Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What is the ratio of potentail difference applied across them so that the magnetic field at their centres is the same?

Options

A. a. 3
B. b. 4
C. c. 6
D. d. 2

Answer

To solve the problem, we need to find the ratio of the potential differences applied across two circular coils (Coil 1 and Coil 2) such that the magnetic field at their centers is the same. Given that the radius of Coil 1 (R1) is twice that of Coil 2 (R2), we can denote R1 = 2R and R2 = R.

### Step-by-Step Solution:

1. **Understanding the Magnetic Field Formula**:
The magnetic field (B) at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0 n I}{2R} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns per unit length,
- \( I \) is the current through the coil,
- \( R \) is the radius of the coil.

2. **Magnetic Field for Both Coils**:
For Coil 1 (with radius R1):
\[ B_1 = \frac{\mu_0 n_1 I_1}{2R_1} \]
For Coil 2 (with radius R2):
\[ B_2 = \frac{\mu_0 n_2 I_2}{2R_2} \]

3. **Setting the Magnetic Fields Equal**:
Since we want the magnetic fields at the centers of both coils to be equal:
\[ B_1 = B_2 \]
Therefore:
\[ \frac{\mu_0 n_1 I_1}{2R_1} = \frac{\mu_0 n_2 I_2}{2R_2} \]

4. **Substituting the Radius Relationships**:
Given that \( R_1 = 2R_2 \) (or \( R_1 = 2R \) and \( R_2 = R \)):
\[ \frac{n_1 I_1}{2(2R)} = \frac{n_2 I_2}{2R} \]
Simplifying this gives:
\[ \frac{n_1 I_1}{4R} = \frac{n_2 I_2}{2R} \]
Cancelling \( R \) from both sides:
\[ \frac{n_1 I_1}{4} = \frac{n_2 I_2}{2} \]
Rearranging gives:
\[ n_1 I_1 = 2n_2 I_2 \]

5. **Finding the Current Ratio**:
If we assume that both coils are made from the same wire, then the number of turns \( n \) is related to the length of the wire and the radius of the coil. The length of the wire for each coil can be expressed as:
\[ L_1 = 2\pi R_1 = 2\pi (2R) = 4\pi R \]
\[ L_2 = 2\pi R_2 = 2\pi R \]
The number of turns \( n \) is inversely proportional to the radius, thus:
\[ n_1 = \frac{L}{4\pi R} \quad \text{and} \quad n_2 = \frac{L}{2\pi R} \]
Therefore, the ratio of the number of turns is:
\[ \frac{n_1}{n_2} = \frac{1/4}{1/2} = \frac{1}{2} \]

6. **Substituting into the Current Ratio**:
From \( n_1 I_1 = 2n_2 I_2 \):
\[ \frac{1}{2} I_1 = 2 \cdot \frac{L}{2\pi R} I_2 \]
Rearranging gives:
\[ I_1 = 4 I_2 \]

7. **Finding the Potential Difference Ratio**:
The potential difference \( V \) across each coil is given by:
\[ V = I \cdot R \]
Thus:
\[ V_1 = I_1 \cdot R_1 \quad \text{and} \quad V_2 = I_2 \cdot R_2 \]
Substituting the values:
\[ V_1 = (4 I_2) \cdot (2R) = 8 I_2 R \]
\[ V_2 = I_2 \cdot R \]
Therefore, the ratio of potential differences is:
\[ \frac{V_1}{V_2} = \frac{8 I_2 R}{I_2 R} = 8 \]

### Conclusion:
The ratio of the potential differences applied across the two coils is:
\[ \frac{V_1}{V_2} = 8 \]