A bar magnet of magnetic moment `3.0 A-m^(2)` is placed in a uniform magnetic induction field of `2xx10^(-5)T`. If each pole of the magnet experiences a force of `6xx10^(-4)N`, the length of the magnet is

Explanation

To find the length of the bar magnet, we can use the relationship between the magnetic moment (M), the magnetic field (B), the force (F) experienced by each pole, and the length (l) of the magnet.

### Step-by-Step Solution:

1. **Understand the Relationship**: The magnetic moment (M) of a bar magnet is given by the formula:
\[ M = m \cdot l \]
where \( m \) is the magnetic pole strength and \( l \) is the length of the magnet.

2. **Force on Each Pole**: The force (F) experienced by each pole of the magnet in a magnetic field (B) is given by:
\[ F = m \cdot B \]
where \( m \) is the magnetic pole strength and \( B \) is the magnetic induction.

3. **Express Pole Strength**: From the force equation, we can express the pole strength \( m \) as:
\[ m = \frac{F}{B} \]

4. **Substitute into the Magnetic Moment Equation**: Now, substitute \( m \) back into the magnetic moment equation:
\[ M = \left(\frac{F}{B}\right) \cdot l \]

5. **Rearranging for Length**: Rearranging this equation to solve for the length \( l \):
\[ l = \frac{M \cdot B}{F} \]

6. **Substituting the Given Values**: Now substitute the given values into the equation:
- Magnetic moment \( M = 3.0 \, \text{A-m}^2 \)
- Magnetic field \( B = 2 \times 10^{-5} \, \text{T} \)
- Force \( F = 6 \times 10^{-4} \, \text{N} \)

Plugging in these values:
\[ l = \frac{3.0 \, \text{A-m}^2 \times 2 \times 10^{-5} \, \text{T}}{6 \times 10^{-4} \, \text{N}} \]

7. **Calculating the Length**:
\[ l = \frac{6.0 \times 10^{-5} \, \text{A-m}^2}{6 \times 10^{-4} \, \text{N}} = 0.1 \, \text{m} \]

### Final Answer:
The length of the magnet is \( 0.1 \, \text{m} \) or \( 10 \, \text{cm} \).

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