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Electromagnetics (Electric Fields (Capacitance of a Coaxial Capacitor…
Electromagnetics
Electric Fields
Capacitance of a Coaxial Capacitor
Gauss' Law to go from charge to electric field strength
Due to symmetry, electric field strength will be same at all points on surface of Gaussian cyclinder and acts perpendicular to all surface elements so parallel to area vector
Energy in a charged capacitor
Electric Energy Density
Capacitance of Parallel Plate Capacitor
calculate E1 due to positive plate
charge enclosed by Gaussian cuboid: ρA = Q
E between plates = E1 + E2 = 2E1
Potential Difference at point P due to continuous line charge
Electric Field produced by infinitely long line charge
Cylindrical Gaussian surface, length L, around line
Capacitance of speherical capacitor
Magnetic Fields
Magnetic Field created by a toroid of N turns
C1 - surface within the centre hole of the toroid - no current passes through this circular surface
C2 - surface through the centre of the toroid itself - N amounts of current I pass downwards through the surface, none flows upwards
C3 - surface around outside of toroid - N amounts of current I pass downwards through the surface, and same amount upwards so net current flow is zero
Inductance of a Toroid with N turns of wire
B is perpendicular to dl
Magnetic Field due to current carrying wire
Magnetic Field due to a solenoid
AB: dl parallel to B so B.dl = BS
AD: B perpendicular to dl so B.dl = 0
BC: B perpendicular to dl so B.dl = 0
DC: outside solenoid so B=0 - ampere's law is zero
length S, n turns per metre
EMF induced in closed circuit travelling with velocity v in uniform magnetic field B
Edge 1: From y = 0 to y = s, B = 0 as relevant part of frame is outside magnetic field
Edges 2 and 3: dl parallel to v therefore v x dl = 0
force on parts of wire that are inside B acts upwards - force is on sides of the wire so doesn't contribute to EMF
Edge 4: dl perpendicular to B and v so v x dl = 1
v x dl parallel to B so B.(v x dl) = 1
