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The voltage is defined as the integral of the electic field along a line joining two points. Herein is derived for your pleasure (i)the voltage due to a time varying magnetic field and (ii)the voltage that develops across a resistor.


In terms of an electromagnetic field, the voltage along a closed line is given by the rate of change of magnetic flux through that circuit. To show this, start with one of Maxwell's equations, Faraday's law

curl E=-δB/δt

Here E is the electric field and B the magnetic field. Now integrate over the surface that the closed circuit describes.

∫curl E.ds=-δ(∫B.ds)/δt

Stoke's Theorem allows us to express the surface integral of the curl of a vector as a closed line integral ∫c of that vector. Furthermore, the surface integral of the magnetic field is simply the magnetic flux Φ through that surface. So the previous equation becomes


To simplify matters, assume that the closed line is a circle of radius R. Then the integral of the electic field over the line is simply 2πRE//, where E// is the component of the electric field along the circle. This, in fact, is the voltage around the circle (since voltage is the integral of electric field around the circuit). Thus,

It follows that the faster the increase in magnetic flux, the more negative the voltage generated. This equation relates to the voltage due to a solenoid.


If there is a steady current flowing through an ohmic resistor, the voltage between the two ends of the resistor is the product of the electric field and the length L of the component.(V=EL). From Ohm's law, we know that the current is proportional to the electric field. The constant of proportionality is the electrical conductivity σ The inverse of σ is the resistivity ρ. Thus we have E=ρL and


where J, the current density, in this case is defined as the current per unit cross-sectional area in the resistor.It is beyond the scope of this w/u to explain the the total resistance R of the resistor is directly proportional to the resistivity and length and inversely proportional to the cross-sectional area (R=ρL/A).The previous equation may be rewritten


Of course the cross-sectional area multiplied by the current density is simply the total current I flowing in the resistor (assuming that J is uniform across the area) One finally has an expression for the voltage in terms of the more easily measured quantities R and I.


Obviously, the greater the resistance encountered by the current the stronger the potential difference (voltage) developed across it.