DEFINITION:
Diamagnetic - substances with weak magnetic moments opposite to the applied magnetic field
Paramagnetic - substances with weak magnetic moments in the same directions as the applied magnetic field
Ferromagnetic - substances with magnetic moments between atoms that align to create strong magnetization, which remains after the external magnetic field is removed.
Magnetic flux through a closed surface is:
Biot - Savart Law
The magnetic field, dB at a point P due to an element of length ds that carries a current I is detailed in the following equation:
where:
r is the distance from the element to point P
vector r is the unit vector = 1
We find the total field at point P by integrating this expression over the entire current distribution
Amperes Law:
The law states the line integral
around any closed path is equal to permeability of free space times the enclosed current.
around any closed path is equal to permeability of free space times the enclosed current.
Gauss' Law for Magnetism states the total magnetic flux over a closed surface integral equals zero:
Magnitude of the magnetic field: where the field lines are concentric with the current carrying wire:
- B = [(uo)(I)]/[2pi(r)]
uo = permeability of free space
Where:
r = distance from a long straight wire carrying an electric current
Magnitude of the field around a solenoid:
- B = [(N)(uo)(I)]/[2pi(r)]
Magnitude of the field around a toroid:
- B = [(N)(uo)(I)]
Magnetic force per unit length between two parallel wires separated by a distance a, carrying current I1 and I2 has a magnitude of:
= FB / legnth
Class Practice examples:
CP 11-06-13
BIOT-SAVART LAW
BIOT-SAVART LAW
1. Calculate the magnetic field outside am infinitely long, straight wire.
*set up in integral to find the Magnetic field (B) at point P
i) write the Biot Savart Law:
ii) solve I(dl) x (r unit vector) = I(dl)(1)sin(a) = I(dl)sin(a)
iii) solve r^2 FIRST, SET AXIS & ORIGIN
r^2 = a^2 + x^2
iv) plug back everything back into the original equation
dB = [(uoI(dl)sin(a)]/[4pi(a^2 + x^2)]
v) get dl in terms of x:
x = l
vi) get r in terms of x
r = 1 unit vector
vii) integrate[dB]dx from 0 -> infinity
where uoI/4pi is a constant
*note that the integral must be multiplied by 2 to encompass the entire infinite wire.
2*(uoI/4pi) integral[dB = [(dx)sin(180)]/(a^2 + x^2)] from 0 -> infinity
2*(uoI/4pi) integral[dB = (dx)/(a^2 + x^2)] from 0 -> infinity
Conclusion: dl is always parallel to I, same direction!
given:
Current density (J) is uniform
R = radius
infinite wire
i) Create an Amperian loop
- SURFACEint[B dot dl] = (uo)(I)
ii) pick an arbitrary direction for the current, I
iii) determine the symmetry of the problem to discuss and describe the magnetic field
- If the magnetic field exists (B) it cannot change as you go around the loop.
- At the origin, the physical situation at one point is the same at another point.
- B is not a function of dl because due to the inherent symmetry of the universe.
iv) solve the equation for the Amperian loop
SURFACEint[B dot dl] = (uo)(I)
[B] * SURFACEint[dlcos(theta)] = (uo)(I)
- Magnetic field (B) must be perpendicular to the current (I) and in the same plane as point P
THEREFORE there can be no right or left component. B is therefore tangent to to the current path.
2. determine if the magnetic field is CW or CCW
this can be accomplished using two different methods.
method 1:
Use the RHR to determine the Current (I) path.
Leave B as +/-
Apply Amperes law
Solve for B (should be positive, magnitudes are always positive)
method 2: apply amperes law
i) select a direction for the magnetic field (CW or CCW)
- If B is in the direction of dl, the angle between +/- B and dl is 0
- If B is opposite of the direction of dl, the angle between +/- B and dl is 180
- If I is in the same direction as B, I is positive
- If I is opposite of the direction of B, I is negative.

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