Question

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 per unit length '$\lambda$' then the value of I is:

Answer 1

REF.Image

Consider a length l of both wires, which would

have mass = $\lambda l$

as force per unit length (f) due to magnetic field

is :-

$f=\frac{{\mu }_0{I}^2}{4\pi d}$, we have for length l;

force = Fm = fL= $\frac{{\mu }_0{I}^2\lambda }{4\pi d}$

also from figure, $d=2Lsin\theta$ and $\frac{Fm}{\lambda lg}=tan\theta$, so:-

$\frac{Fm}{\lambda lg}=\frac{M_0{I}^{2}l}{4\pi \left(2Lsin\theta \right)}\times \frac{1}{\lambda lg}=tan\theta$

$\Rightarrow I^2=\frac{\lambda g.4\pi si{n}^2\theta }{{\mu }_{0}cos\theta }\Rightarrow I=sin\theta \sqrt{\frac{4\pi \lambda g}{{\mu }_{0}cos\theta }}\to \left(Ans\right)$