Two long current carrying thin wires, both with current $I$, are held by insulating threads of length $L$ and are in equilibrium as shown in the figure, with threads making an angle $'\theta'$ with the vertical. If wires have mass $\lambda$ per unit length then, the value of $I$ is : ( $g =$ gravitational acceleration)

Question

Six point charges are kept $60^\circ$ apart from each other on the circumference of a circle of radius $ R $ as shown in figure. The net electric field at the center of the circle is ___________. ($ \varepsilon_0 $ is permittivity of free space)

Answer 1

To solve the problem, we need to determine the current $I$ in each of the thin wires, given that they are held in equilibrium and are separated by an angle $\theta$ due to a magnetic repulsion force and the gravitational force acting on them.

Let's analyze the forces acting on the wires:

Magnetic Force between the Wires: The wires carry currents in opposite directions, and therefore, they repel each other. The magnetic force per unit length between two parallel wires separated by distance $d$ and carrying currents $I$ each is given by Ampere's force law: $$ F_m = \frac{\mu_0 I^2}{2 \pi d} $$ where $F_m$ is the magnetic force, $I$ is the current, $d$ is the distance between wires (which can be approximated using $2L\sin\theta$ since $d \approx 2L\sin\theta$ for small angles), and $\mu_0$ is the permeability of free space.
Gravitational Force: Each wire experiences a gravitational force pulling it downward due to its mass. For wires of mass $\lambda L$ (where $\lambda$ is the mass per unit length), the gravitational force on each wire is: $$ F_g = \lambda L g $$ where $g$ is the acceleration due to gravity.
Equilibrium Condition: In equilibrium, the vertical component of the tension in the threads balancing the gravitational force should equal the gravitational force, and the horizontal component of the tension should balance the magnetic repulsion between the wires: $$ T\sin\theta = \frac{\mu_0 I^2}{2 \pi (2L\sin\theta)} $$ $$ T\cos\theta = \lambda L g $$

We solve these equations for $I$:

From the equilibrium condition, solving for $T$ from the gravitational equation:

$$ T = \frac{\lambda L g}{\cos\theta} $$

Substituting for $T\sin\theta$:

$$ \frac{\lambda L g}{\cos\theta}\sin\theta = \frac{\mu_0 I^2}{4 \pi L\sin\theta} $$

Rearranging and simplifying gives us:

$$ I^2 = \frac{4 \pi \lambda gL \sin^2\theta}{\mu_0 \cos\theta} $$

Taking the square root:

$$ I = 2\sin\theta\sqrt{\frac{\pi \lambda gL}{\mu_0 \cos\theta}} $$

The correct answer, therefore, is:

$2\sin\theta\sqrt{\frac{\pi \lambda gL}{\mu_0 \cos\theta}}$

Answer 2