91影视

We need to find current the upper wire to 鈥渇loat鈥� so as to form an equilateral triangle with the lower wires.

The weight per unit length of the floating wire must be balanced by force per unit length due to the interaction of the current of the floating wire with the other two wires as:

${位}_m=\frac{m}{l}=50\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$m.$}\right.$

The weight per unit length is

$f_g=\frac{F_g}{l}=\frac{mg}{l}={位}_{m}g.$

The forces between the wires on the table and the floating wire are repulsive and act at $30掳$away from the vertical. From the figure, we see that the horizontal components cancel out and that their vertical components add up. since they have the same magnitude of

$F_1=F_2=\frac{{渭}_0}{2\mathrm{蟺}}\frac{I^{2}l}{d},$

The magnitude of the net force which points vertically upwards is simply twice the vertical component of , say, $\stackrel{鈫�}{F_1}$

$F=2F_1\cos {30}^{掳}=2F_1\frac{\sqrt{3}}{2}=F_1\sqrt{3}=\frac{{渭}_0}{2\mathrm{蟺}}\frac{I^{2}l}{d}\sqrt{3}.$

Since we are looking at force per unit length we have :

$f=\frac{F}{l}=\frac{{渭}_0}{2\mathrm{蟺}}\frac{I^2}{d}\sqrt{3}.$

Equating this with weight per unit length we have :

${位}_{mg}=\frac{{渭}_0}{2\mathrm{蟺}}\frac{I^2}{d}\sqrt{3}.$

The current is then

$I=\sqrt{\frac{2{\mathrm{蟺诲位}}_{\mathrm{m}}\mathrm{g}}{{渭}_0\sqrt{3}}}=240A.$