A north pole of strengt` pi A m`, is moved around a circle of radius `10cm` which lies around a long straight conductor carrying a current of `10A`. The work doen is nearly

Options

A. `4mu J`
B. `40 mu J`
C. `400 muJ`
D. `0.4mu J`

Explanation

To solve the problem, we need to calculate the work done when a north pole of strength \( m = \pi \, \text{A m} \) is moved around a circle of radius \( r = 10 \, \text{cm} \) (or \( 0.1 \, \text{m} \)) around a long straight conductor carrying a current \( I = 10 \, \text{A} \).

### Step-by-step Solution:

1. **Identify the Parameters**:
- Pole strength \( m = \pi \, \text{A m} \)
- Radius of the circle \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Current \( I = 10 \, \text{A} \)

2. **Determine the Magnetic Field**:
The magnetic field \( B \) around a long straight conductor carrying current \( I \) at a distance \( r \) is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Substituting Values**:
Substitute \( \mu_0 \) and the known values into the equation for \( B \):
\[ B = \frac{4\pi \times 10^{-7} \times 10}{2 \pi \times 0.1} \]
Simplifying this gives:
\[ B = \frac{4 \times 10^{-6}}{0.2} = 2 \times 10^{-5} \, \text{T} \]

4. **Calculate the Work Done**:
The work done \( W \) when moving a magnetic pole in a magnetic field is given by:
\[ W = m \cdot B \cdot 2\pi r \]
where \( 2\pi r \) is the circumference of the circle.

5. **Substituting Values**:
Substitute \( m \), \( B \), and \( r \) into the equation:
\[ W = \pi \cdot (2 \times 10^{-5}) \cdot (2\pi \cdot 0.1) \]
Simplifying this:
\[ W = \pi \cdot (2 \times 10^{-5}) \cdot (0.2\pi) = 4\pi^2 \times 10^{-6} \]

6. **Calculating the Final Value**:
Using \( \pi \approx 3.14 \):
\[ W \approx 4 \cdot (3.14)^2 \cdot 10^{-6} \]
\[ W \approx 4 \cdot 9.8596 \cdot 10^{-6} \approx 39.4384 \times 10^{-6} \, \text{J} \]
Thus, the work done is approximately:
\[ W \approx 39.44 \, \mu J \approx 40 \, \mu J \]

### Final Answer:
The work done is nearly \( 40 \, \mu J \).