Question

Two long, parallel wires are separated by 5.33 cm and carry currents of 2.47 A and 3.39 A, respectively. Find the magnitude of the magnetic force ? that acts on a 2.23 m length of either wire.

Answer 1

Given : d = 5.33 cm (0.0533m) ; I₁ = 2.47 A ;I₂ = 3.39 A ; L= 2.23 m

Constant : $\mu_0$ = 4 $\pi$ ×10^-7 H/m

Solution:

Force per unit length that acts on either wire is given by:

$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$

$\frac{F}{2.23 m} = \frac{(4\pi*10^{-7})(2.47 A)(3.39 A)}{2(3.14) (0.0533m)}$

$F = \frac{(4\pi*10^{-7})(2.47 A)(3.39 A)}{2(3.14) (0.0533m)} (2.23m)$

= 7 ×10^-5 N

Answer : 7 ×10^-5 N