Question

For what electric field strength would the current in a 1.9-mm-diameter nichrome wire be the same as the current in a 0.50-mm-diameter aluminum wire in which the electric field strength is 0.0080 V/m ?

Express your answer to two significant figures and include the appropriate units.

Answer #1

**we know,**

**rho_nichrome = 10^-6 ohm.m
rho_aluminum = 2.65*10^-8
we know, E = J*rho**

**E ---> electric field
J ---> current density
rho --> resistivity**

**so,
E = (I/A)*rho**

**I = E*A/rho**

**I_nichrome =
E_nochrome*A_nichrome/rho_nichrome**

**and**

**I_aluminum=
E_aluminum*A_aluminum/rho_aluminum**

**given**

**I_nichrome = I_aluminum**

**E_nochrome*A_nichrome/rho_nichrome =
E_aluminum*A_aluminum/rho_aluminum**

**E_nichrome =
E_aluminum*(A_aluminum/A_nichrome)*(rho_nichrome/rho_aluminum)**

**= 0.0080*(0.5/1.9)^2*(10^-6/(2.65*10^-8))**

**= 0.021 V/m
<<<<<<<<<----------------Answer**

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