Question

A conductor of length 20 cm is placed at distance 10 cm from an infinite long parallel conductor carrying constant current of 5 A. The magnitude of work done (in SI units) needed to move the conductor slowly to distance of 20 cm from wire if it carries constant current of 2 A is $\frac{\alpha \mu_0}{\pi}$ln (2). Find $\alpha$.

Answer 1

1

Answer 2

The problem asks us to calculate the work done to move a current-carrying conductor in the magnetic field of another infinite current-carrying conductor and then find the value of $\alpha$ by comparing it with a given expression.

1. Magnetic Field due to the Infinite Wire:

An infinite long straight conductor carrying current $I_1$ produces a magnetic field $B$ at a distance $r$ from it, given by:

$B = \frac{\mu_0 I_1}{2\pi r}$

2. Force on the Second Conductor:

The second conductor of length $L$ carrying current $I_2$ is placed parallel to the infinite wire. The force experienced by this conductor in the magnetic field $B$ is given by $F = I_2 L B$. Since the conductors are parallel, the current $I_2$ is perpendicular to the magnetic field lines (which are concentric circles around the infinite wire).

Substituting the expression for $B$:

$F = I_2 L \left(\frac{\mu_0 I_1}{2\pi r}\right) = \frac{\mu_0 I_1 I_2 L}{2\pi r}$

This force is attractive if the currents are in the same direction and repulsive if in opposite directions. The magnitude of the force is what we need for work done.

3. Work Done to Move the Conductor:

To move the conductor slowly from an initial distance $r_1$ to a final distance $r_2$, work must be done against this magnetic force. The work done $W$ is the integral of the force over the distance:

$W = \int_{r_1}^{r_2} F dr$

Since the force $F$ depends on $r$:

$W = \int_{r_1}^{r_2} \frac{\mu_0 I_1 I_2 L}{2\pi r} dr$

The terms $\mu_0, I_1, I_2, L,$ and $2\pi$ are constants, so we can take them out of the integral:

$W = \frac{\mu_0 I_1 I_2 L}{2\pi} \int_{r_1}^{r_2} \frac{1}{r} dr$

Evaluating the integral:

$W = \frac{\mu_0 I_1 I_2 L}{2\pi} [\ln r]_{r_1}^{r_2}$

$W = \frac{\mu_0 I_1 I_2 L}{2\pi} (\ln r_2 - \ln r_1)$

Using the logarithm property $\ln a - \ln b = \ln(a/b)$:

$W = \frac{\mu_0 I_1 I_2 L}{2\pi} \ln \left(\frac{r_2}{r_1}\right)$

4. Substitute Given Values:

Given:

• Length of the conductor, $L = 20 \text{ cm} = 0.2 \text{ m}$
• Current in the infinite wire, $I_1 = 5 \text{ A}$
• Current in the second conductor, $I_2 = 2 \text{ A}$
• Initial distance, $r_1 = 10 \text{ cm} = 0.1 \text{ m}$
• Final distance, $r_2 = 20 \text{ cm} = 0.2 \text{ m}$

Substitute these values into the work done equation:

$W = \frac{\mu_0 (5 \text{ A}) (2 \text{ A}) (0.2 \text{ m})}{2\pi} \ln \left(\frac{0.2 \text{ m}}{0.1 \text{ m}}\right)$

$W = \frac{\mu_0 (10 \times 0.2)}{2\pi} \ln (2)$

$W = \frac{\mu_0 (2)}{2\pi} \ln (2)$

$W = \frac{\mu_0}{\pi} \ln (2)$

5. Compare with the Given Expression:

The problem states that the magnitude of work done is $\frac{\alpha \mu_0}{\pi}\ln (2)$.

Comparing our calculated work done with the given expression:

$\frac{\mu_0}{\pi} \ln (2) = \frac{\alpha \mu_0}{\pi}\ln (2)$

From this comparison, we can see that:

$\alpha = 1$