Solution to Question 2
We need to find the magnetic flux density at a point $P(0, R, 0)$ due to current flowing between two spheres maintained at potential $V$ and $0$ in a medium of conductivity $\sigma$.
Step 1: Current Calculation
The resistance of a single sphere of radius $r$ embedded in an infinite medium of conductivity $\sigma$ is $R_{sphere} = \frac{1}{4\pi\sigma r}$. Since the separation $2d \gg r$, the interaction term is negligible, and the resistances add in series:
$$ R_{eq} \approx \frac{1}{4\pi\sigma r} + \frac{1}{4\pi\sigma r} = \frac{1}{2\pi\sigma r} $$The total current flowing out of the source sphere is:
$$ I = \frac{V}{R_{eq}} = V \cdot 2\pi\sigma r $$Step 2: Magnetic Field Calculation
We treat the spheres as point sources of current. In an infinite medium, the current spreads radially from the source. However, due to symmetry, at point $P(0, R, 0)$ on the bisector, the magnetic field is the superposition of fields created by the current density vector $\mathbf{J}$.
From the generalized Ampere’s law and symmetry for point electrodes, the magnetic field $B$ at a distance $\rho$ from a wire carrying current is $\mu_0 I / 2\pi \rho$. Here, the geometry is distributed. A known result for the magnetic field produced by current flowing between two points separated by $2d$ at a perpendicular distance $R$ (on the bisector) is derived using the relation $\nabla \times \mathbf{B} = \mu_0 \mathbf{J} = \mu_0 \sigma \mathbf{E}$.
Using the coordinate geometry, the field at $(0,R)$ is given by:
$$ B = \frac{\mu_0 I}{2\pi R} \left( 1 – \frac{d}{\sqrt{R^2 + d^2}} \right) $$However, considering the potential gradient method and the specific geometry given in standard electrodynamics texts for this configuration:
$$ B = \frac{\mu_0 \sigma V r d}{R} \left( \frac{1}{d} – \frac{1}{\sqrt{R^2+d^2}} \right) $$Wait, let us check dimensions. $B \propto \mu_0 \sigma V r$.
Substituting $I = 2\pi \sigma V r$ into the expression:
Rearranging to match the required format:
$$ B = \frac{\mu_0 \sigma V r d}{R} \left( \frac{1}{d} – \frac{1}{\sqrt{R^2+d^2}} \right) $$