Two infinitely long straight parallel wires carry equal currents 'I' in the same direction and are 2 m apart. Magnetic induction at a point which is at same normal distance `(sqrt2) m` from each wire has magnitude

Options

A. `(mu_(0)i)/(2 pi)sqrt(2)`
B. `(mu_(0)i)/(4 pi)sqrt(2)`
C. `(mu_(0)i)/(2sqrt(2)pi)`
D. `(mu_(0)i)/(2 pi)`

Explanation

To solve the problem, we need to find the magnetic induction (magnetic field) at a point that is equidistant from two infinitely long parallel wires carrying equal currents in the same direction. Here’s a step-by-step solution:

### Step 1: Understand the setup
We have two parallel wires separated by a distance of 2 m, and we are interested in the magnetic field at a point that is at a distance of \( \sqrt{2} \, m \) from each wire.

### Step 2: Magnetic field due to a single 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.

### Step 3: Calculate the magnetic field from each wire
Since the point is at a distance of \( \sqrt{2} \, m \) from each wire, we can substitute \( r = \sqrt{2} \) into the formula:
\[ B_1 = \frac{\mu_0 I}{2 \pi \sqrt{2}} \]
for wire 1 and
\[ B_2 = \frac{\mu_0 I}{2 \pi \sqrt{2}} \]
for wire 2.

### Step 4: Determine the direction of the magnetic fields
Since both wires carry current in the same direction, the magnetic fields at the point of interest will be in opposite directions. We can use the right-hand rule to determine the direction of the magnetic fields.

### Step 5: Calculate the net magnetic field
The net magnetic field \( B_{net} \) at the point is the vector sum of \( B_1 \) and \( B_2 \). Since they are in opposite directions, we can write:
\[ B_{net} = B_1 - B_2 = \frac{\mu_0 I}{2 \pi \sqrt{2}} - \frac{\mu_0 I}{2 \pi \sqrt{2}} = 0 \]
However, this is incorrect because we need to consider the components of the magnetic fields.

### Step 6: Resolve the components
The angle \( \theta \) between the line connecting the wires and the line to the point is \( 45^\circ \) (since the point is at equal distances from both wires). The effective magnetic field from each wire at the point can be resolved into components. The vertical components will cancel out, and the horizontal components will add up.

The effective magnetic field from each wire along the line connecting the two wires is:
\[ B_{effective} = B \cos(45^\circ) = \frac{\mu_0 I}{2 \pi \sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{\mu_0 I}{4 \pi} \]

### Step 7: Calculate the total magnetic field
Since there are two wires contributing equally to the magnetic field in the same direction:
\[ B_{net} = 2 \cdot B_{effective} = 2 \cdot \frac{\mu_0 I}{4 \pi} = \frac{\mu_0 I}{2 \pi} \]

### Final Answer
Thus, the magnitude of the magnetic induction at the point is:
\[ B_{net} = \frac{\mu_0 I}{2 \pi} \]

### Conclusion
The correct option is Option 4: \( \frac{\mu_0 I}{2 \pi} \).

---