Two long,straight parallel wires 7.6 cm apart carry currents of equal magnitude I.They attract each other with a force per unit length of 4.3 nN/m.The current I is approximately
A)1.6 mA
B)40 mA
C)37 mA
D)59 mA
E)4.0 mA

Question

Quizplus · Accepted Answer

The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ is given by the formula $$F/L = \frac{\mu_0 I_1 I_2}{2\pi d}$$, where $F/L$ is the force per unit length, $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$), and $d$ is the distance between the wires. Given that the wires attract each other with a force per unit length of $4.3 \, \text{nN/m}$ and are $7.6 \, \text{cm}$ apart, and knowing that the currents are equal ($I_1 = I_2 = I$), we can rearrange the formula to solve for $I$:$$4.3 \times 10^{-9} = \frac{4\pi \times 10^{-7} \cdot I^2}{2\pi \cdot 0.076}$$Solving for $I$ gives a value close to $40 \, \text{mA}$, making option B the correct answer.

Two Long,straight Parallel Wires 7