A deflection magnetometer is placed with its arm along the east-west direction (tan A position) and a short bar magnet is placed symmetrically along its axis at some distance with its north pole pointing towards east. In this position the needle of the magnetometer shows a deflection of `60^(@)`. If we double the distance of the bar magnet, then the deflection will be

Options

A. `sin^(-1)[(sqrt3)/(8)]`
B. `cos^(-1)[(sqrt3)/(8)]`
C. `tan^(-1)[(sqrt3)/(8)]`
D. `cot^(-1)[(sqrt3)/(8)]`

Answer

To solve the problem, we need to analyze the relationship between the deflection angle of the magnetometer and the distance from the bar magnet. Here’s a step-by-step solution:

### Step 1: Understand the Initial Setup
We have a deflection magnetometer placed in the east-west direction with a short bar magnet positioned symmetrically along its axis. The north pole of the bar magnet is pointing towards the east, causing the magnetometer's needle to deflect by an angle of \(60^\circ\).

### Step 2: Determine the Magnetic Field due to the Bar Magnet
The magnetic field \(B\) at a distance \(d\) from a short bar magnet can be expressed as:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^3} \]
where \(m\) is the magnetic moment of the bar magnet, and \(\mu_0\) is the permeability of free space.

### Step 3: Relate the Magnetic Field to the Deflection Angle
According to the tangent law, the magnetic field \(B\) can also be related to the horizontal component of the Earth's magnetic field \(B_H\) and the deflection angle \(\theta\) as:
\[ B = B_H \tan(\theta) \]
For the initial condition where \(\theta = 60^\circ\):
\[ B = B_H \tan(60^\circ) = B_H \sqrt{3} \]

### Step 4: Set Up the Equation for Initial Conditions
From the above relationships, we can equate the two expressions for \(B\):
\[ \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^3} = B_H \sqrt{3} \]

### Step 5: Change the Distance and Analyze the New Deflection
Now, if we double the distance of the bar magnet, the new distance \(d_2 = 2d\). The magnetic field at this new distance will be:
\[ B_2 = \frac{\mu_0}{4\pi} \cdot \frac{2m}{(2d)^3} = \frac{\mu_0}{4\pi} \cdot \frac{2m}{8d^3} = \frac{1}{4} \cdot \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^3} \]
Thus, we have:
\[ B_2 = \frac{1}{4} B \]

### Step 6: Relate the New Magnetic Field to the New Deflection Angle
Using the tangent law again for the new deflection angle \(\theta_2\):
\[ B_2 = B_H \tan(\theta_2) \]
Substituting \(B_2\):
\[ \frac{1}{4} B_H \sqrt{3} = B_H \tan(\theta_2) \]
Dividing both sides by \(B_H\) (assuming \(B_H \neq 0\)):
\[ \frac{\sqrt{3}}{4} = \tan(\theta_2) \]

### Step 7: Find the New Deflection Angle
To find \(\theta_2\), we take the arctangent:
\[ \theta_2 = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \]

### Conclusion
Thus, the new deflection angle when the distance is doubled is:
\[ \theta_2 = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \]