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 $\log_{10} 3 = 0.4771$]:
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 $\log_{10} 3 = 0.4771$]:
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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 $17.6 \times 10^{-5}\text{ J/m}$.
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 $17.6 \times 10^{-5}\text{ J/m}$.
Updated On: Jul 22, 2026
- $17.6 \times 10^{-5}\text{ J m}^{-1}$
- $21.2 \times 10^{-5}\text{ J m}^{-1}$
- $16.8 \times 10^{-5}\text{ J m}^{-1}$
- $14.6 \times 10^{-5}\text{ J m}^{-1}$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
Two parallel wires carrying currents in the same direction attract each other.
We need to calculate the work required per unit length to pull them further apart from a separation of $10\text{ cm}$ to $30\text{ cm}$ against this attractive magnetic force.
Step 2: Key Formula and Approach:
The attractive magnetic force per unit length between two parallel conductors is:
\[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r} \] The work done per unit length to change the separation from $r_1$ to $r_2$ is:
\[ \frac{W}{L} = \int_{r_1}^{r_2} \left(\frac{F}{L}\right) dr = \frac{\mu_0 I_1 I_2}{2\pi} \ln \left( \frac{r_2}{r_1} \right) \] Using $\ln(x) = 2.303 \log_{10}(x)$:
\[ \frac{W}{L} = 2.303 \times \frac{\mu_0 I_1 I_2}{2\pi} \log_{10} \left( \frac{r_2}{r_1} \right) \]
Step 3: Detailed Explanation:
• Identify given values:
$I_1 = 20\text{ A}$, $I_2 = 40\text{ A}$
$r_1 = 10\text{ cm} = 0.1\text{ m}$, $r_2 = 30\text{ cm} = 0.3\text{ m}$
$\frac{r_2}{r_1} = \frac{0.3}{0.1} = 3$
$\frac{\mu_0}{2\pi} = 2 \times 10^{-7}\text{ T m A}^{-1}$
• Calculate work per unit length:
\[ \frac{W}{L} = 2.303 \times \left( 2 \times 10^{-7} \right) \times 20 \times 40 \times \log_{10}(3) \] \[ \frac{W}{L} = 2.303 \times 1.6 \times 10^{-4} \times 0.4771 \] \[ \frac{W}{L} = 3.6848 \times 10^{-4} \times 0.4771 \] \[ \frac{W}{L} \approx 1.758 \times 10^{-4}\text{ J m}^{-1} = 17.58 \times 10^{-5}\text{ J m}^{-1} \] Rounding to one decimal place gives:
\[ \frac{W}{L} \approx 17.6 \times 10^{-5}\text{ J m}^{-1} \]
Step 4: Final Answer:
The work required per unit length is $17.6 \times 10^{-5}\text{ J m}^{-1}$, which corresponds to Option (A).
Two parallel wires carrying currents in the same direction attract each other.
We need to calculate the work required per unit length to pull them further apart from a separation of $10\text{ cm}$ to $30\text{ cm}$ against this attractive magnetic force.
Step 2: Key Formula and Approach:
The attractive magnetic force per unit length between two parallel conductors is:
\[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r} \] The work done per unit length to change the separation from $r_1$ to $r_2$ is:
\[ \frac{W}{L} = \int_{r_1}^{r_2} \left(\frac{F}{L}\right) dr = \frac{\mu_0 I_1 I_2}{2\pi} \ln \left( \frac{r_2}{r_1} \right) \] Using $\ln(x) = 2.303 \log_{10}(x)$:
\[ \frac{W}{L} = 2.303 \times \frac{\mu_0 I_1 I_2}{2\pi} \log_{10} \left( \frac{r_2}{r_1} \right) \]
Step 3: Detailed Explanation:
• Identify given values:
$I_1 = 20\text{ A}$, $I_2 = 40\text{ A}$
$r_1 = 10\text{ cm} = 0.1\text{ m}$, $r_2 = 30\text{ cm} = 0.3\text{ m}$
$\frac{r_2}{r_1} = \frac{0.3}{0.1} = 3$
$\frac{\mu_0}{2\pi} = 2 \times 10^{-7}\text{ T m A}^{-1}$
• Calculate work per unit length:
\[ \frac{W}{L} = 2.303 \times \left( 2 \times 10^{-7} \right) \times 20 \times 40 \times \log_{10}(3) \] \[ \frac{W}{L} = 2.303 \times 1.6 \times 10^{-4} \times 0.4771 \] \[ \frac{W}{L} = 3.6848 \times 10^{-4} \times 0.4771 \] \[ \frac{W}{L} \approx 1.758 \times 10^{-4}\text{ J m}^{-1} = 17.58 \times 10^{-5}\text{ J m}^{-1} \] Rounding to one decimal place gives:
\[ \frac{W}{L} \approx 17.6 \times 10^{-5}\text{ J m}^{-1} \]
Step 4: Final Answer:
The work required per unit length is $17.6 \times 10^{-5}\text{ J m}^{-1}$, which corresponds to Option (A).
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