Question

A long horizontally fixed wire carries a current of $100$ampere. Directly above and parallel to it is a fine wire that carries a current of 20 ampere and weights $0.04$newton per meter. The distance between the two wires for which the upper wire is just supported by magnetic repulsion is:
A. ${10^{ - 2}}mm$
B. ${10^{ - 2}}cm$
C. ${10^{ - 2}}m$
D. ${10^{ - 2}}km$

Answer 1

HintWhen the current is flowing in two parallel wire in opposite direction then there is a mutual force of magnetic repulsion between them. Both wire repel each other with same magnitude of force according to the Newton’s third law of motion (action is equal to reaction).

Formula used:
$\dfrac{F}{l} = \dfrac{{{\mu _0}2{I_A}{I_B}}}{{4\pi r}}$

Complete step-by-step solution:Let the rigid wire be $A$and the current flowing in it be${I_A}$, ${I_A} = 100A$
And, the fine wire be $B$and the current flowing in it be${I_B}$, ${I_B} = 20A$
Distance between two wire = $r$
Length of both wire=$l$
Now, the current is flowing in both wires and wire$A$is rigid whereas the wire$B$is moveable. So to hold the wire$B$directly above the wire$A$, the current must be flowing in opposite direction to each other. So that due to the magnetic field produced between them, the wire$A$would repel wire$B$. So to support the upper wire which have some weight, it should be at some distance from lower wire.
So to just support the upper wire, downward force =upward force
Where, the downward force=weight=$mg$
And, the upward force =magnetic force per unit length at distance $r$=$\dfrac{F}{l}$
Here the force$F$is the force through which the lower wire (rigid) push away the upper wire (moveable).
Hence, magnetic force on upper wire per unit length = weight of upper wire per unit length
That is; $\dfrac{F}{l} = \dfrac{{mg}}{l}$
Force exerted on length$l$ due to the magnetic field is given as:
$F = \dfrac{{{\mu _0}2{I_A}{I_B}l}}{{4\pi r}}$
Where,
${\mu _0} = $Permeability of free space
And, $\dfrac{{{\mu _0}}}{{2\pi }} = 2 \times {10^{ - 7}}$
Now we have,
$\dfrac{F}{l} = \dfrac{{{\mu _0}2{I_A}{I_B}}}{{4\pi r}} = \dfrac{{mg}}{l}$
Simplify the above equation by substituting the value of${I_A}$,${I_B}$,$\dfrac{{{\mu _0}}}{{2\pi }}$ and $\dfrac{{mg}}{l} = 0.04$
We get, $\dfrac{{2 \times {{10}^{ - 7}} \times 100 \times 20}}{r} = 0.04$
Further solving for $r$as
$
  r = \dfrac{{4 \times {{10}^3} \times {{10}^{ - 7}} \times {{10}^2}}}{4} \\
  r = {10^{ - 2}}m \\
 $
Hence, the distance between two wires for which the upper wire is just supported by the magnetic repulsion is${10^{ - 2}}m$. Therefore, the correct option is C.

Note:- This magnetic force of attraction or repulsion is due to the interaction of one of the currents with the magnetic field produced by the other current. For two currents flowing in the same direction, the force is repulsive, while for opposite direction, it is attractive.