downwards
When a current carrying conductor is placed in a B, thus a magnetic force will acts on that conductor
F = BIL
Magnitude : FB = BIL Sinθ
The direction of force can be determine using Right Hand Palm Rule