At the centre of a current - carrying circular coil of radius 5 cm, magnetic field due to earth is `0.5xx10^(-5)"Wb m"^(-2)`. The current flowing through the coil, so that it equals the Earth's magnetic field, is

Options

A. (a)40 A
B. (b)`4 A`
C. (c)0.4 A
D. (d)0.2 A

Answer

To solve the problem of finding the current flowing through a circular coil such that the magnetic field at its center equals the Earth's magnetic field, we can follow these steps:

### Step 1: Understand the formula for the magnetic field at the center of a circular coil
The magnetic field \( B \) at the center of a circular coil carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]
where:
- \( B \) is the magnetic field at the center of the coil,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( I \) is the current in amperes,
- \( R \) is the radius of the coil in meters.

### Step 2: Convert the radius from centimeters to meters
Given that the radius \( R \) of the coil is 5 cm, we convert this to meters:
\[ R = \frac{5 \, \text{cm}}{100} = 0.05 \, \text{m} \]

### Step 3: Set the magnetic field equal to the Earth's magnetic field
We know that the magnetic field due to the Earth is given as:
\[ B_E = 0.5 \times 10^{-5} \, \text{T} \]
To find the current \( I \) such that the magnetic field at the center of the coil equals the Earth's magnetic field, we set:
\[ \frac{\mu_0 I}{2R} = B_E \]

### Step 4: Substitute known values into the equation
Substituting the values we have:
\[ \frac{4\pi \times 10^{-7} \, I}{2 \times 0.05} = 0.5 \times 10^{-5} \]

### Step 5: Simplify the equation
First, simplify the left side:
\[ \frac{4\pi \times 10^{-7} \, I}{0.1} = 0.5 \times 10^{-5} \]
This simplifies to:
\[ 40\pi \times 10^{-7} \, I = 0.5 \times 10^{-5} \]

### Step 6: Solve for \( I \)
Rearranging the equation to solve for \( I \):
\[ I = \frac{0.5 \times 10^{-5}}{40\pi \times 10^{-7}} \]
Calculating the right side:
\[ I = \frac{0.5 \times 10^{-5}}{40 \times 3.14 \times 10^{-7}} \approx \frac{0.5 \times 10^{-5}}{1.256 \times 10^{-5}} \approx 0.398 \, \text{A} \]

### Step 7: Round the answer
Rounding this value gives us approximately:
\[ I \approx 0.4 \, \text{A} \]

### Conclusion
The current flowing through the coil so that it equals the Earth's magnetic field is approximately \( 0.4 \, \text{A} \).