For any arbitrary curved conductor carrying steady current $I$ placed in a uniform magnetic field $\vec{B}$, the net magnetic force depends solely on the vector displacement $\vec{L}$ connecting the initial and final endpoints:

• The wire starts at point P and ends at point Q. With two semicircles each of radius $R$ spanning horizontally, the net displacement vector from P to Q along the horizontal $x$-axis is:
  $$\vec{L} = (2R + 2R)\,\hat{i} = 4R\,\hat{i}$$
• The uniform magnetic field emerges normally out of the plane along the $+z$-axis:
  $$\vec{B} = B\,\hat{k}$$
• Taking the cross product:
  $$\vec{F} = I (4R\,\hat{i} \times B\,\hat{k}) = 4 B I R (\hat{i} \times \hat{k}) = 4 B I R (-\hat{j}) = -4 B I R\,\hat{j}$$
  This corresponds to a magnitude of $4BIR$ directed vertically downward.

• Parameters in SI units:
• Diameter $2r = 1\,\text{cm} \implies r = 0.5\,\text{cm} = 0.005\,\text{m} = 5 \times 10^{-3}\,\text{m}$
• Volume $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.5)^3\,\text{cm}^3 = \frac{\pi}{6}\,\text{cm}^3 = \frac{\pi}{6} \times 10^{-6}\,\text{m}^3$
• Density of copper $\rho = 8.0\,\text{g/cc} = 8000\,\text{kg/m}^3$
• Density of oil $\sigma = 0.8\,\text{g/cc} = 800\,\text{kg/m}^3$
• Effective density $\Delta \rho = \rho - \sigma = 7.2\,\text{g/cc} = 7200\,\text{kg/m}^3$
• The net downward apparent weight is:
  $$W_{\text{app}} = V (\rho - \sigma) g = \left(\frac{\pi}{6} \times 10^{-6}\right) \times 7200 \times 10 = 1200\pi \times 10^{-5} = 1.2\pi \times 10^{-2}\,\text{N}$$
• For equilibrium, the upward electrostatic force balances the apparent weight:
  $$q E = W_{\text{app}} \implies q (600\pi) = 1.2\pi \times 10^{-2}$$
  $$q = \frac{1.2\pi \times 10^{-2}}{600\pi} = \frac{1.2 \times 10^{-2}}{600} = 2 \times 10^{-5}\,\text{C}$$