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E&M I. Griffiths Chapter 7. Example 7.9 Long straight wire with time dependent current. What is the induced electric field?. changing. Recipe Find B(t) using Ampere’s law Find magnetic flux current – d F / dt Find E(t) using Faraday’s law. Amperian loop. Step 1:. Ampere’s law:.
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E&M I Griffiths Chapter 7
Example 7.9 Long straight wire with time dependent current. What is the induced electric field? changing Recipe Find B(t) using Ampere’s law Find magnetic flux current –dF/dt Find E(t) using Faraday’s law Amperian loop Step 1: Ampere’s law:
Step 2: Flux current Direction is consistent with Lenz’s law Step 3: Faraday’s law: E(s,t)
But this is because wire was infinitely long This was an example of a method called the Quasistatic approximation. We used an equation from magneto-statics (Ampere’s law) to find B. Then we used an equation from electrodynamics (Faraday’s law) to find E. This approximation is valid if we can ignore “retardation”. That means that all parts of space see the changes in current instantaneously. That means we have to be sufficiently close to the wire, and the changes have to be sufficiently slow: Speed of light Characteristic time for changes in current
Because B1 is proportional to I1 Coefficient of proportionality, “Mutual Inductance”
2 Purely geometric, and symmetric under exchange 1 2.
Symmetry of mutual inductance has a practical benefit Complicated loop 1 Simple loop 2 The symmetry of M12 means that the flux through loop 2 when there is current I in loop 1 is the same as the flux through loop 1 when the same current I is in loop 2. Subscripts can be dropped M12 = M.
A change in I1, creates a change in B1 by Ampere’s law. The change in B1 induces an EMF in loop 2 by Faraday’s law. A change in I1 induces an EMF in any nearby loop, including in loop 1 (itself). -dF1/dt “Self inductance” Lenz’s law: self EMF opposes change in I1
Close the switch. Current starts tries to start flowing. “Back EMF” appears to oppose this current. Practical examples of mutual and self inductance Voltage transformer: Step up, step down. Wire-wound resistor: No good for high frequencies. Pulse transformer: Transmit voltage pulse without transmitting background voltage. Bias-T: Separate DC power supply from high-frequency signals. Etc, etc.
Magnetic energy. Work is required to push current into an inductor against the back EMF. d = a voltage = potential = potential energy per unit charge Gravitational analogy This work gets stored as potential energy d W = -F dx
d Total work done becomes an integral over current, from 0 to I. The work done depends only on geometry (L) and on the final current I.
As for electric energy, there are different ways to express magnetic energy.
This integral is over the volume of the wire, but it can be extended to all space considering that J = 0 outside the wire. Integral follows the wire with current I By Ampere’s law Product rule #6 div. Thm. Magnetic energy density Fields at infinite boundary are zero
Example 7.10. Coaxial solenoids 1: Radius a, n1 turns per unit length 2: Radius b, n2 turns per unit length If current = I in coil 1, what is magnetic flux F2 through coil 2? The field from coil 1 is non-uniform inside coil 2, which makes this problem hard. But, Symmetry of M! Let current I flow in coil 2 instead. Its field is uniform over coil 1. So if I flows in coil 1, Same M I
Example 7.11. Toroidal coil. N total turns of wire. What is self inductance L? Up arrows are currents in individual wires on inner surface of toroid. These are enclosed by Amperian loop, radius s. Total enclosed current = NI Total magnetic flux through all N turns of toroid = LI
Example 7.12. L-R circuit. Close the switch. What is I(t)? Kirchoff’s law 1st order inhomogenous diff. eq. Initial condition One undetermined coefficient
Example 7.13. While there is current, the coax stores magnetic energy. How much? Energy density, u = Integrate over cylindrical shells Radius s, thickness ds, length l So But
Two recipes to find L. Find flux for given I, then use F = LI Find magnetic energy W for given I, then use W = (1/2) L I2 The second method is a volume integral of a scalar, and it is usually the easier method.
Electrodynamics so far Something missing
Problem: Not necessarily zero Mathematically zero Continuity equation We can fix the problem by adding the opposite of this to Ampere’s law. New Ampere’s law + Maxwell’s correction: A changing E can induce a B-field Acts like a current: “The displacement current”
Old Ampere’s law and a charging capacitor The perimeter is the same, but no usual current penetrates the surface This current pierces the surface bounded by the loop New Ampere’s law Now any surface with the same perimeter gives the same answer
Basic equations to solve any Electrodynamics problem 1. Maxwell’s Equations for the fields Gauss Faraday Ampere + Maxwell 2. Force law Lorentz 3. Plus boundary conditions
Maxwell’s Equations determine how charges affect the fields Charges Fields Lorentz force law tells how fields affect the charges Fields Charges Maxwell’s equations would be symmetric if there were magnetic charges, but there are not any.
Maxwell’s equations include all charges and all currents. In matter, some are bound and some are free. (I.e. some are intrinsic and some are extraneous.) Bound charge Bound current density Polarization current density
Total charge Total current density
Maxwell’s equations in terms of D and H and free charge 8 differential equations for 12 quantities E, B, D, H: Not enough equations!
“Constitutive Equations” are additional relations between E and D, H and B. These depend on the medium. For linear media
To obtain the needed boundary conditions, use integral form of Maxwell’s equations
Normal component of electric induction D is discontinuous by the surface free-charge density. If there is none, then this component of D is continuous. Normal component of magnetic induction B is always continuous across any boundary.
H1 Surface free current density A/m H2 Amperian loop Goes to zero: Displacement current is finite, but loop is infinitesimal or Tangential components of H are discontinuous by free surface current density components that flow perpendicular to them.
Loop Goes to zero: Magnetic flux current is finite, but loop is infinitesimal Tangential component of electric field is continuous at a boundary
The total charge in some volume V is If charge flows out of this volume through the boundary surface Div Theorem Integrands have to be the same at every point if equality holds for any volume V Continuity Equation
Some volume containing charges and currents and fields. If a charge moves, then work is done on it by the field. Rate at which work is done on a single charge by the fields = Rate at which work is done on all the charges in V by the fields = Power density delivered by the fields Using Ampere-Maxwell Eq.
Product rule #6 Faraday’s law Div Theorem Poynting’s Theorem Work done on charges by fields per unit time Decrease in field energy per unit time Rate of field energy loss through boundary
Energy per unit time per unit area Energy flux density Energy per unit time Energy flux Poynting’s Theorem The work dW done on the charges increases their mechanical energy Statement of Energy Conservation
Cylindrical wire with current I Integrate over the cylindrical surface The source of Joule heat is field energy that flows into the wire from outside.