An aluminum “12 gauge” wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field

Question

An aluminum “12 gauge” wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

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Khoii Minh 5 months 2021-09-01T10:51:03+00:00 1 Answers 0 views 0

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    2021-09-01T10:52:40+00:00

    Complete Question

    An aluminum “12 gauge” wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

    I = 1.2 A at time 5 secs.

    Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

    Answer:

    The charge is  Q =2.094 C

    Explanation:

    From the question we are told that

        The diameter of the wire is  d =  0.205cm = 0.00205 \ m

         The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

         The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

           The electric field change is mathematically defied as

             E (t) =  0.0004t^2 - 0.0001 +0.0004

         

    Generally the charge is  mathematically represented as

           Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

    Where A is the area which is mathematically represented as

           A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

     So

           \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

    Therefore

          Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

    substituting values

          Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

         Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

    From the question we are told that t =  5 sec

               Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

              Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

             Q =2.094 C

         

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