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

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given quantities
- Induced current \( I = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A} \)
- Resistance of the ring \( R = 0.5 \, \Omega \)
- Time duration \( \Delta t = 5 \, \text{s} \)

### Step 2: Use the relationship between charge, current, and time
The charge \( \Delta Q \) induced can be calculated using the formula:
\[ \Delta Q = I \times \Delta t \]
Substituting the values:
\[ \Delta Q = (10 \times 10^{-3} \, \text{A}) \times (5 \, \text{s}) = 50 \times 10^{-3} \, \text{C} = 0.05 \, \text{C} \]

### Step 3: Relate charge to change in magnetic flux
The change in magnetic flux \( \Delta \Phi \) can be expressed in terms of the charge and resistance using the formula:
\[ \Delta Q = \frac{\Delta \Phi}{R} \]
Rearranging this gives:
\[ \Delta \Phi = \Delta Q \times R \]

### Step 4: Substitute the values to find the change in magnetic flux
Now substituting the values of \( \Delta Q \) and \( R \):
\[ \Delta \Phi = (0.05 \, \text{C}) \times (0.5 \, \Omega) = 0.025 \, \text{Wb} \]

### Step 5: Convert to milliWebers
To express the result in milliWebers:
\[ \Delta \Phi = 0.025 \, \text{Wb} = 25 \, \text{mWb} \]

### Final Answer
The change in magnetic flux through the ring is \( \Delta \Phi = 25 \, \text{mWb} \).

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