Question

A coil of Nichrome wire is 32.0 m long. The wire has a diameter of 0.330 mm and is at 20.0°C.

(a) If it carries a current of 0.420 A, what is the magnitude of the electric field in the wire?
V/m
(b) If it carries a current of 0.420 A, what is the power delivered to it?
W
(c) If the temperature is increased to 360°C and the potential difference across the wire remains constant, what is the power delivered?
W

Answer 1

Part A

Resistance of a wire is given by:

R = rho*L/A

rho = resistivity of nichrome = 1.50*10^-6 ohm-m

L = 32 m

A = pi*d^2/4 = pi*(0.33*10^-3)^2/4 = 8.55*10^-8 m^2

So,

R = 1.5*10^-6*32/(8.55*10^-8) = 561.40 Ohm

Voltage drop will be, from Ohm's law:

dV = i*R

dV = 0.420*561.40 = 235.79 V

Now electric is given by:

E = dV/L = 235.79/32 = 7.37 V/m

Part B

Power is given by:

P = i^2*R = 0.420^2*561.40 = 99.03 W

Part C

Now when we increase the temperature, then resistance will be

R = R0*(1 + alpha*dT)

alpha = temperature Coefficient for Nichrome = 0.4*10^-3 /C

dT = 360 - 20 = 340 C

R = 561.40*(1 + 0.4*10^-3*340) = 637.75 Ohm

Now Power delivered will be

P = V^2/R = 235.79^2/637.75 = 87.18 W

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