Question

Two long parallel straight conductors separated by $10\text{ cm}$ carrying currents $20\text{ A}$, $40\text{ A}$ in the same direction. The work required per unit length to move the conductors apart to $30\text{ cm}$ is [Take $\log_{10} 3 = 0.4771$]:

• A. $17.6 \times 10^{-5}\text{ J m}^{-1}$
• B. $21.2 \times 10^{-5}\text{ J m}^{-1}$
• C. $16.8 \times 10^{-5}\text{ J m}^{-1}$
• D. $14.6 \times 10^{-5}\text{ J m}^{-1}$

Hint:

For parallel wires, the work done per unit length is simply $2 \times 10^{-7} \times I_1 I_2 \ln\left(\frac{r_2}{r_1}\right)$.
With $I_1 I_2 = 800$ and $\ln 3 \approx 1.1$, we get $1.6 \times 10^{-4} \times 1.1 = 1.76 \times 10^{-4}\text{ J/m}$, which easily converts to $17.6 \times 10^{-5}\text{ J/m}$.

Answer 1

The Correct Option is A

Step 1: Write the attractive force per unit length between the wires. \[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r} \] Since both currents flow the same way, moving the wires apart means doing work against this attraction.
Step 2: Lump the constants together before integrating.
Group everything that doesn't depend on $r$ into one constant $k = \dfrac{\mu_0 I_1 I_2}{2\pi} = (2\times 10^{-7})(20)(40) = 1.6\times 10^{-4}\text{ J/m}$, so $\dfrac{W}{L} = k\ln\left(\dfrac{r_2}{r_1}\right)$.
Step 3: Convert the natural log using the given base-10 value. \[ \ln(3) = 2.303\log_{10}(3) = 2.303\times 0.4771 = 1.0989 \]
Step 4: Multiply out to get the work per unit length. \[ \frac{W}{L} = 1.6\times 10^{-4}\times 1.0989 \] \[ \boxed{\dfrac{W}{L} \approx 17.6\times 10^{-5}\text{ J m}^{-1}} \]