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 \( t = 0.1 \, \text{s} \).

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance \( L = 1 \, \text{H} \)
- Resistance \( R = 10 \, \Omega \)
- EMF of the battery \( E = 50 \, \text{V} \)

2. **Calculate the time constant \( \tau \):**
\[ \tau = \frac{L}{R} = \frac{1 \, \text{H}}{10 \, \Omega} = 0.1 \, \text{s} \]

3. **Determine the steady-state current \( I_0 \):**
\[ I_0 = \frac{E}{R} = \frac{50 \, \text{V}}{10 \, \Omega} = 5 \, \text{A} \]

4. **Find the current \( I(t) \) at \( t = 0.1 \, \text{s} \):**
\[ I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) = 5 \left(1 - e^{-\frac{0.1}{0.1}}\right) = 5 \left(1 - e^{-1}\right) \]
\[ I(0.1) = 5 \left(1 - \frac{1}{e}\right) \]

5. **Calculate \( \frac{di}{dt} \):**
\[ \frac{di}{dt} = \frac{I_0}{\tau} e^{-\frac{t}{\tau}} = \frac{5}{0.1} e^{-1} = 50 e^{-1} \]

6. **Find the voltage across the inductor \( V_L \):**
\[ V_L = L \frac{di}{dt} = 1 \cdot (50 e^{-1}) = 50 e^{-1} \, \text{V} \]

7. **Calculate the rate at which magnetic energy is stored \( \frac{dU}{dt} \):**
- The energy stored in the inductor is given by \( U = \frac{1}{2} L I^2 \).
- The rate of change of energy is:
\[ \frac{dU}{dt} = L I \frac{di}{dt} = 1 \cdot \left(5 \left(1 - \frac{1}{e}\right)\right) \cdot (50 e^{-1}) \]
\[ \frac{dU}{dt} = 250 \left(1 - \frac{1}{e}\right) e^{-1} \]

8. **Calculate the rate at which energy is supplied by the battery \( P \):**
\[ P = E \cdot I = 50 \cdot 5 \left(1 - \frac{1}{e}\right) = 250 \left(1 - \frac{1}{e}\right) \]

9. **Find the ratio of the rate of magnetic energy storage to the rate of energy supplied:**
\[ \text{Ratio} = \frac{\frac{dU}{dt}}{P} = \frac{250 \left(1 - \frac{1}{e}\right) e^{-1}}{250 \left(1 - \frac{1}{e}\right)} = e^{-1} \]

### Final Result:
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 \( t = 0.1 \, \text{s} \) is \( \frac{1}{e} \).