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.
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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
∫cE.dl=-δΦ/δt
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,
V=-δΦ/δt
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.
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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
Vresistor=ρJL
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
Vresistor=RAJ
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.
Vresistor=RI
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Obviously, the greater the resistance encountered by the current the stronger the potential difference (voltage) developed across it.