Two long wires are kept along `x` and `y` axis they carry currents `I&II` respectively in `+ve xx` and `+ve y` direction respectively.Find `vecB` at point `(0,0,d)`

Explanation

To find the magnetic field \( \vec{B} \) at the point \( (0, 0, d) \) due to two long wires carrying currents along the x-axis and y-axis, we can follow these steps:

### Step 1: Understand the Setup
We have two long straight wires:
- Wire 1 is along the x-axis and carries current \( I \) in the positive x-direction.
- Wire 2 is along the y-axis and carries current \( II \) in the positive y-direction.

We need to find the magnetic field at the point \( (0, 0, d) \).

### Step 2: Use the Biot-Savart Law
The magnetic field \( \vec{B} \) due to a long straight wire carrying current \( I \) at a distance \( r \) from the wire is given by the formula:
\[ \vec{B} = \frac{\mu_0 I}{2 \pi r} \hat{n} \]
where \( \mu_0 \) is the permeability of free space, and \( \hat{n} \) is the unit vector in the direction given by the right-hand rule.

### Step 3: Calculate the Magnetic Field from Wire 1
For Wire 1 (along the x-axis):
- The distance from the wire to the point \( (0, 0, d) \) is \( r = 0 \) (along the y-axis).
- The magnetic field \( \vec{B_1} \) at point \( (0, 0, d) \) will be directed in the negative z-direction (using the right-hand rule).

Thus, the magnetic field due to Wire 1 is:
\[ \vec{B_1} = \frac{\mu_0 I}{2 \pi d} (-\hat{j}) \]

### Step 4: Calculate the Magnetic Field from Wire 2
For Wire 2 (along the y-axis):
- The distance from the wire to the point \( (0, 0, d) \) is also \( r = 0 \) (along the x-axis).
- The magnetic field \( \vec{B_2} \) at point \( (0, 0, d) \) will be directed in the negative z-direction as well.

Thus, the magnetic field due to Wire 2 is:
\[ \vec{B_2} = \frac{\mu_0 II}{2 \pi d} (-\hat{i}) \]

### Step 5: Combine the Magnetic Fields
The total magnetic field \( \vec{B} \) at point \( (0, 0, d) \) is the vector sum of \( \vec{B_1} \) and \( \vec{B_2} \):
\[ \vec{B} = \vec{B_1} + \vec{B_2} \]
Substituting the values:
\[ \vec{B} = \frac{\mu_0 I}{2 \pi d} (-\hat{j}) + \frac{\mu_0 II}{2 \pi d} (-\hat{i}) \]
\[ \vec{B} = -\frac{\mu_0 I}{2 \pi d} \hat{j} - \frac{\mu_0 II}{2 \pi d} \hat{i} \]

### Final Result
Thus, the magnetic field \( \vec{B} \) at point \( (0, 0, d) \) is:
\[ \vec{B} = -\frac{\mu_0 I}{2 \pi d} \hat{i} - \frac{\mu_0 II}{2 \pi d} \hat{j} \]