A current carrying circular coil of radius 11 cm is placed with its plane in magnetic meridian. The resultant magnetic induction at the centre of the coil is (Given current in the coil is 7A and `B_(H) = 0.4 `Gauss ]

Options

A. `3 xx 10^(-5) T`
B. `4 xx 10^(-5) T`
C. `4 sqrt( 2) xx 10^(-5) T `
D. `0`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Radius of the coil, \( R = 11 \, \text{cm} = 11 \times 10^{-2} \, \text{m} \)
- Current in the coil, \( I = 7 \, \text{A} \)
- Horizontal magnetic induction, \( B_H = 0.4 \, \text{Gauss} = 0.4 \times 10^{-4} \, \text{Tesla} = 4 \times 10^{-5} \, \text{Tesla} \)

### Step 2: Calculate the magnetic field due to the coil at its center
The formula for the magnetic field \( B \) at the center of a circular coil is given by:
\[ B_2 = \frac{\mu_0}{4\pi} \cdot \frac{I}{R} \]
Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)

Substituting the values:
\[ B_2 = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{7}{11 \times 10^{-2}} \]
The \( 4\pi \) cancels out:
\[ B_2 = 10^{-7} \cdot \frac{7}{11 \times 10^{-2}} = \frac{7 \times 10^{-5}}{11} \]
Calculating \( B_2 \):
\[ B_2 \approx 6.36 \times 10^{-5} \, \text{Tesla} \]

### Step 3: Calculate the resultant magnetic induction at the center of the coil
The resultant magnetic induction \( B_{net} \) is given by:
\[ B_{net} = \sqrt{B_1^2 + B_2^2} \]
Where:
- \( B_1 = B_H = 4 \times 10^{-5} \, \text{Tesla} \)
- \( B_2 \approx 6.36 \times 10^{-5} \, \text{Tesla} \)

Substituting the values:
\[ B_{net} = \sqrt{(4 \times 10^{-5})^2 + (6.36 \times 10^{-5})^2} \]
Calculating \( B_{net} \):
\[ B_{net} = \sqrt{(16 \times 10^{-10}) + (40.45 \times 10^{-10})} \]
\[ B_{net} = \sqrt{56.45 \times 10^{-10}} \approx 7.51 \times 10^{-5} \, \text{Tesla} \]

### Step 4: Final Result
The resultant magnetic induction at the center of the coil is approximately \( 7.51 \times 10^{-5} \, \text{Tesla} \).

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