Question

Wire $AB$ is moved with a velocity 2 m/s. Wire is kept near a long current carrying conductor, as shown. Emf induced in the wire is

Answer 1

2 x 10^{-6} ln(4) V

Answer 2

The problem involves calculating the motional electromotive force (EMF) induced in a conductor moving in a non-uniform magnetic field.

1. Magnetic Field due to the Long Wire:
   A long straight current-carrying conductor produces a magnetic field given by the formula: $B = \frac{\mu_0 I}{2 \pi r}$ where $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$ is the permeability of free space, $I$ is the current (5 A), and $r$ is the perpendicular distance from the wire.
   Using the right-hand thumb rule, for an upward current, the magnetic field to the right of the wire is directed into the page.

2. Motional EMF:
   The wire segment AB has a length $L = 3 \text{ m}$ and is moving with a velocity $v = 2 \text{ m/s}$ upwards. The wire A is at a distance $r_1 = 1 \text{ m}$ from the long wire, and wire B is at $r_2 = r_1 + L = 1 + 3 = 4 \text{ m}$ from the long wire.
   Since the magnetic field $B$ depends on the distance $r$, it is not uniform along the length of the wire AB. Therefore, we must integrate to find the total induced EMF.
   Consider a small segment $dr$ of the wire AB at a distance $r$ from the long current-carrying wire. The magnetic field at this segment is $B(r) = \frac{\mu_0 I}{2 \pi r}$.
   The velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$ (velocity is vertical, magnetic field is into the page) and also perpendicular to the length element $d\vec{r}$ of the wire (velocity is vertical, $d\vec{r}$ is horizontal).
   The induced EMF across this small segment is $d\mathcal{E} = B(r) v dr$.
   The total induced EMF across the wire AB is the integral of $d\mathcal{E}$ from $r_1$ to $r_2$: $\mathcal{E} = \int_{r_1}^{r_2} B(r) v dr$ Substitute $B(r)$: $\mathcal{E} = \int_{r_1}^{r_2} \left(\frac{\mu_0 I}{2 \pi r}\right) v dr$ $\mathcal{E} = \frac{\mu_0 I v}{2 \pi} \int_{r_1}^{r_2} \frac{1}{r} dr$ $\mathcal{E} = \frac{\mu_0 I v}{2 \pi} [\ln r]_{r_1}^{r_2}$ $\mathcal{E} = \frac{\mu_0 I v}{2 \pi} (\ln r_2 - \ln r_1)$ $\mathcal{E} = \frac{\mu_0 I v}{2 \pi} \ln \left(\frac{r_2}{r_1}\right)$

3. Substitute the values:

   • $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$
   • $I = 5 \text{ A}$
   • $v = 2 \text{ m/s}$
   • $r_1 = 1 \text{ m}$
   • $r_2 = 4 \text{ m}$

   $\mathcal{E} = \frac{(4\pi \times 10^{-7}) \times 5 \times 2}{2 \pi} \ln \left(\frac{4}{1}\right)$ $\mathcal{E} = (2 \times 10^{-7} \times 5 \times 2) \ln(4)$ $\mathcal{E} = (20 \times 10^{-7}) \ln(4)$ $\mathcal{E} = 2 \times 10^{-6} \ln(4) \text{ V}$

   Using $\ln(4) = 2 \ln(2) \approx 2 \times 0.693 = 1.386$: $\mathcal{E} = 2 \times 10^{-6} \times 1.386$ $\mathcal{E} = 2.772 \times 10^{-6} \text{ V}$

   Polarity:
   Using the right-hand rule for the force on positive charge carriers $\vec{F} = q(\vec{v} \times \vec{B})$:
   $\vec{v}$ is upwards. $\vec{B}$ is into the page. $\vec{v} \times \vec{B}$ points to the left (towards the long wire).
   This means positive charges are pushed towards point A, making A at a higher potential than B.

The final answer is $2 \times 10^{-6} \ln(4) \text{ V}$.

Explanation of the solution:
The magnetic field due to the long current-carrying wire is non-uniform, varying inversely with distance. The motional EMF is calculated by integrating $B(r)v dr$ over the length of the wire AB. The integral leads to $\mathcal{E} = \frac{\mu_0 I v}{2 \pi} \ln \left(\frac{r_2}{r_1}\right)$. Substituting the given values $I=5 \text{ A}$, $v=2 \text{ m/s}$, $r_1=1 \text{ m}$, and $r_2=4 \text{ m}$ yields the induced EMF.