Question

A system consists of a thin charged wire ring of radius R

and a very long uniformly charged thread oriented along the axis

of the ring, with one of its ends coinciding with the centre of the

ring. The total charge of the ring is equal to q. The charge of the

thread (per unit length) is equal to lemda. Find the interaction force be-

tween the ring and the thread.​

Answer 1

Given that,

Radius = R

Charge of ring = q

Charge per unit length = λ₀

We know that,

The electric field strength due to ring at a point on its axis at distance x from the centre of the ring is

$E(x)=\dfrac{qx}{4\pi\epsilon_{0}(R^2+x^2)^{\frac{3}{2}}}$

And from symmetry vector E at every point on the axis is directed along the x-axis.

Let us consider an element (dx) on thread which carries the charge (λ dx). The electric force experienced by the element in the field of ring.

We need to calculate the interaction force between the ring and the thread

Using formula of force

$dF=(\lambda dx) E(x)$

Put the value of E(x)

$dF=\dfrac{\lambda qx}{4\pi\epsilon_{0}(R^2+x^2)^{\frac{3}{2}}}dx$

On integration

$\int{dF}=\int_{0}^{\infty}{\dfrac{\lambda qx}{4\pi\epsilon_{0}(R^2+x^2)^{\frac{3}{2}}}dx}$

$F= \dfrac{\lambda\ q}{4\pi\epsilon_{0}R}$

Hence, The interaction force between the ring and the thread is $\dfrac{\lambda\ q}{4\pi\epsilon_{0}R}$