mEC CYU 9

1. Forces per Unit Length

There are two forces acting between the parallel strips:

• Electrostatic Force ($F_e’$): Repulsive force due to charge density. The strips act like a capacitor. $$ F_e’ = \frac{1}{2} \epsilon_0 E^2 b = \frac{\epsilon_0 b V_0^2}{2d^2} $$
• Magnetic Force ($F_m’$): Attractive force due to parallel currents. $$ F_m’ = \frac{\mu_0 I^2}{2b} $$

2. Initial Equilibrium

Initially, the net force vanishes. Therefore:

$$ F_e’ = F_{m0}’ \implies \frac{\epsilon_0 b V_0^2}{2d^2} = \frac{\mu_0 I_0^2}{2b} $$

3. New Condition

The resistance is increased by a factor $\eta = \sqrt{3}$. Since voltage $V_0$ is constant, the new current is:

$$ I = \frac{I_0}{\eta} $$

The electrostatic force depends only on voltage, so it remains unchanged ($F_e’$).
The magnetic force depends on $I^2$, so the new magnetic force is:

$$ F_{m,new}’ = \frac{\mu_0 (I_0/\eta)^2}{2b} = \frac{1}{\eta^2} \left( \frac{\mu_0 I_0^2}{2b} \right) = \frac{1}{\eta^2} F_e’ $$

4. Net Force Calculation

The new net force per unit length is the difference between the constant repulsive force and the reduced attractive force:

$$ F_{net}’ = F_e’ – F_{m,new}’ = F_e’ – \frac{1}{\eta^2} F_e’ = F_e’ \left( 1 – \frac{1}{\eta^2} \right) $$

Substituting the expression for $F_e’$:

$$ F_{net}’ = \frac{\epsilon_0 b V_0^2}{2d^2} \left( 1 – \frac{1}{\eta^2} \right) $$

Numerical Substitution:

• $\epsilon_0 = 8.85 \times 10^{-12}$ F/m
• $b = 0.04$ m, $d = 0.004$ m
• $V_0 = 2000$ V
• $\eta = \sqrt{3} \implies \eta^2 = 3$

$$ F_{net}’ = \frac{(8.85 \times 10^{-12})(0.04)(2000)^2}{2(0.004)^2} \left( 1 – \frac{1}{3} \right) $$ $$ F_{net}’ = \frac{(8.85 \times 10^{-12})(0.04)(4 \times 10^6)}{32 \times 10^{-6}} \times \frac{2}{3} $$ $$ F_{net}’ \approx 2.9 \times 10^{-2} \text{ N/m} $$