Question

Two straight wires A and B of length 10 m and 12 m carrying currents of $4.0{\text{A}}$ and $6.0{\text{A}}$ respectively in opposite directions, the lie parallel to each other at a distance of $3.0cm$. The force on a 15 cm section of wire B near its centre is
(A) $2.4 \times {10^{ - 5}}N$, attractive
(B) $2.4 \times {10^{ - 5}}N$, repulsive
(C) $1.2 \times {10^{ - 5}}N$, attractive
(D) $1.2 \times {10^{ - 5}}N$, repulsive

Answer 1

Hint : The wires can be assumed to be infinitely long wires since their separation is little compared to their lengths. Two current carrying wires attract each other when they are in the same direction.

Formula used: In this solution we will be using the following formula;
$F = BIL$, where $F$ is the magnetic force on a wire, $B$ is the magnetic field around the wire due to an external agent (another wire for example), $I$ is the current flowing through the wire of interest, and $L$ is the length of interest of the wire.
$B = \dfrac{{{\mu _0}I}}{{2\pi r}}$, where $B$ is the magnetic field created by a current carrying wire, $I$ us the current flowing through the wire, $r$ is the distance from the wire.

Complete step by step answer
When two current carrying wires are at close proximity to each other, they attract or repel each other with a force proportional to the magnetic field each create at the location of the other wire.
The magnetic force on wire B due to the magnetic field of wire A can be given as
${F_B} = {B_A}{I_B}{L_B}$ where ${F_B}$ is the force on wire B, ${B_A}$ is the field due to wire A, ${I_B}$ and ${L_B}$ are the current and length (or of interest) of wire B respectively.
Now, because the wires are so close together compared to the lengths of the wires, we can assume that wire A is infinitely long.
Hence the magnetic field is
${B_A} = \dfrac{{{\mu _0}{I_A}}}{{2\pi r}}$ where ${\mu _0}$ is permeability of free space,${I_A}$ is the current on wire A, and $r$ is the distance of interest from the wire, in this case the distance of wire B
${B_A} = \dfrac{{4\pi \times {{10}^{ - 7}} \times 4.0}}{{2\pi \times 0.03}}$
By computation,
${B_A} = 2.67 \times {10^{ - 5}}T$
For the force of a 15cm section on B
${F_B} = {B_A}{I_B}{L_B} = 2.67 \times {10^{ - 5}} \times 6 \times 0.15$
By computation, we get the force as
${F_B} = 2.4 \times {10^{ - 5}}N$
Since they were in the opposite direction, hence it is repulsive.
So the correct option is B.

Note
For clarity, in general, a repulsive force is generated when two currents are in opposite directions. This is because the field due to one wire would have created a force in a direction towards itself on a current flowing in the same direction according to the right hand rule. But this becomes reversed as it moves in the opposite direction. Since $I$ becomes $ - I$ with respect to one another.