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 EMF induced in the uniform metal rod moving with a uniform velocity \( v \) parallel to a long straight wire carrying a current \( I \), we can follow these steps:

### Step 1: Understand the magnetic field around the wire
The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying 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 2: Consider a small element of the rod
Let’s consider a small element of the rod of length \( dx \) at a distance \( x \) from the wire. The magnetic field at this point is:
\[ B(x) = \frac{\mu_0 I}{2 \pi x} \]

### Step 3: Calculate the induced EMF in the small element
The induced EMF \( dE \) in the small element \( dx \) moving with velocity \( v \) is given by:
\[ dE = B \cdot v \cdot dx \]
Substituting the expression for \( B \):
\[ dE = \left(\frac{\mu_0 I}{2 \pi x}\right) v \, dx \]

### Step 4: Integrate to find the total EMF
To find the total EMF \( E \) induced in the entire rod, we need to integrate \( dE \) from \( r_1 \) to \( r_2 \):
\[ E = \int_{r_1}^{r_2} dE = \int_{r_1}^{r_2} \left(\frac{\mu_0 I v}{2 \pi x}\right) dx \]
This integral can be simplified:
\[ E = \frac{\mu_0 I v}{2 \pi} \int_{r_1}^{r_2} \frac{1}{x} \, dx \]

### Step 5: Solve the integral
The integral of \( \frac{1}{x} \) is \( \ln x \):
\[ E = \frac{\mu_0 I v}{2 \pi} \left[ \ln x \right]_{r_1}^{r_2} \]
This evaluates to:
\[ E = \frac{\mu_0 I v}{2 \pi} \left( \ln r_2 - \ln r_1 \right) \]
Using the properties of logarithms, we can write this as:
\[ E = \frac{\mu_0 I v}{2 \pi} \ln \left( \frac{r_2}{r_1} \right) \]

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

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