A straight wire 5m long is placed parallel to an infinitely long straight wire carrying current 5A. The distance between the two wires is 10 cm and the 5m long wire is carrying current 2A. The force on the shorter wire is :

Question

A straight wire 5m long is placed parallel to an infinitely long straight wire carrying current 5A. The distance between the two wires is 10 cm and the 5m long wire is carrying current 2A. The force on the shorter wire is : (A) (μ0/4π)×103 N (B) 103 N (C) (μ0/4π)×10-7 N (D) 10-7 N

Tardigrade Admin · Accepted Answer

F = (Fm/I) l = (μ0/4 π) × (2 I1 I2/r) × l = (μ0/4 π) × (2 × 5 × 2/0.1) × 5 = (μ0/4 π ) × 103 N.

Q. A straight wire 5m long is placed parallel to an infinitely long straight wire carrying current 5A. The distance between the two wires is 10 cm and the 5m long wire is carrying current 2A. The force on the shorter wire is :

$\frac{μ _{0}}{4 π}$ $\times 1 0^{3} N$

$1 0^{3} N$

$\frac{μ _{0}}{4 π}$ $\times 1 0^{- 7} N$

$1 0^{- 7} N$

F = $(F_{m} / I) l = \frac{μ _{0}}{4 π} \times \frac{2 I _{1} I _{2}}{r} \times l = \frac{μ _{0}}{4 π} \times \frac{2 \times 5 \times 2}{0.1} \times 5 = \frac{μ _{0}}{4 π} \times 1 0^{3}$ N.

Q. A straight wire 5m long is placed parallel to an infinitely long straight wire carrying current 5A. The distance between the two wires is 10 cm and the 5m long wire is carrying current 2A. The force on the shorter wire is :

4πμ0​​×103N

103N

4πμ0​​×10−7N

10−7N

F = (Fm​/I)l=4πμ0​​×r2I1​I2​​×l=4πμ0​​×0.12×5×2​×5=4πμ0​​×103 N.

$\frac{μ _{0}}{4 π}$ $\times 1 0^{3} N$

$1 0^{3} N$

$\frac{μ _{0}}{4 π}$ $\times 1 0^{- 7} N$

$1 0^{- 7} N$

F = $(F_{m} / I) l = \frac{μ _{0}}{4 π} \times \frac{2 I _{1} I _{2}}{r} \times l = \frac{μ _{0}}{4 π} \times \frac{2 \times 5 \times 2}{0.1} \times 5 = \frac{μ _{0}}{4 π} \times 1 0^{3}$ N.