A self-induced wmf in a solenoid of inductance L changes in time as`varepsilon=varepsilon_0e^(-kz).` Assuming the charge is finite.find the total charge that passes a point in the wire of the solenoid.

Options

A. `varepsilon_0/(Lk^2)`
B. `(2varepsilon_0)/(Lk^2)`
C. `varepsilon_0/(2Lk^2)`
D. `(3varepsilon_0)/(Lk^2)`

Answer

To solve the problem of finding the total charge that passes a point in the wire of a solenoid with a self-induced EMF given by \( \varepsilon = \varepsilon_0 e^{-kz} \), we can follow these steps:

### Step 1: Relate EMF to Inductance and Current
The self-induced EMF in a solenoid can be expressed as:
\[ \varepsilon = -L \frac{di}{dt} \]
Given that \( \varepsilon = \varepsilon_0 e^{-kz} \), we can equate the two expressions:
\[ \varepsilon_0 e^{-kz} = -L \frac{di}{dt} \]

### Step 2: Rearranging the Equation
Rearranging the equation to isolate \( di \):
\[ \frac{di}{dt} = -\frac{\varepsilon_0 e^{-kz}}{L} \]

### Step 3: Integrate to Find Current
We can now integrate both sides with respect to time \( t \):
\[ \int di = -\frac{\varepsilon_0}{L} \int e^{-kt} dt \]
This gives:
\[ i = -\frac{\varepsilon_0}{L} \cdot \left(-\frac{1}{k} e^{-kt}\right) + C \]
where \( C \) is the constant of integration. Thus, we have:
\[ i = \frac{\varepsilon_0}{Lk} e^{-kt} + C \]

### Step 4: Relate Current to Charge
We know that current \( i \) is related to charge \( q \) by:
\[ i = \frac{dq}{dt} \]
Substituting for \( i \):
\[ \frac{dq}{dt} = \frac{\varepsilon_0}{Lk} e^{-kt} + C \]

### Step 5: Integrate to Find Total Charge
Now we can integrate to find the total charge \( q \):
\[ \int dq = \int \left(\frac{\varepsilon_0}{Lk} e^{-kt} + C\right) dt \]
This results in:
\[ q = -\frac{\varepsilon_0}{Lk^2} e^{-kt} + Ct + Q_0 \]
where \( Q_0 \) is another constant of integration.

### Step 6: Evaluate Total Charge
Assuming the charge is finite and considering the limits of integration (from \( t = 0 \) to \( t = \infty \)), we can evaluate the charge:
\[ q = \left[ -\frac{\varepsilon_0}{Lk^2} e^{-kt} \right]_{0}^{\infty} + C\left[t\right]_{0}^{\infty} \]
As \( t \to \infty \), \( e^{-kt} \to 0 \), and at \( t = 0 \), \( e^{-k \cdot 0} = 1 \):
\[ q = -\frac{\varepsilon_0}{Lk^2}(0 - 1) + C(\infty - 0) \]
The term with \( C \) becomes infinite unless \( C = 0 \). Thus, we find:
\[ q = \frac{\varepsilon_0}{Lk^2} \]

### Final Result
The total charge that passes a point in the wire of the solenoid is:
\[ q = \frac{\varepsilon_0}{Lk^2} \]