posted by smiley on .
The current-density magnitude in a certain circular wire is J = (2.75 × 10
where r is the radial distance out to the wire’s radius of 3.00 mm. The potential applied to
the wire (end to end) is 60.0 V. How much energy is converted to thermal energy in
k=2.75•10¹º A/m⁴ R=0.003 m
i=∫JdA=∫k•r²•2•π•r•dr= k•π•R²/2 =3.5 A
The rate of thermal energy generation is P=i•V=3.5•60=210 W.
Assuming a steady rate, the thermal energy generated in 2h=7200 s is
Q=P •Δt=210•7200= 1.5112 •10⁶ J