A square coil of side `25cm` having `1000` turns is rotated with a uniform speed in a magnetic field about axis perpendicular to the direction of the field. At an instant `t`, the e.m.f. induced in the coil is `e=200 sin 100pi t`. The magnetic induction is

Options

A. 0.02 T
B. `10^(-3)T`
C. 0.1 T
D. 0.01 T

Answer

To solve the problem, we need to find the magnetic induction (B) given the induced EMF (e) in a square coil. Let's break it down step by step.

### Step 1: Understand the given information
- The square coil has a side length of \(25 \, \text{cm}\) (which we will convert to meters).
- The coil has \(1000\) turns.
- The induced EMF is given by the equation:
\[ e = 200 \sin(100 \pi t) \]

### Step 2: Convert the side length to meters
The side length of the square coil in meters is:
\[ \text{Side length} = 25 \, \text{cm} = 0.25 \, \text{m} \]

### Step 3: Calculate the area of the coil
The area \(A\) of the square coil is given by:
\[ A = \text{side}^2 = (0.25 \, \text{m})^2 = 0.0625 \, \text{m}^2 \]

### Step 4: Identify the angular frequency
From the given EMF equation:
\[ e = 200 \sin(100 \pi t) \]
we can identify that the angular frequency \(\omega\) is:
\[ \omega = 100 \pi \, \text{rad/s} \]

### Step 5: Use the formula for induced EMF
The induced EMF \(e\) in a coil can be expressed as:
\[ e = N \cdot A \cdot B \cdot \omega \cdot \sin(\omega t) \]
where:
- \(N\) = number of turns = \(1000\)
- \(A\) = area of the coil = \(0.0625 \, \text{m}^2\)
- \(B\) = magnetic induction (which we need to find)
- \(\omega\) = angular frequency = \(100 \pi \, \text{rad/s}\)

### Step 6: Set up the equation
From the induced EMF formula, we can equate:
\[ 200 = 1000 \cdot 0.0625 \cdot B \cdot (100 \pi) \]

### Step 7: Simplify the equation
Calculating the constants:
\[ 1000 \cdot 0.0625 = 62.5 \]
Thus, the equation becomes:
\[ 200 = 62.5 \cdot B \cdot (100 \pi) \]

### Step 8: Solve for B
Rearranging the equation to solve for \(B\):
\[ B = \frac{200}{62.5 \cdot (100 \pi)} \]
Calculating the denominator:
\[ 62.5 \cdot 100 \pi = 6250 \pi \]
Thus:
\[ B = \frac{200}{6250 \pi} \]

### Step 9: Calculate the value of B
Now we can calculate \(B\):
\[ B = \frac{200}{6250 \cdot 3.14} \approx \frac{200}{19625} \approx 0.0102 \, \text{T} \]

### Final Answer
The magnetic induction \(B\) is approximately:
\[ B \approx 0.01 \, \text{T} \]