Three long, straight and parallel wires carrying currents are arranged as shown in the figure. The wire C which carries a current of 50 A is so replaced that it experiences no force. The distance of wire C from wire A is

Question

Three long, straight and parallel wires carrying currents are arranged as shown in the figure. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9d4de496e6e83f7f967d3f2aca7953a7-.png" /> The wire C which carries a current of 50 A is so replaced that it experiences no force. The distance of wire C from wire A is (A) 9 cm (B) 7 cm (C) 5 cm (D) 3 cm

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e5afbbff430cc6088ff433b1bf459d54-.png" /> (FA C/l)=(FB C/l) (μ0/4 π) ⋅ (2 × 15 × 50/x) =(μ0/4 π) × (2 × 50 × 10/15-x) (15/x)=(10/(15-x)) 225-15 x=10 x ⇒ 25 x=255 or x=9 cm

Q. Three long, straight and parallel wires carrying currents are arranged as shown in the figure.

The wire $C$ which carries a current of $50 A$ is so replaced that it experiences no force. The distance of wire $C$ from wire $A$ is

9 cm

7 cm

5 cm

3 cm

$\frac{F _{A C}}{l} = \frac{F _{BC}}{l}$
$\frac{μ _{0}}{4 π} \cdot \frac{2 \times 15 \times 50}{x}$
$= \frac{μ _{0}}{4 π} \times \frac{2 \times 50 \times 10}{15 - x}$
$\frac{15}{x} = \frac{10}{( 15 - x )}$
$225 - 15 x = 10 x$
$\Rightarrow 25 x = 255$
or $x = 9 c m$

Q. Three long, straight and parallel wires carrying currents are arranged as shown in the figure. The wire C which carries a current of 50A is so replaced that it experiences no force. The distance of wire C from wire A is