Question

Two straight parallel wires, both carrying 10 A current in the same direction, attract each other with a force of current $1 \times {10^{ - 3}}N$. If both currents are doubled, the force of attraction will be-
a) $1 \times {10^{ - 3}}N$
b) $2 \times {10^{ - 3}}N$
c) $4 \times {10^{ - 3}}N$
d) $0.25 \times {10^{ - 3}}N$

Answer 1

Hint: Whenever two parallel current carrying conductors are placed nearby to each other then there is force of attraction or replusion between the conductors and here we will use the general relation between force between two parallel conducting wires according to given conditions.

Complete answer:
Look there is two current carrying conductor. A current flowing through a metal wire creates a magnetic field. This magnetic field can be used to produce electric current in another wire.
The strength of the magnetic field created by a current depends on the amount of current flowing through the wire and the distance between the wires.

When two parallel wires are carrying currents in the same direction, the currents attract each other and if current flows in opposite direction, then they are repelling each other. Now we are clear about the concept. Then let’s go for next.

First find the magnetic field. We know the formula
$\vec B = \frac{{\mu _0^{}I}}{{2\pi r}}\hat z$
here
$B = {\text{ magnetic field, }}{\mu _0} = {\text{ parmeability, }}$
$r = {\text{distance form the wire, }}$
$I = {\text{ current}, {\hat z}} = {\text{a unit vector towards into plane}}$ .

So, we get the magnetic field. Now find out the force due to this magnetic on the second wire. Here the formula is $F = ilB$ where $i = {\text{current, l = length of the wire ,B = magnetic field}}$ . now if we put the value of B in the equation of force.
Then, $F = il \times \left( {\frac{{\mu _0^{}I}}{{2\pi r}}} \right)$ .

So, we get a nice-looking formula. But carefully look at question that we have to find force per unit length.

So, ${F_n} = \frac{{il \times \left( {\frac{{\mu _0^{}I}}{{2\pi r}}} \right)}}{l} = \frac{{{\mu _0}iI}}{{2\pi r}}$ . here notices a thing that $F\propto iI$. So, if we if we double the current on both wire the force become four times. Let’s find out the force of attraction per unit length when current becomes double.

That is $4 \times {10^{ - 3}}N$ because force of normal condition is given $1 \times {10^{ - 3}}N$ .
Hence option c is the correct option.

Note: If you run electricity through a wire, you create a magnetic field. This field can produce a force on another wire nearby. The magnitude of this force is proportional to the current in the first wire, the distance between the wires, and the permeability of the surrounding material. This force can be used to move objects or to generate electricity.