Prove that the current at which a fuse will melt is independent of the length of its wire.

Explanation

Let there be one fuse wire of radius r and length l. The resistivity of material of wire is `rho`. Now, the resistance of wire can be written as follows:
`R= rho (l)/(pi r^(2)`
Let I be the amount of current flowing through this wire. The rate at which heat will be generated can be written as follows:
`H= I^(2) R rArr H = I^(2) (rho l)/(pi r^(2))` ...(i)
When a steady state temperature is attained, the rate of joules becomes equal to the rate of radiation. Let e be the emissivity of the material of the wire and T the absolute temperature. Then, according to Stefan.s law of radiation. Rate of radiation `= e sigma A T^(4)`. Here, A is the surface area of the body. Surface area of fuse wire is `A= 2pi rl`. Hence, the rate of radiation can be written as follows:
Rate of radiation `= 2 e sigma pi r lT^(4)`...(ii)
In case of the fuse wire, a steady state temperature is usually very high in comparison to its surrounding. So we can neglect the rate of heat absorption from the nearby objects. Hence, for a steady state we can equate the rate of radiation to rate of joules heating.
`I^(2) (rho l)/(pi r^(2)) =2e sigma pi r l T^(4) rArr T^(4) = (I^(2) rho)/(2e sigma pi^(2) r^(3))`
We can see that the relation of temperature with the amount of current flowing through it is independent of the length of fuse wire. Note that when for a certain value of I, temperature T reaches the melting point, the fuse will blow.