Two straight long conductors `AOB` and `COD` are perpendicular to each other and carry currents `I_(1)` and `I_(2)`. The magnitude of the magnetic induction at a point `P` at a distance `d` from the point `o` in a direction perpendicular to the plane `ABCD` is :

Options

A. `(mu_(0))/(2pid)(I_(1)+I_(2))`
B. `(mu_(0))/(4pid)(I_(1)+I_(2))`
C. `(mu_(0))/(2pid)sqrt(I_(2)^(2)+I_(2)^(2))`
D. `(mu_(0))/(2pid)((I_(1)I_(2))/(I_(1)+I_(2)))`

Explanation

To solve the problem of finding the magnitude of the magnetic induction at point P due to two perpendicular long conductors carrying currents \( I_1 \) and \( I_2 \), we can follow these steps:

### Step 1: Understand the Configuration
We have two long straight conductors, \( AOB \) and \( COD \), that are perpendicular to each other. The currents flowing through them are \( I_1 \) in conductor \( AOB \) and \( I_2 \) in conductor \( COD \). The point \( P \) is located at a distance \( d \) from the intersection point \( O \) of the two conductors.

### Step 2: Use Biot-Savart Law
According to the Biot-Savart Law, the magnetic field \( B \) due to a long straight conductor at a distance \( d \) from it is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi d} \]
where \( \mu_0 \) is the permeability of free space.

### Step 3: Calculate Magnetic Fields Due to Each Conductor
1. **Magnetic Field due to Conductor \( AOB \)**:
\[ B_{AOB} = \frac{\mu_0 I_1}{2 \pi d} \]
This magnetic field will be directed according to the right-hand rule.

2. **Magnetic Field due to Conductor \( COD \)**:
\[ B_{COD} = \frac{\mu_0 I_2}{2 \pi d} \]
This magnetic field will also be directed according to the right-hand rule.

### Step 4: Determine the Directions of the Magnetic Fields
- The direction of \( B_{AOB} \) will be out of the plane (let's assume it is in the positive \( x \)-direction).
- The direction of \( B_{COD} \) will be into the plane (let's assume it is in the negative \( y \)-direction).

### Step 5: Calculate the Resultant Magnetic Field
Since the two magnetic fields are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the resultant magnetic field \( B_{net} \):
\[ B_{net} = \sqrt{B_{AOB}^2 + B_{COD}^2} \]
Substituting the expressions for \( B_{AOB} \) and \( B_{COD} \):
\[ B_{net} = \sqrt{\left(\frac{\mu_0 I_1}{2 \pi d}\right)^2 + \left(\frac{\mu_0 I_2}{2 \pi d}\right)^2} \]

### Step 6: Factor Out Common Terms
Factoring out the common terms:
\[ B_{net} = \frac{\mu_0}{2 \pi d} \sqrt{I_1^2 + I_2^2} \]

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