Two long parallel straight conductors carry current `i_(1)` and `i_(2)(i_(1)gti_(2))`. When the currents are in the same directions, the magnetic field at a point midway between the wires is `20muT`. If the direction of `i_(2)` is reversed, the field becomes `50muT`. The ratio of the current `i_(1)//i_(2)` is :

Options

A. `5//2`
B. `7//3`
C. `4//3`
D. `5//3`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the magnetic field due to two long parallel conductors
When two long parallel conductors carry currents \(I_1\) and \(I_2\), the magnetic field at a point midway between them can be calculated using the formula for the magnetic field due to a long straight conductor:
\[ B = \frac{\mu_0 I}{2 \pi d} \]
where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space, \(I\) is the current, and \(d\) is the distance from the wire to the point where the magnetic field is being measured.

### Step 2: Set up the equations for the two scenarios
1. **When the currents are in the same direction:**
The net magnetic field at point P (midway) is given as \(20 \, \mu T\):
\[ B = \frac{\mu_0 I_1}{2 \pi d} - \frac{\mu_0 I_2}{2 \pi d} = 20 \, \mu T \]
Simplifying this gives:
\[ \frac{\mu_0}{2 \pi d} (I_1 - I_2) = 20 \, \mu T \quad \text{(Equation 1)} \]

2. **When the current \(I_2\) is reversed:**
The net magnetic field at point P is now \(50 \, \mu T\):
\[ B = \frac{\mu_0 I_1}{2 \pi d} + \frac{\mu_0 I_2}{2 \pi d} = 50 \, \mu T \]
Simplifying this gives:
\[ \frac{\mu_0}{2 \pi d} (I_1 + I_2) = 50 \, \mu T \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
Now we have two equations:
1. \(\frac{\mu_0}{2 \pi d} (I_1 - I_2) = 20\)
2. \(\frac{\mu_0}{2 \pi d} (I_1 + I_2) = 50\)

Let \(k = \frac{\mu_0}{2 \pi d}\). Then we can rewrite the equations as:
1. \(k(I_1 - I_2) = 20\)
2. \(k(I_1 + I_2) = 50\)

### Step 4: Divide the two equations
Dividing Equation 1 by Equation 2:
\[ \frac{I_1 - I_2}{I_1 + I_2} = \frac{20}{50} = \frac{2}{5} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 5(I_1 - I_2) = 2(I_1 + I_2) \]
Expanding this:
\[ 5I_1 - 5I_2 = 2I_1 + 2I_2 \]
Rearranging terms:
\[ 5I_1 - 2I_1 = 5I_2 + 2I_2 \]
This simplifies to:
\[ 3I_1 = 7I_2 \]

### Step 6: Find the ratio of currents
Thus, the ratio of the currents \( \frac{I_1}{I_2} \) is:
\[ \frac{I_1}{I_2} = \frac{7}{3} \]

### Final Answer
The ratio of the currents \(I_1\) to \(I_2\) is \( \frac{7}{3} \).

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