Question

Magnetic force on current carrying wire
4
Two long and parallel wires are at a distance of 0.1 m
and a current of 5 A is flowing in each of these wires.
The force per unit length due to these wires will be
(1) 5× 10-N/m
(2) 5 x 10-N/m
(3) 2.5 10 N/m
(4) 2.5 x 10 N/m​

Answer 1

I can't understand this problem I am so sorry

Answer 2

Answer:

Two parallel conductors carrying equal currents of 5A  are placed 0.1 m apart. The force per unit length between the two conductors is $5*10^{-5} N/m$

Explanation:

• From the Biot-Savart law, we know there will be a magnetic field due to current-carrying conductors, that magnetic field exerts a force on the conductor

• suppose that a conductor A with  current $I_{A}$ and another conductor B carrying current $I_{B}$ are separated by a distance $r$ the force on a length $L$ of the conductor A due to conductor B is

                                 $F_{AB}=I_ALB_{B}$

• From Ampere's law, the magnetic field at the conductor B is

                                $B_{B}=\mu _0I_B/2\pi r$

• so the force between the conductors

                               $F_{AB}=\mu _0I_BI_{A} L/2\pi r$

from the question, we have

currents in both the conductors   $I_{A}=5A$, $I_{B} =5A$

the distance between the conductors $r=0.1m$

asked for force per unit length $L=1m$

value of $\mu _0=4\pi *10^{-7}$

⇒

$F_{AB}=\frac{4\pi *10^{-7} *5*5}{2\pi *0.1} =5*10^{-5}N/m$