Due to straight current carrying conductor, a null point occurred at p on east of the conductor. The net magnetic induction at a point Q which is half the distance of 'p on north of the conductor is

Options

A. zero
B. `B_H`
C. `sqrt(2)B_H`
D. `sqrt(5) B_H `

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Setup
We have a straight current-carrying conductor, and a null point occurs at point P, which is located to the east of the conductor. The magnetic field at point P is balanced, meaning the magnetic field due to the conductor is equal to the magnetic field due to the Earth at that point.

### Step 2: Define the Magnetic Field at Point P
Let the magnetic induction at point P (which is the null point) be denoted as \( B_H \). This is the horizontal component of the magnetic field due to the Earth.

### Step 3: Determine the Position of Point Q
Point Q is located to the north of the conductor and is at half the distance of point P from the conductor. If we denote the distance from the conductor to point P as \( r \), then the distance from the conductor to point Q is \( \frac{r}{2} \).

### Step 4: Calculate the Magnetic Induction at Point Q
The magnetic induction due to a straight current-carrying conductor is inversely proportional to the distance from the conductor. Therefore, if the magnetic induction at point P is \( B_H \), the magnetic induction at point Q, denoted as \( B_{I'} \), can be calculated as follows:

\[ B_{I'} = \frac{B_H}{\frac{r}{2}} = 2B_H \]

### Step 5: Combine the Magnetic Fields
At point Q, we have two components of magnetic induction:
1. The magnetic induction due to the conductor, \( B_{I'} = 2B_H \) (acting towards the south).
2. The Earth's magnetic induction, \( B_H \) (acting towards the north).

To find the net magnetic induction at point Q, we will use the Pythagorean theorem since these two magnetic fields are perpendicular to each other:

\[ B_Q = \sqrt{(B_{I'})^2 + (B_H)^2} \]

Substituting the values we have:

\[ B_Q = \sqrt{(2B_H)^2 + (B_H)^2} \]
\[ B_Q = \sqrt{4B_H^2 + B_H^2} \]
\[ B_Q = \sqrt{5B_H^2} \]
\[ B_Q = \sqrt{5} B_H \]

### Conclusion
Thus, the net magnetic induction at point Q is \( \sqrt{5} B_H \).

### Final Answer
The correct option is \( \sqrt{5} B_H \).

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