Two wires are held perpendicular to the plane of paper and are 5 m apart. They carry currents of 2.5 A and 5 A in same direction. Then, the magnetic field strength (B) at a point midway between the wires will be

Options

A. `(mu_(0))/(4pi)T`
B. `(mu_(0))/(2pi)T`
C. `(3mu_(0))/(2pi)T`
D. `(3mu_(0))/(4pi)T`

Answer

To find the magnetic field strength (B) at a point midway between two wires carrying currents in the same direction, we can follow these steps:

### Step 1: Identify the setup
We have two parallel wires that are 5 m apart. The first wire carries a current of \(I_1 = 2.5 \, \text{A}\) and the second wire carries a current of \(I_2 = 5 \, \text{A}\). The point of interest is located midway between the two wires, which means it is 2.5 m from each wire.

### Step 2: Use the formula for the magnetic field due to a long straight current-carrying wire
The magnetic field \(B\) at a distance \(r\) from a long straight wire carrying current \(I\) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \(\mu_0\) is the permeability of free space (\(\mu_0 \approx 4\pi \times 10^{-7} \, \text{T m/A}\)).

### Step 3: Calculate the magnetic field due to the first wire
For the first wire (2.5 A):
- Distance from the point to the wire, \(r = 2.5 \, \text{m}\)
\[ B_1 = \frac{\mu_0 \cdot 2.5}{2 \pi \cdot 2.5} = \frac{\mu_0}{4 \pi} \]

### Step 4: Calculate the magnetic field due to the second wire
For the second wire (5 A):
- Distance from the point to the wire, \(r = 2.5 \, \text{m}\)
\[ B_2 = \frac{\mu_0 \cdot 5}{2 \pi \cdot 2.5} = \frac{\mu_0}{\pi} \]

### Step 5: Determine the direction of the magnetic fields
Using the right-hand rule:
- For the first wire (2.5 A), the magnetic field at the midpoint will be directed into the plane of the paper (let's denote this as negative).
- For the second wire (5 A), the magnetic field at the midpoint will be directed out of the plane of the paper (let's denote this as positive).

### Step 6: Calculate the net magnetic field
Since the magnetic fields are in opposite directions, we can find the net magnetic field by subtracting the smaller from the larger:
\[ B_{\text{net}} = B_2 - B_1 = \frac{\mu_0}{\pi} - \frac{\mu_0}{4\pi} = \frac{4\mu_0}{4\pi} - \frac{\mu_0}{4\pi} = \frac{3\mu_0}{4\pi} \]

### Step 7: Substitute \(\mu_0\) to find the numerical value
Substituting \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\):
\[ B_{\text{net}} = \frac{3(4\pi \times 10^{-7})}{4\pi} = 3 \times 10^{-7} \, \text{T} = 3 \times 10^{-7} \, \text{T} \]

### Final Answer
The magnetic field strength at the point midway between the wires is:
\[ B_{\text{net}} = 3 \times 10^{-7} \, \text{T} \]

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