The magnetic field at a point midway between two parallel long wires carrying currents in the same direction is `10 mu T`. If the direction of the smaller current among them is reversed, the field becomes `30 mu T` . The ratio of the larger to the smaller current in them is

Options

A. `3 : 1`
B. `2 : 1`
C. `4 : 1`
D. `3 :2`

Explanation

To solve the problem, we need to analyze the magnetic fields produced by two parallel wires carrying currents in the same direction, and then how the situation changes when the direction of the smaller current is reversed.

### Step-by-Step Solution:

1. **Define the currents and magnetic field**:
- Let the currents in the two wires be \( I_1 \) (larger current) and \( I_2 \) (smaller current).
- The magnetic field at a point due to a long straight wire is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi d} \]
- Where \( \mu_0 \) is the permeability of free space, \( I \) is the current, and \( d \) is the distance from the wire to the point of interest.

2. **Magnetic field when currents are in the same direction**:
- The magnetic field at the midpoint between the two wires (let's call it point P) when both currents are in the same direction is:
\[ B = B_1 - B_2 = \frac{\mu_0 I_1}{2 \pi D} - \frac{\mu_0 I_2}{2 \pi D} \]
- Given that this magnetic field is \( 10 \, \mu T \):
\[ \frac{\mu_0}{2 \pi D} (I_1 - I_2) = 10 \, \mu T \quad \text{(1)} \]

3. **Magnetic field when the smaller current is reversed**:
- When \( I_2 \) is reversed, the magnetic field at point P becomes:
\[ B = B_1 + B_2 = \frac{\mu_0 I_1}{2 \pi D} + \frac{\mu_0 I_2}{2 \pi D} \]
- Given that this magnetic field is \( 30 \, \mu T \):
\[ \frac{\mu_0}{2 \pi D} (I_1 + I_2) = 30 \, \mu T \quad \text{(2)} \]

4. **Set up the equations**:
- From equation (1):
\[ I_1 - I_2 = \frac{10 \, \mu T \cdot 2 \pi D}{\mu_0} \]
- From equation (2):
\[ I_1 + I_2 = \frac{30 \, \mu T \cdot 2 \pi D}{\mu_0} \]

5. **Solve the equations**:
- Let \( x = \frac{2 \pi D}{\mu_0} \).
- Then we have:
\[ I_1 - I_2 = 10x \quad \text{(3)} \]
\[ I_1 + I_2 = 30x \quad \text{(4)} \]
- Add equations (3) and (4):
\[ (I_1 - I_2) + (I_1 + I_2) = 10x + 30x \]
\[ 2I_1 = 40x \implies I_1 = 20x \]
- Subtract equation (3) from (4):
\[ (I_1 + I_2) - (I_1 - I_2) = 30x - 10x \]
\[ 2I_2 = 20x \implies I_2 = 10x \]

6. **Find the ratio of the currents**:
- The ratio of the larger current to the smaller current is:
\[ \frac{I_1}{I_2} = \frac{20x}{10x} = 2 \]

### Final Answer:
The ratio of the larger current to the smaller current is \( 2:1 \).