3.4 O HM ’ S L AW · Part 2

and the resistance of the combination R C is [from Eq. ( . )], I ( . ) since V/I = R , the resistance of either of the slabs. Thus, doubling the length of a conductor doubles the resistance. In general, then resistance is proportional to length, l ∝ ( . ) Next, imagine dividing the slab into two by cutting it lengthwise so that the slab can be considered as a combination of two identical slabs of length l , but each having a cross sectional area of A / [Fig. . (c)]. For a given voltage V across the slab, if I is the current through the entire slab, then clearly the current flowing through each of the two half-slabs is I / . Since the potential difference across the ends of the half-slabs is V , i.e., the same as across the full slab, the resistance of each of the half-slabs R is . ( / ) I I ( . ) Thus, halving the area of the cross-section of a conductor doubles the resistance. In general, then the resistance R is inversely proportional to the cross-sectional area, A ∝ ( . ) Combining Eqs. ( . ) and ( . ), we have l A ∝ ( . ) and hence for a given conductor l A ρ ( . ) where the constant of proportionality ρ depends on the material of the conductor but not on its dimensions. ρ is called resistivity . Using the last equation, Ohm’s law reads I l I A ρ ( . ) Current per unit area (taken normal to the current), I / A , is called current density and is denoted by j . The SI units of the current density are A/m . Further, if E is the magnitude of uniform electric field in the conductor whose length is l , then the potential difference V across its ends is El . Using these, the last equation reads