A coil of inductance `1 H` and resistance `10Omega` is connected to a resistanceless battery of emf `50 V` at time `t=0`. Calculate the ratio of rthe rate which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at `t=0.1s`.

Explanation

To solve the problem, we need to calculate the ratio of the rate at which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at \( t = 0.1 \, s \).

### Step 1: Identify the given values
- Inductance \( L = 1 \, H \)
- Resistance \( R = 10 \, \Omega \)
- EMF of the battery \( V_0 = 50 \, V \)
- Time \( t = 0.1 \, s \)

### Step 2: Calculate the time constant \( \tau \)
The time constant \( \tau \) for an RL circuit is given by:
\[ \tau = \frac{L}{R} \]
Substituting the values:
\[ \tau = \frac{1 \, H}{10 \, \Omega} = 0.1 \, s \]

### Step 3: Calculate the current \( I(t) \) at \( t = 0.1 \, s \)
The current in the circuit as a function of time is given by:
\[ I(t) = \frac{V_0}{R} \left(1 - e^{-\frac{t}{\tau}}\right) \]
Substituting the values:
\[ I(0.1) = \frac{50 \, V}{10 \, \Omega} \left(1 - e^{-\frac{0.1}{0.1}}\right) \]
\[ I(0.1) = 5 \left(1 - e^{-1}\right) \]

### Step 4: Calculate the potential difference across the inductor \( V_L(t) \)
The potential difference across the inductor at time \( t \) is given by:
\[ V_L(t) = V_0 e^{-\frac{t}{\tau}} \]
Substituting the values:
\[ V_L(0.1) = 50 \, V \cdot e^{-1} \]

### Step 5: Calculate the rate at which magnetic energy is stored in the coil
The rate at which magnetic energy is stored in the coil is given by:
\[ P_{magnetic} = V_L(t) \cdot I(t) \]
Substituting the expressions we found:
\[ P_{magnetic} = (50 \, e^{-1}) \cdot (5(1 - e^{-1})) \]
\[ P_{magnetic} = 250 e^{-1} (1 - e^{-1}) \]

### Step 6: Calculate the rate at which energy is supplied by the battery
The rate at which energy is supplied by the battery is given by:
\[ P_{battery} = V_0 \cdot I(t) \]
Substituting the values:
\[ P_{battery} = 50 \cdot 5(1 - e^{-1}) = 250(1 - e^{-1}) \]

### Step 7: Calculate the ratio of the two powers
Now, we can find the ratio:
\[ \text{Ratio} = \frac{P_{magnetic}}{P_{battery}} = \frac{250 e^{-1} (1 - e^{-1})}{250(1 - e^{-1})} \]
The \( 250(1 - e^{-1}) \) terms cancel out:
\[ \text{Ratio} = e^{-1} \]

### Step 8: Calculate the numerical value
Since \( e^{-1} \approx 0.3679 \):
\[ \text{Ratio} \approx 0.37 \]

### Final Answer
The ratio of the rate at which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at \( t = 0.1 \, s \) is approximately \( 0.37 \).

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