A uniform metal rod is moving with a uniform velocity v parallel to a long straight wire carrying a current I. The rod is perpendicular to the wire with its ends at distance `r_(1) and r_(2)` (with `r_(2) gt r_(1)`) form it. The E.M.F. induced in the rod at that instant is

Options

A. zero
B. `(mu_(0)Iv)/(2pi)ln((r_(2))/(r_(1)))`
C. `(mu_(0)Iv)/(2pi)ln((r_(1))/(r_(2)))`
D. `(mu_(0)Iv)/(4pi)(1-(r_(1))/(r_(2)))`

Answer

To find the E.M.F. induced in a uniform metal rod moving parallel to a long straight wire carrying a current, we can follow these steps:

### Step 1: Understand the setup
We have a uniform metal rod of length \( L \) moving with a uniform velocity \( v \) parallel to a long straight wire carrying a current \( I \). The rod is perpendicular to the wire, with its ends at distances \( r_1 \) and \( r_2 \) from the wire, where \( r_2 > r_1 \).

### Step 2: Determine the magnetic field due to the wire
The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying a current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 \) is the permeability of free space.

### Step 3: Calculate the change in magnetic flux
As the rod moves, the magnetic flux \( \Phi \) through the rod changes. The magnetic flux through the rod can be expressed as:
\[ \Phi = B \cdot A \]
where \( A \) is the area swept by the rod as it moves. The area \( A \) can be expressed as:
\[ A = L \cdot d \]
where \( d \) is the distance the rod moves in time \( dt \), which is \( v \cdot dt \).

### Step 4: Express the change in flux
The change in flux \( d\Phi \) can be expressed as:
\[ d\Phi = B \cdot dA = B \cdot (L \cdot v \cdot dt) \]
Substituting for \( B \):
\[ d\Phi = \frac{\mu_0 I}{2 \pi r} \cdot (L \cdot v \cdot dt) \]

### Step 5: Integrate to find the total change in flux
To find the total change in flux as the rod moves from \( r_1 \) to \( r_2 \), we integrate:
\[ \Phi = \int_{r_1}^{r_2} \frac{\mu_0 I}{2 \pi r} L \cdot v \, dr \]
This gives:
\[ \Phi = \frac{\mu_0 I L v}{2 \pi} \int_{r_1}^{r_2} \frac{1}{r} \, dr = \frac{\mu_0 I L v}{2 \pi} \left[ \ln r \right]_{r_1}^{r_2} \]
\[ \Phi = \frac{\mu_0 I L v}{2 \pi} \left( \ln r_2 - \ln r_1 \right) = \frac{\mu_0 I L v}{2 \pi} \ln \left( \frac{r_2}{r_1} \right) \]

### Step 6: Find the induced E.M.F.
The induced E.M.F. \( \epsilon \) in the rod is given by Faraday's law of electromagnetic induction:
\[ \epsilon = -\frac{d\Phi}{dt} \]
Since \( \Phi \) depends on \( t \) through the motion of the rod, we differentiate:
\[ \epsilon = -\frac{d}{dt} \left( \frac{\mu_0 I L v}{2 \pi} \ln \left( \frac{r_2}{r_1} \right) \right) \]
Since \( \mu_0, I, L, v \) are constants, we have:
\[ \epsilon = \frac{\mu_0 I L v}{2 \pi} \cdot \frac{1}{r_2} \cdot \frac{dr_2}{dt} - \frac{1}{r_1} \cdot \frac{dr_1}{dt} \]
However, since \( r_1 \) and \( r_2 \) are fixed distances, we can simplify this to:
\[ \epsilon = \frac{\mu_0 I v}{2 \pi} \ln \left( \frac{r_2}{r_1} \right) \]

### Final Result
Thus, the induced E.M.F. in the rod is:
\[ \epsilon = \frac{\mu_0 I v}{2 \pi} \ln \left( \frac{r_2}{r_1} \right) \]