Two long parallel wires are 30 cm apart carrying currents 10 A and 15 A respectively in the same direction. The force acting over a length of 5 m of the wires is

Question

Two long parallel wires are 30 cm apart carrying currents 10 A and 15 A respectively in the same direction. The force acting over a length of 5 m of the wires is (A) 5× 10- 4 N , (attraction) (B) 1× 10- 4N , (attraction) (C) 5× 10- 4N , (repulsion) (D) 1× 10- 4N , (repulsion)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-xkegy4jdf5oy.png" /> Force acting between two parallel wire F=(μ 0 i1 i2/2 π r)l F=((2 × 1 0- 7) (10) (15)/0.3)× 5 F=5× 10- 4 N

Q. Two long parallel wires are $30 c m$ apart carrying currents $10 A$ and $15 A$ respectively in the same direction. The force acting over a length of $5 m$ of the wires is

$5 \times 1 0^{- 4} N$ , (attraction)

$1 \times 1 0^{- 4} N$ , (attraction)

$5 \times 1 0^{- 4} N$ , (repulsion)

$1 \times 1 0^{- 4} N$ , (repulsion)

Force acting between two parallel wire
$F = \frac{μ _{0} i _{1} i _{2}}{2 π r} l$
$F = \frac{( 2 \times 1 0 ^{- 7} ) ( 10 ) ( 15 )}{0.3} \times 5$
$F = 5 \times 1 0^{- 4} N$

Q. Two long parallel wires are 30cm apart carrying currents 10A and 15A respectively in the same direction. The force acting over a length of 5m of the wires is

5×10−4N , (attraction)

1×10−4N , (attraction)

5×10−4N , (repulsion)

1×10−4N , (repulsion)