A coil having an inductance `L` and a resistance `R` is connected to a battery of `emf epsilon`. Find the time elapsed before, the magnetic field energy stored in the circuit reaches half its maximum value.

Options

A. `R/L ln [sqrt2/(sqrt-1)]`
B. `L/Rln [(sqrt2-1)/sqrt2]`
C. `L/R ln (sqrt2/(sqrt2-1))`
D. `R/L ln [(sqrt2-1)/sqrt2]`

Explanation

To solve the problem of finding the time elapsed before the magnetic field energy stored in the circuit reaches half its maximum value, we can follow these steps:

### Step 1: Understand the maximum energy stored in the inductor
The maximum current \( I_0 \) in the circuit when the switch is closed can be given by Ohm's law:
\[ I_0 = \frac{\epsilon}{R} \]
The maximum energy \( U \) stored in the magnetic field of the inductor is given by:
\[ U = \frac{1}{2} L I_0^2 \]

### Step 2: Determine the condition for half the maximum energy
We want to find the time \( t \) when the energy \( U' \) stored in the inductor is half of the maximum energy:
\[ U' = \frac{1}{2} U = \frac{1}{2} \left( \frac{1}{2} L I_0^2 \right) = \frac{1}{4} L I_0^2 \]

### Step 3: Relate energy to current
The energy stored in the inductor at any time \( t \) can be expressed in terms of the current \( I(t) \):
\[ U' = \frac{1}{2} L I^2 \]
Setting this equal to our expression for half the maximum energy:
\[ \frac{1}{2} L I^2 = \frac{1}{4} L I_0^2 \]
We can cancel \( \frac{1}{2} L \) from both sides (assuming \( L \neq 0 \)):
\[ I^2 = \frac{1}{2} I_0^2 \]
Taking the square root gives:
\[ I = \frac{I_0}{\sqrt{2}} \]

### Step 4: Find the expression for current as a function of time
The current \( I(t) \) in an RL circuit is given by:
\[ I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \( \tau = \frac{L}{R} \) is the time constant.

### Step 5: Set the two expressions for current equal
Now we set the expression for current equal to \( \frac{I_0}{\sqrt{2}} \):
\[ \frac{I_0}{\sqrt{2}} = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{\sqrt{2}} = 1 - e^{-\frac{t}{\tau}} \]

### Step 6: Solve for \( e^{-\frac{t}{\tau}} \)
Rearranging gives:
\[ e^{-\frac{t}{\tau}} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} \]

### Step 7: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ -\frac{t}{\tau} = \ln\left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) \]
Thus,
\[ t = -\tau \ln\left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) \]

### Step 8: Substitute \( \tau \)
Substituting \( \tau = \frac{L}{R} \):
\[ t = -\frac{L}{R} \ln\left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) \]

### Step 9: Final expression
The final expression for the time \( t \) when the energy reaches half of its maximum value is:
\[ t = \frac{L}{R} \ln\left(\frac{\sqrt{2}}{\sqrt{2} - 1}\right) \]