A particle carrying a charge equal to `100` times the charge on an electron is rotating per second in a circular path of radius `0.8 metre`. The value of the magnetic field produced at the centre will be (`mu_(0)=` permeability for vacuum)

Options

A. `(10^(-7))/(mu_(0))`
B. `10^(-17)mu_(0)`
C. `10^(-6)mu_(0)`
D. `10^(-7)mu_(0)`

Explanation

To solve the problem step by step, we will calculate the magnetic field produced at the center of a circular path by a charged particle.

### Step 1: Determine the charge of the particle
The charge \( Q \) of the particle is given as 100 times the charge of an electron. The charge of an electron \( e \) is approximately \( 1.6 \times 10^{-19} \) coulombs. Therefore, the charge of the particle can be calculated as:
\[ Q = 100 \times e = 100 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-17} \text{ coulombs} \]

### Step 2: Calculate the current \( I \)
The particle is rotating in a circular path, which means it creates a current. The current \( I \) can be calculated using the formula:
\[ I = \frac{Q}{T} \]
where \( T \) is the time period. Since the particle is rotating once per second, \( T = 1 \) second. Thus,
\[ I = \frac{1.6 \times 10^{-17}}{1} = 1.6 \times 10^{-17} \text{ amperes} \]

### Step 3: Use the formula for the magnetic field \( B \)
The magnetic field \( B \) at the center of a circular loop carrying current can be calculated using the formula:
\[ B = \frac{\mu_0 I}{2\pi r} \]
where \( \mu_0 \) is the permeability of free space (approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \)) and \( r \) is the radius of the circular path. Given that \( r = 0.8 \) meters, we can substitute the values:
\[ B = \frac{(4\pi \times 10^{-7}) \times (1.6 \times 10^{-17})}{2\pi \times 0.8} \]

### Step 4: Simplify the expression
Now, we simplify the expression:
\[ B = \frac{4 \times 10^{-7} \times 1.6 \times 10^{-17}}{2 \times 0.8} \]
Calculating the denominator:
\[ 2 \times 0.8 = 1.6 \]
Now substituting back:
\[ B = \frac{4 \times 10^{-7} \times 1.6 \times 10^{-17}}{1.6} \]
The \( 1.6 \) cancels out:
\[ B = 4 \times 10^{-7} \times 10^{-17} = 4 \times 10^{-24} \text{ T} \]

### Conclusion
Thus, the magnetic field produced at the center of the circular path is:
\[ B = 4 \times 10^{-24} \text{ T} \]