A wire X of length 50 cm carrying a current of 2 A is placed parallel to a long wire Y of length 5 m. The wire Y carries a current of 3 A. The distance between two wires is 5 cm and currents flow in the same direction. The force acting on the wire Y is :

Question

A wire X of length 50 cm carrying a current of 2 A is placed parallel to a long wire Y of length 5 m. The wire Y carries a current of 3 A. The distance between two wires is 5 cm and currents flow in the same direction. The force acting on the wire Y is : (A) 1.2 × 10-5 N directed towards wire X. (B) 1.2 × 10-4 N directed away from wire X. (C) 1.2 × 10-4 N directed towards wire X. (D) 2.4 × 10-5 N directed towards wire X.

Tardigrade Admin · Accepted Answer

Force of interaction = I 1 ℓ1 B 12 =(μ0 I 1 I 2/2 π r ) ℓ1 =(4 π × 10-7 × 6 × 0.5/2 π × 5 × 10-2) =1.2 × 10-5 text towards X

Q. A wire $X$ of length $50 c m$ carrying a current of 2 A is placed parallel to a long wire $Y$ of length $5 m$ . The wire Y carries a current of $3 A$ . The distance between two wires is $5 c m$ and currents flow in the same direction. The force acting on the wire $Y$ is :

$1.2 \times 1 0^{- 5} N$ directed towards wire $X$ .

$1.2 \times 1 0^{- 4} N$ directed away from wire $X$ .

$1.2 \times 1 0^{- 4} N$ directed towards wire $X$ .

$2.4 \times 1 0^{- 5} N$ directed towards wire $X$ .

Force of interaction $= I_{1} ℓ_{1} B_{12}$
$= \frac{μ _{0} I _{1} I _{2}}{2 π r} ℓ_{1}$
$= \frac{4 π \times 1 0 ^{- 7} \times 6 \times 0.5}{2 π \times 5 \times 1 0 ^{- 2}}$
$= 1.2 \times 1 0^{- 5} towards X$

Q. A wire X of length 50cm carrying a current of 2 A is placed parallel to a long wire Y of length 5m. The wire Y carries a current of 3A. The distance between two wires is 5cm and currents flow in the same direction. The force acting on the wire Y is :

1.2×10−5N directed towards wire X.

1.2×10−4N directed away from wire X.

1.2×10−4N directed towards wire X.

2.4×10−5N directed towards wire X.