Question

A wire AB is carrying a steady current 12 A and is lying on the table. Another wire CD carrying 5 A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the directions of the current flowing in CD with respect to that in AB [Take the value of $g = 10m{s^{ - 2}}$]

Answer 1

From Ampere’s circuital law we know that when current passes through any wire, some magnetic field is being created. Again, when a charged particle passes through some magnetic field it feels some force which is known as Lorentz force. So, when two current carrying wires are kept parallel then they both experience some attractive or repulsive force due to the magnetic field created by the other one.

Formula Used:
Force per unit length experienced by each wire is given by:
$\dfrac{F}{l} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2{I_1}{I_2}}}{r}$ (1)
Where,
F is the total force experienced,
l is the length of the wire,
${\mu _0}$ is the permeability of vacuum, ${\mu _0} = 4\pi \times {10^{ - 7}}T.m{A^{ - 1}}$,
${I_1}$ denotes current through the first wire,
${I_2}$ denotes current through second wire,
r is the separation distance between two wires.

Weight of a body of mass m is given as:
$W = mg$ (2)
Where,
m is the mass of the body,
g is acceleration due to gravity.

Complete step by step answer:
Given:
1. Current carrying by the first wire AB is ${I_1} = 12A$.
2. Current carrying by the second wire CD is ${I_2} = 5A$.
3. Distance between two wires is $r = 1mm = {10^{ - 3}}m$.
4. Value of gravitational acceleration is given as $g = 10m{s^{ - 2}}$.

To find: Mass per unit length of the wires.

Step 1
Here, for the wire CD to remain suspended, its weight must be nullified by the opposite directional force. Hence, it must be$F = W$. Now, use eq.(2) in eq.(1) to get the expression of mass per unit length as:
$\dfrac{W}{l} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2{I_1}{I_2}}}{r} \\\ \Rightarrow \dfrac{{mg}}{l} = \dfrac{{{\mu _0}}}{{2\pi }}\dfrac{{{I_1}{I_2}}}{r} \\\ \therefore \dfrac{m}{l} = \dfrac{{{\mu _0}{I_1}{I_2}}}{{2\pi gr}} \\\$ (3)

Step 2
Now, put the values of ${I_1}$,${I_2}$, g, r and ${\mu _0}$in eq.(3) to get $\dfrac{m}{l}$as:
$\dfrac{m}{l} = \dfrac{{\left( {4\pi \times {{10}^{ - 7}}T.m{A^{ - 1}}} \right) \times 12A \times 5A}}{{2\pi \times 10m{s^{ - 2}} \times {{10}^{ - 3}}m}} \\\ \therefore \dfrac{m}{l} = 1.2 \times {10^{ - 3}}kg{m^{ - 1}} \\\$

Final answer:
The mass per unit length of the wire CD is $1.2 \times {10^{ - 3}}kg{m^{ - 1}}$.

Note: In the given question the value of current passing through the wire is given but nothing is mentioned about their direction. A student might get confused thinking about the current direction in both the wires. For the CD wire to suspend the force must be repulsive, i.e. it must be negative. Notice in eq.(1), the negative sign can only occur if one of the current be negative, i.e. they must be opposite in direction.