In an atom the electron has a time period of `0.16 xx 10^( -15)`s in a circular orbit of radius 0.5 `A^(0)`. The magnetic induction at the centre of the orbit will be (in tesla)

Options

A. 12.56
B. 125.6
C. 1.256
D. 25.12

Answer

To solve the problem of finding the magnetic induction at the center of the orbit of an electron in an atom, we can follow these steps:

### Step 1: Identify the given values
- Time period (T) = \(0.16 \times 10^{-15}\) seconds
- Radius (r) = \(0.5 \, \text{Å} = 0.5 \times 10^{-10} \, \text{m}\)
- Charge of electron (e) = \(1.6 \times 10^{-19} \, \text{C}\)

### Step 2: Calculate the current (I) produced by the moving electron
The current due to a moving charge can be calculated using the formula:
\[ I = \frac{e}{T} \]
Substituting the values:
\[ I = \frac{1.6 \times 10^{-19} \, \text{C}}{0.16 \times 10^{-15} \, \text{s}} = \frac{1.6}{0.16} \times 10^{-4} = 10 \times 10^{-4} = 1.0 \times 10^{-3} \, \text{A} \]

### Step 3: Use the formula for magnetic induction (B) at the center of the orbit
The magnetic induction at the center of the circular orbit can be calculated using the formula:
\[ B = \frac{\mu_0 I}{2r} \]
Where \(\mu_0\) (the permeability of free space) is given by:
\[ \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \]
Substituting the values into the equation:
\[ B = \frac{(4\pi \times 10^{-7}) (1.0 \times 10^{-3})}{2(0.5 \times 10^{-10})} \]

### Step 4: Simplify the expression
Calculating the denominator:
\[ 2(0.5 \times 10^{-10}) = 1.0 \times 10^{-10} \]
Thus,
\[ B = \frac{(4\pi \times 10^{-7}) (1.0 \times 10^{-3})}{1.0 \times 10^{-10}} = (4\pi \times 10^{3}) \, \text{T} \]

### Step 5: Calculate the value of B
Using \(\pi \approx 3.14\):
\[ B \approx 4 \times 3.14 \times 10^{3} = 12.56 \, \text{T} \]

### Conclusion
The magnetic induction at the center of the orbit is approximately \(12.56 \, \text{T}\).

### Final Answer
The magnetic induction at the center of the orbit will be \(12.56 \, \text{T}\).

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