Two long conductors, separated by a distance d carry currents I1 , and I2 in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3d . The new value of force between them is

Question

Two long conductors, separated by a distance d carry currents I1 , and I2 in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3d . The new value of force between them is (A)  -2F  (B)  (F/3)  (C)  -(2F/3)  (D)  -(F/3)

Tardigrade Admin · Accepted Answer

Magnitude of magnetic force per unit length between parallel current carrying wires,

F = \frac{μ _{0}}{2 π} \frac{i _{1} i _{2}}{r}

Initially,

F = \frac{μ _{0}}{2 π} \frac{i _{1} i _{2}}{d}

As both the wires carry current in same direction, this force is attractive.
Finally

i_{1}^{'} = 2 i

and

r = 3 d

So,

F^{'} = \frac{μ _{0}}{2 π} \frac{( 2 i _{1} ) i _{2}}{3 d}

= \frac{2}{3} F

Since, the direction of current is reversed in one of the wire, hence

F^{'}

would be repulsive in nature.

∴ F^{'} = - \frac{2}{3} F

$- 2 F$

$\frac{F}{3}$

$- \frac{2 F}{3}$

$- \frac{F}{3}$

−2F

3F​

−32F​

−3F​

$- 2 F$

$\frac{F}{3}$

$- \frac{2 F}{3}$

$- \frac{F}{3}$