A magnet is taken towards a conducting ring in such a way that a constant current of `10mA` is induced in it. The total resistance of the ring is `0.5Omega`. In `5s`, the magnetic flux through the ring changes by

Options

A. `0.25mWb`
B. `25mWb`
C. `50mWb`
D. `15mWb`

Explanation

To solve the problem, we will use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a closed loop is equal to the rate of change of magnetic flux through the loop.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Induced current (I) = 10 mA = \(10 \times 10^{-3}\) A
- Resistance of the ring (R) = 0.5 Ω
- Time interval (Δt) = 5 s

2. **Use Ohm's Law to Find Induced EMF:**
According to Ohm's Law, the induced emf (ε) can be calculated using the formula:
\[ \text{emf} = I \times R \]
Substituting the known values:
\[ \text{emf} = (10 \times 10^{-3} \, \text{A}) \times (0.5 \, \Omega) = 5 \times 10^{-3} \, \text{V} = 5 \, \text{mV} \]

3. **Apply Faraday's Law:**
Faraday's law states that:
\[ \text{emf} = -\frac{\Delta \Phi}{\Delta t} \]
Rearranging this gives us:
\[ \Delta \Phi = -\text{emf} \times \Delta t \]
Since we are interested in the magnitude, we can ignore the negative sign:
\[ \Delta \Phi = \text{emf} \times \Delta t \]

4. **Substitute the Values:**
Now, substituting the values we have:
\[ \Delta \Phi = (5 \times 10^{-3} \, \text{V}) \times (5 \, \text{s}) = 25 \times 10^{-3} \, \text{Wb} \]

5. **Convert to Mill Weber:**
\[ \Delta \Phi = 25 \, \text{mWb} \]

### Final Answer:
The total change in magnetic flux through the ring in 5 seconds is \(25 \, \text{mWb}\).

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