A "bar" magnet of length `16 cm` has a pole strength of `500` milli amp.`m`. The angle at which it should be placed to the direction of external magnetic field of induction `2.5` gauses so that it may experience a torque of `sqrt(3)xx10^(-5)Nm` is

Options

A. `pi`
B. `pi//2`
C. `pi//3`
D. `pi//6`

Explanation

To solve the problem, we need to find the angle at which a bar magnet experiences a specified torque when placed in an external magnetic field. We will follow these steps:

### Step 1: Understand the given parameters
- Length of the bar magnet, \( l = 16 \, \text{cm} = 0.16 \, \text{m} \)
- Pole strength, \( p = 500 \, \text{mA} \cdot \text{m} = 500 \times 10^{-3} \, \text{A} \cdot \text{m} \)
- Magnetic field induction, \( B = 2.5 \, \text{Gauss} = 2.5 \times 10^{-4} \, \text{T} \)
- Torque, \( \tau = \sqrt{3} \times 10^{-5} \, \text{Nm} \)

### Step 2: Calculate the magnetic moment \( M \)
The magnetic moment \( M \) of a bar magnet is given by the formula:
\[ M = p \cdot l \]
Substituting the values:
\[ M = (500 \times 10^{-3} \, \text{A} \cdot \text{m}) \cdot (0.16 \, \text{m}) = 80 \times 10^{-5} \, \text{A} \cdot \text{m}^2 = 0.08 \, \text{A} \cdot \text{m}^2 \]

### Step 3: Relate torque to magnetic moment and angle
The torque \( \tau \) experienced by a magnetic dipole in a magnetic field is given by:
\[ \tau = M \cdot B \cdot \sin(\theta) \]
Where \( \theta \) is the angle between the magnetic moment and the magnetic field.

### Step 4: Substitute the known values into the torque equation
From the torque equation, we can rearrange it to find \( \sin(\theta) \):
\[ \sin(\theta) = \frac{\tau}{M \cdot B} \]
Substituting the known values:
\[ \sin(\theta) = \frac{\sqrt{3} \times 10^{-5}}{(0.08) \cdot (2.5 \times 10^{-4})} \]

### Step 5: Calculate the denominator
Calculating \( M \cdot B \):
\[ M \cdot B = 0.08 \cdot 2.5 \times 10^{-4} = 0.02 \times 10^{-4} = 2 \times 10^{-6} \]

### Step 6: Calculate \( \sin(\theta) \)
Now substituting back into the equation for \( \sin(\theta) \):
\[ \sin(\theta) = \frac{\sqrt{3} \times 10^{-5}}{2 \times 10^{-6}} = \frac{\sqrt{3}}{2} = 0.866 \]

### Step 7: Find the angle \( \theta \)
To find \( \theta \), we take the inverse sine:
\[ \theta = \sin^{-1}(0.866) = 60^\circ \]

### Final Answer
The angle at which the bar magnet should be placed to experience the specified torque is \( \theta = 60^\circ \).

---