The resistance of a circular coil of 50 turns and 10 cm diameter is `5Omega`. What must be the potential difference across the ends of the coil so as to nullify the earth's magnetic field (H=0.314 gauss) at the centre of the coil? How should the coil be placed to achieve this result?

Options

A. 0.5 V with plane of coil normal to the magnetic meridian
B. 0.5 V with plane of ocil in the magnetic meridian
C. 0.25 V with plane of coil normal to the magnetic meridian
D. 0.25 V with plane of coil in the magnetic meridian

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Given Information
We have a circular coil with:
- Number of turns (n) = 50
- Diameter = 10 cm, hence the radius (r) = 10 cm / 2 = 5 cm = 0.05 m
- Resistance (R) = 5 Ω
- Earth's magnetic field (H) = 0.314 gauss = 0.314 × 10^-4 T (since 1 gauss = 10^-4 T)

### Step 2: Use the Formula for Magnetic Field at the Center of the Coil
The magnetic field (B) at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2\pi n I}{R} \]
Where:
- \( \mu_0 \) (permeability of free space) = \( 4\pi \times 10^{-7} \, T \cdot m/A \)
- n = number of turns
- I = current through the coil
- R = radius of the coil

### Step 3: Rearranging the Formula to Solve for Current (I)
To nullify the Earth's magnetic field, the magnetic field produced by the coil must equal the Earth's magnetic field (H). Thus, we set B = H:
\[ H = \frac{\mu_0 n I}{2R} \]
Rearranging for I gives:
\[ I = \frac{2HR}{\mu_0 n} \]

### Step 4: Substitute the Values into the Equation
Substituting the known values:
- \( H = 0.314 \times 10^{-4} \, T \)
- \( R = 0.05 \, m \)
- \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \)
- \( n = 50 \)

Now substituting these values:
\[ I = \frac{2 \times (0.314 \times 10^{-4}) \times 0.05}{4\pi \times 10^{-7} \times 50} \]

### Step 5: Calculate the Current (I)
Calculating the numerator:
\[ 2 \times (0.314 \times 10^{-4}) \times 0.05 = 3.14 \times 10^{-6} \]

Calculating the denominator:
\[ 4\pi \times 10^{-7} \times 50 \approx 6.2832 \times 10^{-5} \]

Now, calculating I:
\[ I = \frac{3.14 \times 10^{-6}}{6.2832 \times 10^{-5}} \approx 0.050 \, A \]

### Step 6: Calculate the Potential Difference (V)
Using Ohm's law, \( V = I \times R \):
\[ V = 0.050 \, A \times 5 \, \Omega = 0.25 \, V \]

### Step 7: Determine the Placement of the Coil
To nullify the Earth's magnetic field, the plane of the coil should be placed normal (perpendicular) to the magnetic meridian.

### Final Answer
The potential difference required across the ends of the coil is **0.25 V**, and the coil should be placed **normal to the magnetic meridian**.

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