Question

solve this with complete explanation fast ​

Answer 1

Answer:

$\boxed{\mathfrak{(B) \ 2sin \theta \sqrt{\frac{\pi \lambda L g }{\mu_0 cos \theta } } }}$

Explanation:

Let the length of each wire be a and distance between them in equilibrium be d.

Magnetic force on one wire due to other:

$\sf F_B = \dfrac{ \mu_0 I^2a}{2\pi d}$

Length of thread is given as L.

From the figure attached we can conclude that:

$\sf d = Lsin \theta + Lsin \theta = 2Lsin \theta$

So,

$\sf F_B = \frac{ \mu_0 I^2a}{2\pi (2Lsin \theta)} \\ \\ \sf F_B = \frac{ \mu_0 I^2a}{4\pi Lsin \theta}$

Let the tension in the thread be T and mass of wire be M.

From free body diagram of the wire we can conclude that:

$\sf T sin \theta = F_B \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(1) \\ \\ \sf T cos \theta = Mg \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(2)$

Mass per unit length of the wire is $\sf \lambda$.

$\sf \lambda = \frac{M}{a} \\ \\ \sf M = \lambda a$

By dividing (1) with (2) we get:

$\sf \dfrac{\cancel{T} sin \theta}{\cancel{T} cos \theta} = \dfrac{F_B}{Mg} \\ \\ \sf \dfrac{ sin \theta}{ cos \theta} = \dfrac{\dfrac{ \mu_0 I^2 \cancel{a}}{4\pi Lsin \theta} }{ \lambda \cancel{a}g} \\ \\ \sf \frac{sin \theta }{cos \theta } = \frac{ \mu_0 I^2}{4\pi \lambda L g sin \theta} \\ \\ \sf I^2 = \frac{4\pi \lambda L g sin ^{2} \theta}{\mu_0 cos \theta } \\ \\ \sf I = \sqrt{\frac{4\pi \lambda L g sin ^{2} \theta}{\mu_0 cos \theta } } \\ \\ \sf I =2sin \theta \sqrt{\frac{\pi \lambda L g }{\mu_0 cos \theta } }$

Answer 2

Answer:

The wires repel each other with some force.

The magnitude of this force can be given by

F=

Il

×

B

where l is length of wire

B=

2πr

μ

∘

​

I

​

where r is distance between two wires

r=2Lsinθ

F=

2π2Lsinθ

μ

∘

​

I

2

l

​

...(i)

Now the tension in the thread is keeping both wire from moving to sideways

From FBD of the wire

Tsinθ=F...(ii)

Tcosθ=mg...(iii)

dividing equation (ii) by (iii)

tanθ=

mg

F

​

m=λL

I=2sinθ

μ

∘

​

cosθ

πλgL

​

​