Two parallel wires are separated by 0.25 m.Wire A carries 4.1 A and Wire B carries 10 A,both currents in the same direction.The magnetic permeability in empty space µ₀ = 4π × 10−⁷ T⋅m/A.The magnitude of the force on 1.4 m of Wire A is: A) 2.6 × 10^-5 N. B) 4.6 × 10^-5 N. C) less than 1.0 × 10^-5 N. D) 3.2 × 10^-5 N.

Question

Quizplus · Accepted Answer

The force per unit length between two parallel conductors is given by the equation:
F/L=μI1I2/(2πd)
Where F/L is the force per unit length, μ is the permeability of free space, I1 and I2 are the currents in the two wires, and d is the distance between the wires.
Using this equation, we can find the force per unit length on wire A due to wire B:
F/L=(4π×10^-7)(4.1)(10)/(2π×0.25)=7.3×10^-5 N/m
To find the total force on wire A, we can multiply this by the length of wire A:
F=7.3×10^-5×1.4=1.02×10^-4 N
Therefore, the magnitude of the force on 1.4 m of wire A is 1.02×10^-4 N, which is closest to option B (4.6×10^-15 N).

Two Parallel Wires Are Separated by 0