A very long wire carrying a current `I = 5.0 A` is bent at right angles. Find the magnetic induction at a point lying on a perpendicular normal to the plane of the wire drawn through the Point of bending at a distance `l = 35 cm` from it.

Explanation

To solve the problem of finding the magnetic induction at a point lying on a perpendicular normal to the plane of the wire drawn through the point of bending at a distance \( l = 35 \, \text{cm} \) from it, we can follow these steps:

### Step 1: Understand the Configuration
We have a long straight wire that is bent at right angles. The magnetic field due to a straight wire carrying current can be calculated using the Biot-Savart law or Ampere's law. The point of interest is located at a distance \( l \) from the bending point of the wire.

### Step 2: Magnetic Field due to a 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, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

### Step 3: Calculate the Magnetic Field from Each Segment
Since the wire is bent at right angles, we can treat the two segments of the wire separately.

1. **For the first segment (straight wire)**:
- The distance from the point of bending to the point of interest is \( l = 35 \, \text{cm} = 0.35 \, \text{m} \).
- The magnetic field due to this segment at the point is:
\[ B_1 = \frac{\mu_0 I}{2\pi l} = \frac{(4\pi \times 10^{-7}) \times 5.0}{2\pi \times 0.35} \]
Simplifying this gives:
\[ B_1 = \frac{(4 \times 10^{-7}) \times 5.0}{2 \times 0.35} = \frac{20 \times 10^{-7}}{0.7} = \frac{20 \times 10^{-7}}{7} \approx 2.857 \times 10^{-7} \, \text{T} \]

2. **For the second segment (perpendicular wire)**:
- The magnetic field due to the second segment at the point of interest is zero because the distance is effectively zero (the point lies on the perpendicular bisector of the segment).

### Step 4: Combine the Magnetic Fields
Since the magnetic field from the second segment is zero, the total magnetic induction at the point is simply the magnetic field from the first segment:
\[ B_{\text{total}} = B_1 + B_2 = B_1 + 0 = B_1 \approx 2.857 \times 10^{-7} \, \text{T} \]

### Final Answer
The magnetic induction at the point is approximately:
\[ B \approx 2.857 \times 10^{-7} \, \text{T} \]