There is one square loop of area 0.5 `m^2` coplanar with a long straight wire in which a current of 20 ampere is flowering. Distance of the loop from the wire is 30. If current in the wire changes from 20 A to 0 in time 0.5 a. What will be the average emf induced in the loop ?

Answer

To solve the problem of finding the average emf induced in a square loop due to a changing current in a nearby straight wire, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Area of the square loop, \( A = 0.5 \, m^2 \)
- Initial current in the wire, \( I_1 = 20 \, A \)
- Final current in the wire, \( I_2 = 0 \, A \)
- Distance from the wire to the loop, \( d = 30 \, cm = 0.3 \, m \)
- Time duration for the change in current, \( \Delta t = 0.5 \, s \)

2. **Calculate the Change in Current**:
\[ \Delta I = I_2 - I_1 = 0 - 20 = -20 \, A \]

3. **Determine the Rate of Change of Current**:
\[ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{-20 \, A}{0.5 \, s} = -40 \, A/s \]

4. **Calculate the Magnetic Field (B) due to the Wire**:
The magnetic field at a distance \( d \) from a long straight wire carrying current \( I \) is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2I}{d} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \).

Substituting the values:
\[ B = \frac{(4\pi \times 10^{-7})}{4\pi} \cdot \frac{2 \times 20}{0.3} = \frac{10^{-7} \cdot 40}{0.3} = \frac{4 \times 10^{-6}}{0.3} = \frac{4 \times 10^{-6}}{0.3} \approx 1.33 \times 10^{-5} \, T \]

5. **Calculate the Induced EMF**:
According to Faraday's law of electromagnetic induction, the induced emf (\( \mathcal{E} \)) in the loop is given by:
\[ \mathcal{E} = -A \cdot \frac{dB}{dt} \]
Since \( B \) is changing due to the changing current, we can express \( \frac{dB}{dt} \) in terms of \( \frac{dI}{dt} \):
\[ \frac{dB}{dt} = \frac{\mu_0}{4\pi} \cdot \frac{2}{d} \cdot \frac{dI}{dt} \]
Substituting the values:
\[ \frac{dB}{dt} = \frac{(4\pi \times 10^{-7})}{4\pi} \cdot \frac{2}{0.3} \cdot (-40) = 10^{-7} \cdot \frac{2 \cdot (-40)}{0.3} \]
\[ = -\frac{8 \times 10^{-7}}{0.3} = -2.67 \times 10^{-6} \, T/s \]

Now substituting into the emf equation:
\[ \mathcal{E} = -0.5 \cdot (-2.67 \times 10^{-6}) = 1.33 \times 10^{-6} \, V \]

### Final Answer:
The average emf induced in the loop is approximately:
\[ \mathcal{E} \approx 1.33 \times 10^{-6} \, V \]