Physics - Magnetism Question with Solution | TestHub

Let the force per unit length on wire C due to wire A be $F_{CA}$ and due to wire B be $F_{CB}$.

Since the currents in all wires are in the same direction, the forces between them are attractive. Thus, wire A pulls wire C to the left, and wire B pulls wire C to the right.

For wire C to experience no net force, the magnitudes of these two forces must be equal:

$F_{CA}=F_{CB}$

$\frac{{\mu }_0{I}_A{I}_C}{2\pi x}=\frac{{\mu }_0{I}_B{I}_C}{2\pi (15-x)}$

Canceling the common terms, we get:

$\frac{I_A}{x}=\frac{I_B}{15-x}$

Substituting the given values $I_A=15$ A and $I_B=10$ A:

$\frac{15}{x}=\frac{10}{15-x}$

$\frac{3}{x}=\frac{2}{15-x}$

Cross-multiplying yields:

$3(15-x)=2x$

$45-3x=2x$

$5x=45$

$x=9$ cm

Answer: $9$