Question

Two long parallel wires are separated by 11 cm. One of the wires carries a current of 54 A and the other carries a current of 45 A. Determine the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current.

Answer 1

Explanation:

It is given that,

The separation between two parallel wires, r = 11 cm = 0.11 m

Current in wire 1, $q_1=54\ A$

Current in wire 2, $q_2=45\ A$

Length of wires, l = 4.3 m

We need to find the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current. The magnetic force per unit length is given by :

$\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 54\times 45\times 4.3}{2\pi \times 0.11}\\\\F=0.0189\ N$

So, the magnetic force on a 4.3 m length of the wire  on both of currents is F=0.0189 N.