A short magnet oscillation in vibration magnetometer with a frequency 10Hz. A downward current of 15A is established in a long vertical wire placed 20cm to the West of the magnet. The new frequency of the short magnet is (the horizontal of the component os earth's magnetic field is `12mu`)

Options

A. 4Hz
B. 2.5Hz
C. 9Hz
D. 15Hz

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the magnetic field due to the current-carrying wire.
The formula for the magnetic field \( B \) at a distance \( a \) from a long straight current-carrying wire is given by:

\[ B = \frac{\mu_0 I}{2 \pi a} \]

Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)
- \( I = 15 \, \text{A} \) (current)
- \( a = 20 \, \text{cm} = 0.2 \, \text{m} \) (distance from the wire)

Substituting the values into the formula:

\[ B = \frac{(4\pi \times 10^{-7}) \times 15}{2 \pi \times 0.2} \]

### Step 2: Simplify the expression.
The \( \pi \) in the numerator and denominator cancels out:

\[ B = \frac{(4 \times 10^{-7}) \times 15}{2 \times 0.2} \]
\[ B = \frac{60 \times 10^{-7}}{0.4} \]
\[ B = 150 \times 10^{-7} \, \text{T} = 1.5 \times 10^{-5} \, \text{T} = 15 \, \mu T \]

### Step 3: Calculate the new magnetic field.
The total magnetic field \( B' \) acting on the magnet is the sum of the horizontal component of the Earth's magnetic field \( B_e \) and the magnetic field due to the wire \( B \):

Given that \( B_e = 12 \, \mu T \):

\[ B' = B_e + B = 12 \, \mu T + 15 \, \mu T = 27 \, \mu T \]

### Step 4: Relate the frequencies to the magnetic fields.
The frequency of oscillation \( f \) of a magnet in a magnetic field is proportional to the square root of the magnetic field:

\[ f \propto \sqrt{B} \]

Let \( f_0 = 10 \, \text{Hz} \) be the initial frequency and \( f' \) be the new frequency. The relationship can be expressed as:

\[ \frac{f'}{f_0} = \sqrt{\frac{B'}{B_e}} \]

### Step 5: Substitute the values.
Substituting the known values:

\[ \frac{f'}{10} = \sqrt{\frac{27 \, \mu T}{12 \, \mu T}} = \sqrt{\frac{27}{12}} = \sqrt{\frac{9}{4}} = \frac{3}{2} \]

### Step 6: Calculate the new frequency.
Now, we can find \( f' \):

\[ f' = 10 \times \frac{3}{2} = 15 \, \text{Hz} \]

### Final Answer:
The new frequency of the short magnet is \( 15 \, \text{Hz} \).

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