Currents of 10 A and 2 A are passed through two parallel wires A and B respectively in opposite directions. If the wire is uniformly long and the length of wire B is 2 m. The force on conductor B at separation 10 cm from A will be :

Question

Currents of 10 A and 2 A are passed through two parallel wires A and B respectively in opposite directions. If the wire is uniformly long and the length of wire B is 2 m. The force on conductor B at separation 10 cm from A will be : (A)  40× 10-7N  (B) 8 π × 10-7N  (C)  8× 10-5N  (D)  4× 10-5N

Tardigrade Admin · Accepted Answer

Here, i1=10 A, i2=2 A r=10 cm =0.1 m , l=2 m Using the relation F=(μ0 i1 i2 l/2 π R) or F=(4 π × 10-7 × 10 × 2 × 2/2 π × 0.10) So, F=8 × 10-5 N

Q. Currents of $10 A$ and $2 A$ are passed through two parallel wires $A$ and $B$ respectively in opposite directions. If the wire is uniformly long and the length of wire $B$ is $2 m$ . The force on conductor $B$ at separation $10 c m$ from A will be :

$40 \times 1 0^{- 7} N$

8 $π$ $\times 1 0^{- 7} N$

$8 \times 1 0^{- 5} N$

$4 \times 1 0^{- 5} N$

Here, $i_{1} = 10 A, i_{2} = 2 A$
$r = 10 c m = 0.1 m, l = 2 m$
Using the relation $F = \frac{μ _{0} i _{1} i _{2} l}{2 π R}$
or $F = \frac{4 π \times 1 0 ^{- 7} \times 10 \times 2 \times 2}{2 π \times 0.10}$
So, $F = 8 \times 1 0^{- 5} N$

Q. Currents of 10A and 2A are passed through two parallel wires A and B respectively in opposite directions. If the wire is uniformly long and the length of wire B is 2m. The force on conductor B at separation 10cm from A will be :

40×10−7N

8 π ×10−7N

8×10−5N

4×10−5N