Two long parallel straight wires A and B carrying currents of 4.0 A and 5.0 A in same direction separated by a distance of 4 × 10-2 m. The force on a 0.20 m section of wire A is

Question

Two long parallel straight wires A and B carrying currents of 4.0 A and 5.0 A in same direction separated by a distance of 4 × 10-2 m. The force on a 0.20 m section of wire A is (A) 2 × 10-5 N towards B (B) 2 × 10-5 N away from B (C) 2 × 10-5 N perpendicular to B (D) 2 × 10-5 N parallel to B

Tardigrade Admin · Accepted Answer

I1=4 A , I2=5 A , d=4 × 10-2 m , l=0.20 m Force on a current carrying wire, F =(μ0 I1 I2 l/2 π d) =(4 π × 10-7 × 4 × 5/2 × π × 4 × 10-2) × 0.20 =2 × 10-5 N towards B

Q. Two long parallel straight wires $A$ and $B$ carrying currents of $4.0 A$ and $5.0 A$ in same direction separated by a distance of $4 \times 1 0^{- 2} m$ . The force on a $0.20 m$ section of wire $A$ is

$2 \times 1 0^{- 5} N$ towards $B$

$2 \times 1 0^{- 5} N$ away from $B$

$2 \times 1 0^{- 5} N$ perpendicular to $B$

$2 \times 1 0^{- 5} N$ parallel to $B$

$I_{1} = 4 A, I_{2} = 5 A, d = 4 \times 1 0^{- 2} m, l = 0.20 m$
Force on a current carrying wire,
$F = \frac{μ _{0} I _{1} I _{2} l}{2 π d}$
$= \frac{4 π \times 1 0 ^{- 7} \times 4 \times 5}{2 \times π \times 4 \times 1 0 ^{- 2}} \times 0.20$
$= 2 \times 1 0^{- 5} N$ towards $B$

Q. Two long parallel straight wires A and B carrying currents of 4.0A and 5.0A in same direction separated by a distance of 4×10−2m. The force on a 0.20m section of wire A is

2×10−5N towards B

2×10−5N away from B

2×10−5N perpendicular to B

2×10−5N parallel to B