The current-density magnitude in a certain cicular wire is J= (2.75*10^10A/m^4)r^, where r is the radial distance out to the wire 's radius of 3.00 mm. The potential applied to the wire is 60.0 V. how much energy is converted to thermal enery in 2 hours?
To calculate the amount of energy converted to thermal energy in 2 hours, we need to determine the power dissipated by the wire and then multiply it by the duration.
The power dissipated by the wire can be calculated using the formula:
P = ∫ J * E * dV
Where:
P is the power dissipated,
J is the current density vector,
E is the electric field vector, and
dV is the volume element.
Since J = (2.75 * 10^10 A/m^4)r^, we can rewrite it in terms of radial distance r:
J = (2.75 * 10^10 A/m^4)r^ * (1/m) = (2.75 * 10^10 A/m^3)r^
The electric field E can be determined using the potential difference (V) applied to the wire:
E = -∇V
Since the wire is cylindrical and uniform, the electric field is radial and given by:
E = - dV/dr
Therefore, E = - dV/dr = - d(60.0 V)/dr = 0 V/m
Since the electric field E is zero, the power dissipated by the wire is also zero.
Hence, there is no energy converted to thermal energy in 2 hours as the wire does not dissipate any power.