A helium nucleus makes a full rotation in a circle of radius 0.8 metre in two seconds. The value of the magnetic field B at the centre of the circle will be

Options

A. `(10^(-19))/(mu_(0))`
B. `10^(-19)mu_(0)`
C. `2xx10^(-10)mu_(0)`
D. `(2xx10^(-10))/(mu_(0))`

Answer

To find the magnetic field \( B \) at the center of the circle due to a helium nucleus making a full rotation, we can follow these steps:

### Step 1: Identify the parameters
- Radius of the circle \( r = 0.8 \) m
- Time for one full rotation \( T = 2 \) s
- Charge of a helium nucleus \( q = 2 \times 1.6 \times 10^{-19} \) C (since a helium nucleus has 2 protons)

### Step 2: Calculate the current \( i \)
The current \( i \) can be calculated using the formula:
\[ i = \frac{q}{T} \]
Substituting the values:
\[ q = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \]
\[ i = \frac{3.2 \times 10^{-19} \, \text{C}}{2 \, \text{s}} = 1.6 \times 10^{-19} \, \text{A} \]

### Step 3: Use the formula for the magnetic field \( B \)
The magnetic field at the center of a circular loop carrying current is given by:
\[ B = \frac{\mu_0 i}{2r} \]
Where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

### Step 4: Substitute the values into the formula
Substituting the values of \( \mu_0 \), \( i \), and \( r \):
\[ B = \frac{4\pi \times 10^{-7} \, \text{T m/A} \times 1.6 \times 10^{-19} \, \text{A}}{2 \times 0.8 \, \text{m}} \]
\[ B = \frac{4\pi \times 10^{-7} \times 1.6 \times 10^{-19}}{1.6} \]
\[ B = 4\pi \times 10^{-7} \times 10^{-19} \, \text{T} \]
\[ B = 4\pi \times 10^{-26} \, \text{T} \]

### Step 5: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ B \approx 4 \times 3.14 \times 10^{-26} \, \text{T} \approx 12.56 \times 10^{-26} \, \text{T} \]
\[ B \approx 1.256 \times 10^{-25} \, \text{T} \]

### Final Answer
The value of the magnetic field \( B \) at the center of the circle is approximately \( 1.256 \times 10^{-25} \, \text{T} \).

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