A square loop of side `22cm` is converted into circular loop in `0.4s`. A uniform magnetic field of `0.2T`directed normal to the loop, then the induced in the loop is

Options

A. `6.6xx10^(-3)V`
B. `6.6xx10^(-5)V`
C. `6.6xx10^(-4)V`
D. `6.6xx10^(-8)V`

Answer

To solve the problem step by step, we will calculate the induced electromotive force (emf) in the loop as it changes from a square shape to a circular shape in a uniform magnetic field.

### Step 1: Calculate the Area of the Square Loop
The side of the square loop is given as \(22 \, \text{cm}\).

\[ \text{Area of the square loop} = \text{side}^2 = (22 \, \text{cm})^2 = 484 \, \text{cm}^2 \]

### Step 2: Convert the Area to Square Meters
To convert the area from square centimeters to square meters, we use the conversion factor \(1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2\).

\[ \text{Area in m}^2 = 484 \, \text{cm}^2 \times 10^{-4} = 0.0484 \, \text{m}^2 \]

### Step 3: Calculate the Radius of the Circular Loop
Since the square loop is converted into a circular loop, the circumference of the square loop will equal the circumference of the circular loop.

\[ \text{Circumference of the square loop} = 4 \times \text{side} = 4 \times 22 \, \text{cm} = 88 \, \text{cm} \]

The circumference of the circular loop is given by:

\[ \text{Circumference of the circular loop} = 2 \pi r \]

Setting these equal gives:

\[ 88 \, \text{cm} = 2 \pi r \]

Solving for \(r\):

\[ r = \frac{88 \, \text{cm}}{2 \pi} = \frac{88 \times 10^{-2} \, \text{m}}{2 \pi} \approx \frac{0.88}{6.2832} \approx 0.140 \, \text{m} \]

### Step 4: Calculate the Area of the Circular Loop
Now we can calculate the area of the circular loop using the radius found:

\[ \text{Area of the circular loop} = \pi r^2 = \pi (0.140)^2 \approx 0.0616 \, \text{m}^2 \]

### Step 5: Calculate the Change in Magnetic Flux
The change in magnetic flux (\(\Delta \Phi\)) is given by:

\[ \Delta \Phi = B \cdot (\text{Area of circular loop} - \text{Area of square loop}) = 0.2 \, \text{T} \cdot (0.0616 \, \text{m}^2 - 0.0484 \, \text{m}^2) \]

Calculating the change in area:

\[ \Delta \Phi = 0.2 \cdot (0.0616 - 0.0484) = 0.2 \cdot 0.0132 = 0.00264 \, \text{Wb} \]

### Step 6: Calculate the Induced EMF
The induced emf (\( \mathcal{E} \)) can be calculated using Faraday's law of electromagnetic induction:

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

Given that \(\Delta t = 0.4 \, \text{s}\):

\[ \mathcal{E} = -\frac{0.00264 \, \text{Wb}}{0.4 \, \text{s}} = -0.0066 \, \text{V} = -6.6 \, \text{mV} \]

Thus, the induced emf in the loop is approximately \(6.6 \, \text{mV}\).

### Final Answer
The induced emf in the loop is \(6.6 \, \text{mV}\).

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