The time ($t$) it takes for a fuse wire to heat up and melt depends on the heat generated by the current passing through it and the heat required to melt the wire's material.
1. Heat Generated ($H$): According to Joule's law, the heat generated in a conductor is given by $H \propto I^2 R t$, where $I$ is the current, $R$ is the resistance, and $t$ is the time.
2. Resistance ($R$): The resistance of a fuse wire is calculated as $R = \frac{\rho l}{A}$, where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area. Since the area $A = \pi r^2$, we have $R \propto \frac{l}{r^2}$.
3. Heat Required for Melting ($H_{melt}$): The heat required to melt the fuse wire is proportional to its mass ($m$). The mass is given by $m = \text{density} \times \text{Volume}$. The volume ($V$) of the wire is $V = A \cdot l = \pi r^2 l$. Therefore, $H_{melt} \propto m \propto V \propto r^2 l$.
4. Time Calculation: The fuse melts when the heat generated equals the heat required for melting ($H = H_{melt}$). So, we have the proportionality:
$I^2 R t \propto r^2 l$
Substituting the proportionality for $R$ ($R \propto \frac{l}{r^2}$):
$I^2 \left( \frac{l}{r^2} \right) t \propto r^2 l$
Now, we solve for $t$:
$t \propto \frac{r^2 l}{I^2 (l/r^2)}$
$t \propto \frac{r^2 l \cdot r^2}{I^2 l}$
$t \propto \frac{r^4 l}{I^2 l}$
The length term $l$ cancels out:
$t \propto \frac{r^4}{I^2}$
The question asks for the dependence when passing the *maximum current*. This implies $I$ is a fixed value. Therefore, the time of heating ($t$) is proportional to the fourth power of the radius ($r^4$) and has no dependence on the length ($l^0$).
$t \propto r^4 l^0$
A uniform time-varying magnetic field exists in a circular region of radius $R$, directed perpendicular into the plane of the paper, increasing at a constant rate $\alpha$. A straight conducting rod of length $2R$ is placed exactly along the diameter of the circular region (passing through the centre). Find the induced emf across the rod.

There is a ring of radius $r$ having linear charge density $\lambda$ and rotating with a uniform angular velocity $\omega$. The magnitude of the magnetic field produced by this ring at its own centre would be
($\mu_0 = \text{permeability of air}$)