A compass needle whose magnetic moment is `60Am^2` pointing geograhic north at a certain place, where the horizontal component of earth's magnetic field is `40muWbm^-2` experiences a torque `1*2xx10^-3Nm`. What is the declination of the place?

Options

A. `20^(@)`
B. `45^(@)`
C. `60^(@)`
D. `30^(@)`

Explanation

To solve the problem step by step, we need to find the declination of the place using the given values. Here’s how we can do it:

### Step 1: Understand the Torque Formula
The torque (\( \tau \)) experienced by a compass needle in a magnetic field is given by the formula:
\[ \tau = m \cdot H \cdot \sin(\theta) \]
where:
- \( m \) is the magnetic moment of the compass needle,
- \( H \) is the horizontal component of the Earth's magnetic field,
- \( \theta \) is the angle of declination.

### Step 2: Rearrange the Formula
We can rearrange the formula to solve for \( \sin(\theta) \):
\[ \sin(\theta) = \frac{\tau}{m \cdot H} \]

### Step 3: Substitute the Given Values
From the problem, we have:
- \( \tau = 1.2 \times 10^{-3} \, \text{Nm} \)
- \( m = 60 \, \text{Am}^2 \)
- \( H = 40 \, \mu\text{Wb/m}^2 = 40 \times 10^{-6} \, \text{Wb/m}^2 \)

Now we can substitute these values into the rearranged formula:
\[ \sin(\theta) = \frac{1.2 \times 10^{-3}}{60 \times 40 \times 10^{-6}} \]

### Step 4: Calculate the Denominator
First, calculate the denominator:
\[ 60 \times 40 \times 10^{-6} = 2400 \times 10^{-6} = 2.4 \times 10^{-3} \]

### Step 5: Calculate \( \sin(\theta) \)
Now substitute the denominator back into the equation:
\[ \sin(\theta) = \frac{1.2 \times 10^{-3}}{2.4 \times 10^{-3}} = \frac{1.2}{2.4} = \frac{1}{2} \]

### Step 6: Determine \( \theta \)
Since \( \sin(\theta) = \frac{1}{2} \), we know that:
\[ \theta = 30^\circ \]

### Conclusion
The declination of the place is \( 30^\circ \).

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