Two long parallel straight wires X and Y separated by a distance of $5\,\text{cm}$ in air carry currents of $10\,\text{A}$ and $5\,\text{A}$ respectively in opposite directions. Calculate the magnitude and direction of the force on a $20\,\text{cm}$ length of the wire Y.

$F=4\times10^{-5}\,\text{N}$, directed away from X (repulsive).

CBSE Class 12 · 2004 · Magnetic Effects of CurrentShort Answer

Two long parallel straight wires X and Y separated by a distance of

5 cm

in air carry currents of

10 A

and

5 A

respectively in opposite directions. Calculate the magnitude and direction of the force on a

20 cm

length of the wire Y.

Answer: $F = 4 \times 1 0^{- 5} N$ , directed away from X (repulsive).

The force per unit length between the wires is

\frac{F}{ℓ} = \frac{μ _{0} I _{X} I _{Y}}{2 π r}

F = \frac{μ _{0} I _{X} I _{Y}}{2 π r} ℓ = \frac{( 4 π \times 1 0 ^{- 7} ) ( 10 ) ( 5 )}{2 π ( 5 \times 1 0 ^{- 2} )} \times (20 \times 1 0^{- 2})

F = 2 \times 1 0^{- 7} \times \frac{50}{5 \times 1 0 ^{- 2}} \times 0.20 = 4 \times 1 0^{- 5} N

Since the currents are in opposite directions the force is repulsive, so wire Y is pushed away from X.

CBSE Class 12 Board · 2004 · Subjective