Respuesta :

MissPhiladelphia
Electric field strength = resistivity of copper x current density
where
p= 1.72 x 10^-8 ohm meter
diameter = 2.05mm=.00205 m
current = 2.75 A
get first the current density:
current density = current/ cross section area
find the cross section area
cross section area = pi.(d/2)^2;  
cross section = 3.3 006x10-6 m^2
substitute the values 
current density = 2.75A/3.3006x 10-6m^2
current density=35.55 x1 0^2 A/m^2
Electric field stregnth =1.72 x 10^-8 ohm meter x 35.55 x10^2 A/m^2
Electric field stregnth= 46.415 Volts/m

The electric field strength of copper is 46.415 V/m.


onyebuchinnaji

The electric field strength of  the wire is 5.37 x 10⁻⁴ T.

Electric field strength of  the wire

The electric field strength of  the wire is calculated using Biot-Savart law as shown below;

[tex]B = \frac{\mu_0 I}{2\pi r}[/tex]

where;

  • [tex]\mu _0[/tex] is permeability of free space
  • I is the current
  • r is the radius of the wire

The electric field strength of  the wire is calculated as follows;

[tex]B = \frac{4\pi \times 10^{-7} \times 2.75}{2\pi \times 1.025 \times 10^{-3}} \\\\B = 5.37 \times 10^{-4} \ T[/tex]

Learn more about electric field in a wire here: https://brainly.com/question/14543998

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