The current-density magnitude in a certain circular wire isJ = (2.90 1010A/m4)r^{2}, where r is theradial distance out to the wire's radius of 3.00 mm. The potential applied to the wire (end toend) is 60.0 V. How much energy is converted to thermal energy in1.60 h?

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The current-density magnitude in a certain circular wire isJ = (2.90 1010A/m4)r^{2}, where r is theradial distance out to the wire&#039;s radius of 3.00 mm. The potential applied to the wire (end toend) is 60.0 V. How much energy is converted to thermal energy in1.60 h?

avortarF · Accepted Answer

wow...this looks like fun.Here&#039;s the idea:    Power =energy/time.       And for a resistor(the wire)   P = VISo... you need to first find the current in the wire; then youhave the power dissipated (i.e. turned to heat) by the wire; thenyou can multiply power by time to get the energy.First, the current: you are given current density, J, asa function of r. So you have to integrate over r to get thecurrent.      dI = J dA = K r^{2} (2?rdr)       where K is 2.90 x1010A/m4 and the dA is a thinstrip of length 2?r and width dr.Now you can get I:      integrate    2 ? K r3dr   from r = 0 to r = 3 x 10-3mCalc a value for I by using K.Then calc Power using I and V = 60V   Power is in watts, which areJoules/sec     mult power, in watts, by time insec ( your time is 5760 sec)

The current-density magnitude in a certain circular wire isJ = (2.90 1010A/m4)r^{2}, where r is theradial distance out to the wire's radius of 3.00 mm. The potential applied to the wire (end toend) is 60.0 V. How much energy is converted to thermal energy in1.60 h?

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