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. 3
B. 4
C. 6
D. 2

Explanation

To solve the problem, we need to find the ratio of potential differences applied across two circular coils, given that the radius of the first coil (r1) is twice that of the second coil (r2).

### Step-by-Step Solution:

1. **Define the Radii**:
Let the radius of coil 2 be \( r \). Then, the radius of coil 1 will be \( r_1 = 2r \).

2. **Magnetic Field Formula**:
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 \( R \) is the radius of the coil.

3. **Magnetic Fields for Both Coils**:
For coil 1:
\[ B_1 = \frac{\mu_0 I_1}{2r_1} = \frac{\mu_0 I_1}{4r} \]
For coil 2:
\[ B_2 = \frac{\mu_0 I_2}{2r_2} = \frac{\mu_0 I_2}{2r} \]

4. **Setting the Magnetic Fields Equal**:
Since we want the magnetic fields at the centers of both coils to be equal:
\[ B_1 = B_2 \]
This gives us:
\[ \frac{\mu_0 I_1}{4r} = \frac{\mu_0 I_2}{2r} \]
Simplifying this, we can cancel \( \mu_0 \) and \( r \):
\[ \frac{I_1}{4} = \frac{I_2}{2} \]
Cross-multiplying leads to:
\[ I_1 = 2I_2 \]

5. **Resistance of the Coils**:
The resistance \( R \) of a wire is given by:
\[ R = \frac{\rho L}{A} \]
Since both coils are made from the same wire, the total length of the wire is the same. The length of wire for coil 1 is \( L_1 = 2\pi r_1 = 4\pi r \) and for coil 2 is \( L_2 = 2\pi r_2 = 2\pi r \).

6. **Resistance of Each Coil**:
The resistance for coil 1:
\[ R_1 = \frac{\rho (4\pi r)}{A} \]
The resistance for coil 2:
\[ R_2 = \frac{\rho (2\pi r)}{A} \]

7. **Applying Ohm's Law**:
The potential difference \( V \) across a coil is given by Ohm's law:
\[ V = IR \]
For coil 1:
\[ V_1 = I_1 R_1 = I_1 \left(\frac{\rho (4\pi r)}{A}\right) \]
For coil 2:
\[ V_2 = I_2 R_2 = I_2 \left(\frac{\rho (2\pi r)}{A}\right) \]

8. **Finding the Ratio of Potential Differences**:
Now, substituting \( I_1 = 2I_2 \) into the equations for \( V_1 \) and \( V_2 \):
\[ V_1 = (2I_2) \left(\frac{\rho (4\pi r)}{A}\right) = \frac{8\pi \rho r I_2}{A} \]
\[ V_2 = I_2 \left(\frac{\rho (2\pi r)}{A}\right) = \frac{2\pi \rho r I_2}{A} \]

Now, taking the ratio:
\[ \frac{V_1}{V_2} = \frac{\frac{8\pi \rho r I_2}{A}}{\frac{2\pi \rho r I_2}{A}} = \frac{8}{2} = 4 \]

### Final Answer:
The ratio of potential differences applied across the two coils is:
\[ \frac{V_1}{V_2} = 4:1 \]