A square loop of side 20 cm is placed on a plane and magnetic field intensity of 0.1 T is applied at an angle 30 with the plane of the loop. Magnetic field is changed to zero uniformly within 0.8 second. Calculate the emf induced in the loop.

Answer

To solve the problem of calculating the induced EMF in the square loop, we will follow these steps:

### Step 1: Calculate the area of the square loop
The side length of the square loop is given as 20 cm. We need to convert this into meters for standard SI units.

\[ L = 20 \, \text{cm} = 20 \times 10^{-2} \, \text{m} = 0.2 \, \text{m} \]

The area \( A \) of the square loop is given by:

\[ A = L^2 = (0.2 \, \text{m})^2 = 0.04 \, \text{m}^2 \]

### Step 2: Calculate the magnetic flux through the loop
The magnetic field intensity \( B \) is given as 0.1 T, and it is applied at an angle of 30 degrees with respect to the plane of the loop. The angle \( \theta \) between the magnetic field and the area vector of the loop is \( 90^\circ - 30^\circ = 60^\circ \).

The magnetic flux \( \Phi \) through the loop is given by the formula:

\[ \Phi = B \cdot A \cdot \cos(\theta) \]

Substituting the values:

\[ \Phi = 0.1 \, \text{T} \cdot 0.04 \, \text{m}^2 \cdot \cos(60^\circ) \]

Since \( \cos(60^\circ) = \frac{1}{2} \):

\[ \Phi = 0.1 \cdot 0.04 \cdot \frac{1}{2} = 0.002 \, \text{Wb} = 2 \times 10^{-3} \, \text{Wb} \]

### Step 3: Determine the change in magnetic flux
The magnetic field is changed to zero uniformly over a time interval of 0.8 seconds. Therefore, the final magnetic flux \( \Phi_2 \) will be:

\[ \Phi_2 = 0 \, \text{Wb} \]

The change in magnetic flux \( \Delta \Phi \) is given by:

\[ \Delta \Phi = \Phi_2 - \Phi_1 = 0 - 2 \times 10^{-3} = -2 \times 10^{-3} \, \text{Wb} \]

### Step 4: Calculate the induced EMF
The induced EMF \( \mathcal{E} \) can be calculated using Faraday's law of electromagnetic induction, which states:

\[ \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} \]

Substituting the values we have:

\[ \mathcal{E} = -\frac{-2 \times 10^{-3} \, \text{Wb}}{0.8 \, \text{s}} = \frac{2 \times 10^{-3}}{0.8} = 2.5 \times 10^{-3} \, \text{V} \]

### Final Answer
The induced EMF in the loop is:

\[ \mathcal{E} = 2.5 \times 10^{-3} \, \text{V} \, \text{or} \, 2.5 \, \text{mV} \]

---