A magnetic needle of pole strength `'m'` is privoted at its centre. Its `N-`pole is pulled eastward by a string. Then the horizontal force required to produce a deflection of `theta` from magnetic meridian `(B_(H) "horizontal componet of earths magnetic field")`

Options

A. `mB cos theta`
B. `mB sin theta`
C. `mB tan theta `
D. `mB cot theta`

Explanation

To solve the problem of finding the horizontal force required to produce a deflection of angle \( \theta \) from the magnetic meridian for a magnetic needle of pole strength \( m \), we can follow these steps:

### Step 1: Understand the Setup
We have a magnetic needle that is pivoted at its center. The north pole of the needle is being pulled eastward by a string. The magnetic needle will deflect from its original position in the magnetic meridian by an angle \( \theta \).

### Step 2: Identify Forces Acting on the Needle
1. The magnetic needle experiences a magnetic force due to the Earth's magnetic field, denoted as \( B_H \) (the horizontal component of the Earth's magnetic field).
2. An external force \( F \) is applied horizontally to pull the north pole of the needle eastward.

### Step 3: Calculate the Torque Due to Magnetic Force
The torque \( \tau_m \) due to the magnetic force acting on the north pole can be expressed as:
\[ \tau_m = m \cdot B_H \cdot L \cdot \sin(\theta) \]
Where:
- \( m \) = pole strength of the magnetic needle
- \( B_H \) = horizontal component of the Earth's magnetic field
- \( L \) = distance from the pivot point to the north pole of the needle

### Step 4: Calculate the Torque Due to External Force
The torque \( \tau_F \) due to the external force \( F \) is given by:
\[ \tau_F = F \cdot L \cdot \cos(\theta) \]
Where:
- \( F \) = external force applied
- \( L \) = distance from the pivot point to the north pole of the needle

### Step 5: Set Up the Torque Balance Equation
At equilibrium, the torques due to the magnetic force and the external force must be equal:
\[ \tau_m = \tau_F \]
Substituting the expressions for torque, we get:
\[ m \cdot B_H \cdot L \cdot \sin(\theta) = F \cdot L \cdot \cos(\theta) \]

### Step 6: Simplify the Equation
We can cancel \( L \) from both sides (assuming \( L \neq 0 \)):
\[ m \cdot B_H \cdot \sin(\theta) = F \cdot \cos(\theta) \]

### Step 7: Solve for the External Force \( F \)
Rearranging the equation to solve for \( F \):
\[ F = \frac{m \cdot B_H \cdot \sin(\theta)}{\cos(\theta)} \]
Using the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we can rewrite \( F \) as:
\[ F = m \cdot B_H \cdot \tan(\theta) \]

### Final Answer
Thus, the horizontal force required to produce a deflection of \( \theta \) from the magnetic meridian is:
\[ F = m \cdot B_H \cdot \tan(\theta) \]