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

Options

A. `0.5 m`
B. `0.3 m`
C. `0.2 m`
D. `0.1 m`

Explanation

To find the length of the bar magnet, we can follow these steps:

### Step 1: Understand the relationship between force, magnetic moment, and magnetic field.
The force experienced by each pole of the magnet in a magnetic field is given by the formula:
\[ F = m \cdot B \]
where:
- \( F \) is the force on each pole,
- \( m \) is the pole strength,
- \( B \) is the magnetic induction (magnetic field).

### Step 2: Relate magnetic moment to pole strength and length.
The magnetic moment \( M \) of the magnet is related to the pole strength \( m \) and the length \( l \) of the magnet by the formula:
\[ M = m \cdot l \]
From this, we can express pole strength \( m \) as:
\[ m = \frac{M}{l} \]

### Step 3: Substitute the expression for pole strength into the force equation.
Substituting \( m \) from the previous step into the force equation gives:
\[ F = \left(\frac{M}{l}\right) \cdot B \]

### Step 4: Rearrange the equation to solve for length \( l \).
Rearranging the equation to solve for \( l \) gives:
\[ l = \frac{M \cdot B}{F} \]

### Step 5: Plug in the known values.
We know:
- \( M = 3.0 \, \text{A-m}^2 \)
- \( B = 2 \times 10^{-5} \, \text{T} \)
- \( F = 6 \times 10^{-4} \, \text{N} \)

Substituting these values into the equation:
\[ l = \frac{3.0 \, \text{A-m}^2 \cdot 2 \times 10^{-5} \, \text{T}}{6 \times 10^{-4} \, \text{N}} \]

### Step 6: Calculate the length \( l \).
Calculating the numerator:
\[ 3.0 \cdot 2 \times 10^{-5} = 6.0 \times 10^{-5} \, \text{A-m}^2 \]
Now substituting into the equation for \( l \):
\[ l = \frac{6.0 \times 10^{-5}}{6 \times 10^{-4}} \]

Calculating this gives:
\[ l = \frac{6.0}{6} \times 10^{-5 + 4} = 1.0 \times 10^{-1} = 0.1 \, \text{m} \]

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

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