EMI O2

1. Material Parameters:

We are given a wire with:

• Mass: $m$
• Density: $d$
• Resistivity: $\rho$

Let the length of the wire be $l$ and its cross-sectional area be $a$.

Volume of the wire $V = \frac{m}{d}$. Also, $V = l \cdot a$.

$$ a = \frac{m}{d \cdot l} $$

2. Resistance of the Loop:

The resistance $R$ of the wire is given by:

$$ R = \rho \frac{l}{a} = \rho \frac{l}{(m / dl)} = \frac{\rho d l^2}{m} $$

3. Induced EMF and Current:

The loop is placed in a magnetic field $B = \beta t$. The induced EMF is determined by the rate of change of flux.

$$ \varepsilon = \frac{d\Phi}{dt} = A_{\text{loop}} \frac{dB}{dt} $$

Where $A_{\text{loop}}$ is the geometric area enclosed by the wire loop (not the cross-section $a$).

$$ \frac{dB}{dt} = \beta \implies \varepsilon = A_{\text{loop}} \beta $$

The induced current $I$ is:

$$ I = \frac{\varepsilon}{R} = \frac{A_{\text{loop}} \beta}{(\rho d l^2 / m)} = \frac{m \beta}{\rho d} \left( \frac{A_{\text{loop}}}{l^2} \right) $$

4. Maximizing the Current:

To maximize $I$, we need to maximize the ratio $\frac{A_{\text{loop}}}{l^2}$. This is an isoperimetric problem: for a fixed perimeter $l$, which shape encloses the maximum area?

The shape that maximizes area for a given perimeter is a Circle.

For a circle of perimeter $l$:

$$ l = 2\pi r \implies r = \frac{l}{2\pi} $$ $$ A_{\text{loop}} = \pi r^2 = \pi \left( \frac{l}{2\pi} \right)^2 = \frac{l^2}{4\pi} $$

5. Final Calculation:

Substitute the maximum area ratio back into the current equation:

$$ I_{\text{max}} = \frac{m \beta}{\rho d} \left( \frac{1}{4\pi} \right) = \frac{\beta m}{4\pi \rho d} $$