Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that O is their common point for the two. The wires carry `l_(1)` and `l_(2)` currents, respectively. Point P is lying at distance d from 0 along a direction perpendicular to the plane containing the wire. The magnetic field at the point P will be

Options

A. `(mu_(0))/(2pid)((I_(1))/(I_(2)))`
B. `(mu_(0))/(2pid)(I_(1)+I_(2))`
C. `(mu_(0))/(2pid)(I_(1)^(2)+I_(2)^(2))`
D. `(mu_(0))/(2pid)(I_(1)^(2)+I_(2)^(2))^(1//2)`

Answer

To find the magnetic field at point P due to the two perpendicular conducting wires carrying currents \( I_1 \) and \( I_2 \), we can follow these steps:

### Step 1: Understand the Configuration
We have two long, straight, identical conducting wires AOB and COD placed at right angles to each other. The currents in the wires are \( I_1 \) (in wire AOB) and \( I_2 \) (in wire COD). The point P is located at a distance \( d \) from the common point O, perpendicular to the plane formed by the wires.

### Step 2: Magnetic Field Due to a Long Straight 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 at Point P
1. **Magnetic Field due to Wire AOB (current \( I_1 \)):**
- The distance from wire AOB to point P is \( d \).
- The direction of the magnetic field due to wire AOB at point P is given by the right-hand rule. If the current \( I_1 \) is flowing upwards, the magnetic field at point P will be in the positive y-direction (J cap).
- Therefore, the magnetic field \( B_1 \) at point P due to wire AOB is:
\[ B_1 = \frac{\mu_0 I_1}{2\pi d} \hat{j} \]

2. **Magnetic Field due to Wire COD (current \( I_2 \)):**
- The distance from wire COD to point P is also \( d \).
- If the current \( I_2 \) is flowing towards the observer (out of the plane), the magnetic field at point P will be in the negative x-direction (−I cap).
- Therefore, the magnetic field \( B_2 \) at point P due to wire COD is:
\[ B_2 = -\frac{\mu_0 I_2}{2\pi d} \hat{i} \]

### Step 4: Find the Net Magnetic Field
The net magnetic field \( B_{net} \) at point P is the vector sum of \( B_1 \) and \( B_2 \):
\[ B_{net} = B_1 + B_2 = \frac{\mu_0 I_1}{2\pi d} \hat{j} - \frac{\mu_0 I_2}{2\pi d} \hat{i} \]

### Step 5: Calculate the Magnitude of the Net Magnetic Field
To find the magnitude of the net magnetic field \( |B_{net}| \):
\[ |B_{net}| = \sqrt{ \left( \frac{\mu_0 I_1}{2\pi d} \right)^2 + \left( -\frac{\mu_0 I_2}{2\pi d} \right)^2 } \]
\[ = \frac{\mu_0}{2\pi d} \sqrt{ I_1^2 + I_2^2 } \]

### Final Answer
Thus, the magnetic field at point P is:
\[ |B_{net}| = \frac{\mu_0}{2\pi d} \sqrt{ I_1^2 + I_2^2 } \]