# Physics E&M

A slab of metal of volume V is made into a rod of length L. The rod carries current I when the electric field inside is E.Find the resistivity of the metal?

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1. NVM again sorry I solved it.

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2. The cross sectional area of the rod is A = V/L. The resistance of the rod is
R = (resistivity)*L/A
= (resistivity)*L^2/V

I = V/R = E*L/R,
= E*L*V/(resistivity)*L^2
= E*V/[resistivity*L]

Solve for resisitivity

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3. The rod is now stretched so that its length is doubled. If the electric field remains the same, what is the new current I' in the rod? The answer to the first part was (EV)/(IL)

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4. If you double the length, the area decreases by a factor 1/2. In this case you keep the E field the same and not the voltage. Voltage doubles. Resistance increases by a factor of 4. the current therefore decreases by a factor of two.

You can also see this from your formula. resistivity, E, and V are constant, so I is inversely proportional to L.

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5. Thank you phew I got it done on time

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6. A piece of copper is made into a rod with a square cross-section. The side of the square is 2.00 centimeters. The resistivity of copper is 1.7 \cdot 10^{-8} \Omega\cdot {\rm m}. An unknown electric field E, directed along the rod, creates a current of 12.0 amperes through the rod. Find the magnitude of E.

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7. kjlk

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8. 1)
let:
V = change in V
E = electric field
I = current
L = length of rod
A = square cross-section
R = resistance

given equations (from physic):

E = V / L

I = V / R

R = rho * L /A

by that:
E*L =V
IR =V

R = rho *L /A

EL = I * rho * L /A

E = I * rho /A

E = 12 * 1.7e-8 / (2/100)^2
= 5.1e-4

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