A coil of cross-sectional area `5xx10^(-4)m^(2)` and having number of turns 1000 is placed prependicular to a magnetic field of `10^(-2)`T. The coil is connected to a galvanometer of resistance `500 Omega` . The induced charge generated in the coil on rotating it through an angle of `pi` radian will be

Options

A. `10 muC`
B. `20 muC`
C. `50 muC`
D. `100 muC`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the initial magnetic flux (Φ_i)
The initial magnetic flux (Φ_i) through the coil when it is perpendicular to the magnetic field can be calculated using the formula:
\[ \Phi_i = B \cdot A \cdot \cos(\theta) \]
Where:
- \(B = 10^{-2} \, T\) (magnetic field strength)
- \(A = 5 \times 10^{-4} \, m^2\) (cross-sectional area of the coil)
- \(\theta = 0\) (since the coil is perpendicular to the magnetic field)

Substituting the values:
\[ \Phi_i = (10^{-2}) \cdot (5 \times 10^{-4}) \cdot \cos(0) = (10^{-2}) \cdot (5 \times 10^{-4}) \cdot 1 = 5 \times 10^{-6} \, Wb \]

### Step 2: Calculate the final magnetic flux (Φ_f)
When the coil is rotated through an angle of \(\pi\) radians, the angle becomes 180 degrees, and the final magnetic flux (Φ_f) can be calculated as:
\[ \Phi_f = B \cdot A \cdot \cos(\theta) \]
Where:
- \(\theta = \pi\) (180 degrees)

Substituting the values:
\[ \Phi_f = (10^{-2}) \cdot (5 \times 10^{-4}) \cdot \cos(\pi) = (10^{-2}) \cdot (5 \times 10^{-4}) \cdot (-1) = -5 \times 10^{-6} \, Wb \]

### Step 3: Calculate the change in magnetic flux (ΔΦ)
The change in magnetic flux (ΔΦ) is given by:
\[ \Delta \Phi = \Phi_f - \Phi_i \]
Substituting the values:
\[ \Delta \Phi = (-5 \times 10^{-6}) - (5 \times 10^{-6}) = -10 \times 10^{-6} \, Wb = -1 \times 10^{-5} \, Wb \]

### Step 4: Calculate the induced EMF (ε)
The induced EMF (ε) can be calculated using Faraday's law of electromagnetic induction:
\[ \epsilon = -\frac{\Delta \Phi}{\Delta t} \]
Assuming the rotation happens in a time interval \(\Delta t\), we can express it as:
\[ \epsilon = \frac{1 \times 10^{-5}}{\Delta t} \]

### Step 5: Calculate the induced current (I)
Using Ohm's law, the induced current (I) can be calculated as:
\[ I = \frac{\epsilon}{R} \]
Where:
- \(R = 500 \, \Omega\) (resistance of the galvanometer)

Substituting the values:
\[ I = \frac{1 \times 10^{-5}/\Delta t}{500} = \frac{2 \times 10^{-8}}{\Delta t} \, A \]

### Step 6: Calculate the induced charge (Q)
The induced charge (Q) can be calculated using the formula:
\[ Q = I \cdot \Delta t \]
Substituting the expression for I:
\[ Q = \left(\frac{2 \times 10^{-8}}{\Delta t}\right) \cdot \Delta t = 2 \times 10^{-8} \, C \]

### Final Answer
The induced charge generated in the coil on rotating it through an angle of \(\pi\) radians is:
\[ Q = 2 \times 10^{-8} \, C \]

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