Question

. Current of equal magnitude flows through two long parallel wires having separation of 1.35 cm. If the force per unit length on each of the wires in 4.76 ´ 10-2 N, what must be I ? [Ans: 56.7 A]

Answer 1

Given that,

The separation between two parallel wires, r = 1.35 cm

The force per unit length of each wire is $4.76\times 10^{-2}\ N$

To find,

The value of current i.e. I.

Solution,

The force per unit length acting between two parallel wires is given by :

$\dfrac{F}{l}=\dfrac{\mu_o I^2}{2\pi r}$

Putting all the value, we get :

$I=\sqrt{\dfrac{2\pi rF}{\mu_o l}} \\\\I=\sqrt{\dfrac{2\pi \times 1.35\times 10^{-2}\times 4.76\times 10^{-2}}{4\pi \times 10^{-7}}} \\\\I=56.68\ A$

or

I = 56.7 A

So, the current flowing in each wire is 56.7 A.