a heater wire length 50cm and a 1mm^2 in cross section carries a current of 2A when connected across a 2 V battery. what is the resistivity of the wire?
Given:
L = 50 cm = 0.50 m.
A = 1mm^2 = 10^-6 m^2.
E = 2 Volts.
I = 2 Amps.
R = E/I = 2/2 = 1 Ohm.
R = p*L/A.
1 = p*0.5/10^-6,
p = 2*10^-6 Ohm-meters = Resistivity.
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To find the resistivity of the wire, we can use the formula:
Resistance (R) = (Resistivity (ρ) * Length (L)) / Cross-sectional Area (A)
In this case, we are given:
Length (L) = 50 cm = 0.5 m (since 1 m = 100 cm)
Cross-sectional Area (A) = 1 mm^2 = 1 x 10^-6 m^2 (since 1 m^2 = 1 x 10^6 mm^2)
Current (I) = 2 A
Voltage (V) = 2 V
First, we need to calculate the resistance using Ohm's Law:
Resistance (R) = Voltage (V) / Current (I)
R = 2 V / 2 A
R = 1 Ω (Ohm)
Now, rearrange the resistivity formula to solve for resistivity (ρ):
ρ = (R * A) / L
Substitute the given values:
ρ = (1 Ω * 1 x 10^-6 m^2) / 0.5 m
Simplify the calculation:
ρ = 2 x 10^-6 Ω·m
Therefore, the resistivity of the wire is 2 x 10^-6 Ω·m.