Physics Solution 2

Solution 2: Current in a Composite Spherical Shell

1. Analyzing the Resistance Model

The current flows from one pole of the sphere to the other. The spherical shell can be viewed as a series circuit consisting of the main copper body and the equatorial strip.

Because the strip completely circles the equator, all current flowing from the top hemisphere to the bottom hemisphere must pass through it. Therefore, the total resistance $R_{total}$ is:

$R_{total} = R_{copper\_parts} + R_{strip}$

When the strip is changed from Aluminium to Iron, only the $R_{strip}$ component changes. The rest of the sphere remains constant.

2. Calculating Resistance of the Strip

The strip has a width $b$ (which is the length $L$ along the direction of current flow) and a cross-sectional area $A$ perpendicular to current flow.

Length of current path through strip: $L = b$
Cross-sectional area: $A = \text{Circumference} \times \text{thickness} = 2\pi r t$

Thus, the resistance of the strip is:

$R_{strip} = \rho \frac{b}{2\pi r t}$

3. Deriving the Current for the Iron Strip

Let $R_0$ be the resistance of the copper parts. Initially (with Aluminium), the total resistance $R_1$ is:

$$R_1 = \frac{V}{I_1} = R_0 + \rho_{Al} \frac{b}{2\pi r t}$$

With the Iron strip, the new resistance $R_2$ is:

$$R_2 = R_0 + \rho_{Fe} \frac{b}{2\pi r t}$$

We can express $R_2$ in terms of $R_1$ by subtracting the Aluminium contribution and adding the Iron contribution:

$$R_2 = R_1 – \rho_{Al} \frac{b}{2\pi r t} + \rho_{Fe} \frac{b}{2\pi r t} = \frac{V}{I_1} + (\rho_{Fe} – \rho_{Al})\frac{b}{2\pi r t}$$

The new current $I_2$ is $V / R_2$:

$I_2 = \frac{V}{\frac{V}{I_1} + (\rho_{Fe} – \rho_{Al})\frac{b}{2\pi r t}}$