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Question: An aluminum wire carrying a current has a diameter of 0.84mm. The electric field in the wire is 0.49...

An aluminum wire carrying a current has a diameter of 0.84mm. The electric field in the wire is 0.49V/m. What is,
(a) the current carried in the wire
(b) the potential difference between two points in the wire 12.0m apart
(c) the resistance of 12 m length of the wire
Specific resistance of aluminum is 2.75×108Ωm2.75\times {{10}^{8}}\Omega m

Explanation

Solution

The specific resistance of the above material as well as the diameter of area of the cross section of the wire is given to us. Hence we can determine the resistance of the wire across its varying length. Hence we can determine the current in the wire, potential difference between any two points and the resistance of the wire as well.
Formula used:
E=VdE=\dfrac{V}{d}
R=ρlAR=\dfrac{\rho l}{A}
I=VRI=\dfrac{V}{R}

Complete step-by-step answer:
In the above question it is given to us that the diameter of the wire is d=0.84mm. hence the area of cross section of the above wire is,
A=π(d2)2 ⇒A=π(0.84×10−32)2 ⇒A=π(0.42×10−3)2 ⇒A=3.14×0.1764×10−6 ⇒A=0.553×10−6m2 \begin{aligned} & A=\pi {{\left( \dfrac{d}{2} \right)}^{2}} \\\ & \Rightarrow A=\pi {{\left( \dfrac{0.84\times {{10}^{-3}}}{2} \right)}^{2}} \\\ & \Rightarrow A=\pi {{(0.42\times {{10}^{-3}})}^{2}} \\\ & \Rightarrow A=3.14\times 0.1764\times {{10}^{-6}} \\\ & \Rightarrow A=0.553\times {{10}^{-6}}{{m}^{2}} \\\ \end{aligned}
Let us say the electric field across the wire of length d is E. Hence the potential V across the length d is given by,
E=VdE=\dfrac{V}{d}
In the above question we are asked to determine the potential difference across a length of 12m between two points. Hence from the above expression we get it equal to,
E=Vd ⇒V=Ed ⇒V=0.49V/m×12m ⇒V=0.49V/m×12m ⇒V=5.88V \begin{aligned} & E=\dfrac{V}{d} \\\ & \Rightarrow V=Ed \\\ & \Rightarrow V=0.49V/m\times 12m \\\ & \Rightarrow V=0.49V/m\times 12m \\\ & \Rightarrow V=5.88V \\\ \end{aligned}
Hence the potential difference between two points in the wire 12.0m apart is 5.88V. The resistance (R)of the above wire of aluminum of length l and specific resistance ρ=2.75×108Ωm\rho =2.75\times {{10}^{8}}\Omega m is given by,
R=ρlAR=\dfrac{\rho l}{A}
Hence the resistance of wire of length 12m is numerically equal to,
R=ρlA ⇒R=2.75×108Ωm×12m0.553×10−6m2=59.6×1014Ω \begin{aligned} & R=\dfrac{\rho l}{A} \\\ & \Rightarrow R=\dfrac{2.75\times {{10}^{8}}\Omega m\times 12m}{0.553\times {{10}^{-6}}{{m}^{2}}}=59.6\times {{10}^{14}}\Omega \\\ \end{aligned}
Now let us determine the current in the wire. We know the potential difference across the length of 12m as well as the resistance of the wire of length 12m. Therefore from ohms law the current in the wire (II) is equal to,
I=VR ⇒I=5.88V59.6×1014Ω=0.098×10−14A=9.8×10−16A \begin{aligned} & I=\dfrac{V}{R} \\\ & \Rightarrow I=\dfrac{5.88V}{59.6\times {{10}^{14}}\Omega }=0.098\times {{10}^{-14}}A=9.8\times {{10}^{-16}}A \\\ \end{aligned}
Hence the current in the wire is equal to 9.8×10−16A9.8\times {{10}^{-16}}A.

Note:
In the above solution we stated that the current in the circuit is the same as that of the current across the wire of length 12m. This is because we can consider the entire remaining portion of the wire to be connected in series. And since current in series remains the same, we could imply the current across the entire wire is also the same.