A long solenoid contains 1000 turns/cm and an alternating current of peak value 1A is flowing in it. A search coil of area of cross-section `1 x× 10^(–4) m^(2)` and having 50 turns is placed inside the solenoid with its plane perpendicular to the axis of the solenoid. A peak voltage of `2pi^(2)xx10^(-2)V`V is produced in the search coil. The frequency of current in the solenoid will be –

Options

A. 50 Hz
B. 100 Hz
C. 500 Hz
D. 1000 Hz

Answer

To solve the problem, we need to find the frequency of the alternating current in the solenoid based on the given parameters. Let's break it down step by step.

### Step 1: Identify the given values
- Number of turns in the solenoid, \( N_s = 1000 \, \text{turns/cm} = 10^5 \, \text{turns/m} \)
- Peak current, \( I_{peak} = 1 \, \text{A} \)
- Area of the search coil, \( A = 1 \times 10^{-4} \, \text{m}^2 \)
- Number of turns in the search coil, \( N_c = 50 \)
- Peak voltage induced in the search coil, \( V_{peak} = 2\pi^2 \times 10^{-2} \, \text{V} \)

### Step 2: Calculate the magnetic field inside the solenoid
The magnetic field \( B \) inside a solenoid is given by:
\[ B = \mu_0 N_s I \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \) (permeability of free space).

Substituting the values:
\[ B = (4\pi \times 10^{-7}) \times (10^5) \times (1) = 4\pi \times 10^{-2} \, \text{T} \]

### Step 3: Calculate the magnetic flux \( \Phi \) through the search coil
The magnetic flux \( \Phi \) through the search coil is given by:
\[ \Phi = N_c B A \]
Substituting the values:
\[ \Phi = 50 \times (4\pi \times 10^{-2}) \times (1 \times 10^{-4}) = 50 \times 4\pi \times 10^{-6} \, \text{Wb} \]

### Step 4: Calculate the induced EMF \( \mathcal{E} \)
The induced EMF in the search coil can also be expressed using Faraday's law of electromagnetic induction:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
For an alternating current, this can be expressed as:
\[ \mathcal{E} = V_{peak} = N_c \frac{dB}{dt} A \]
The rate of change of magnetic field \( \frac{dB}{dt} \) can be related to the frequency \( f \):
\[ \frac{dB}{dt} = 2\pi f B \]
Thus,
\[ V_{peak} = N_c A (2\pi f B) \]

### Step 5: Rearranging to find frequency \( f \)
Rearranging the equation gives:
\[ f = \frac{V_{peak}}{2\pi N_c A B} \]

### Step 6: Substitute the known values
Substituting the known values:
\[ f = \frac{2\pi^2 \times 10^{-2}}{2\pi \times 50 \times (1 \times 10^{-4}) \times (4\pi \times 10^{-2})} \]
Simplifying:
\[ f = \frac{2\pi^2 \times 10^{-2}}{2\pi \times 50 \times 4\pi \times 10^{-6}} \]
\[ f = \frac{10^{-2}}{50 \times 4 \times 10^{-6}} = \frac{10^{-2}}{2 \times 10^{-4}} = 50 \, \text{Hz} \]

### Final Answer
The frequency of the current in the solenoid is \( f = 50 \, \text{Hz} \).

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