Magnetic field induction at centre of circular coil of radius 5 cm and carrying a current 0.9 A is (in S.I. units) (`in_(0)`=absolute permittivity of air S.I. units, velocity of light `= 3xx10^(8)`m/s)

Options

A. `1/(in_(0)10^(16))`
B. `(10^(16))/(in_(0))`
C. `(in_(0))/(10^(16))`
D. `10^(16)in_(0)`

Answer

To find the magnetic field induction (B) at the center of a circular coil of radius 5 cm carrying a current of 0.9 A, we can follow these steps:

### Step 1: Convert the radius to meters
The radius given is 5 cm. We need to convert this to meters for our calculations.
\[ r = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \]

### Step 2: Use the formula for magnetic field at the center of a circular coil
The formula for the magnetic field induction at the center of a circular coil is given by:
\[ B = \frac{\mu_0 I}{2r} \]
where:
- \(B\) is the magnetic field induction,
- \(\mu_0\) is the permeability of free space,
- \(I\) is the current in amperes,
- \(r\) is the radius of the coil in meters.

### Step 3: Determine the value of \(\mu_0\)
We know that:
\[ \mu_0 = \frac{1}{\epsilon_0 c^2} \]
where:
- \(\epsilon_0\) is the permittivity of free space,
- \(c\) is the speed of light in vacuum, given as \(3 \times 10^8 \, \text{m/s}\).

### Step 4: Substitute the values into the formula
We need the value of \(\epsilon_0\) which is approximately \(8.85 \times 10^{-12} \, \text{F/m}\). Now we can calculate \(\mu_0\):
\[ \mu_0 = \frac{1}{(8.85 \times 10^{-12}) \times (3 \times 10^8)^2} \]
Calculating \(c^2\):
\[ c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \]
Now substituting this back:
\[ \mu_0 = \frac{1}{(8.85 \times 10^{-12}) \times (9 \times 10^{16})} \]
Calculating \(\mu_0\):
\[ \mu_0 \approx 4\pi \times 10^{-7} \, \text{T m/A} \]

### Step 5: Substitute all known values into the magnetic field formula
Now substituting \(I = 0.9 \, \text{A}\), \(r = 5 \times 10^{-2} \, \text{m}\), and \(\mu_0\):
\[ B = \frac{(4\pi \times 10^{-7}) \times 0.9}{2 \times (5 \times 10^{-2})} \]

### Step 6: Calculate the magnetic field
Calculating the above expression:
\[ B = \frac{(4\pi \times 10^{-7} \times 0.9)}{(1 \times 10^{-1})} \]
\[ B = (4\pi \times 10^{-6} \times 0.9) \]
\[ B \approx 3.6 \times 10^{-6} \, \text{T} \]

### Final Answer
Thus, the magnetic field induction at the center of the circular coil is approximately:
\[ B \approx 3.6 \times 10^{-6} \, \text{T} \]