An electric current passes through a long straight copper wire. At a distance of 5 cm from the straight wire, the magnetic field is B. What is the magnetic field at a distance of 20 cm from the wire ?

Options

A. `B/6`
B. `B/4`
C. `B/3`
D. `B/2`

Explanation

To solve the problem, we will use the formula for the magnetic field around a long straight wire, which is given by Ampere's law. The magnetic field (B) at a distance (r) from a long straight wire carrying current (I) is given by:

\[ B = \frac{\mu_0 I}{2 \pi r} \]

Where:
- \( \mu_0 \) is the permeability of free space (a constant),
- \( I \) is the current flowing through the wire,
- \( r \) is the distance from the wire.

### Step-by-step Solution:

1. **Identify the Given Values**:
- At a distance \( r_1 = 5 \) cm, the magnetic field is \( B_1 = B \).
- We need to find the magnetic field \( B_2 \) at a distance \( r_2 = 20 \) cm.

2. **Understand the Relationship**:
- The magnetic field is inversely proportional to the distance from the wire. This means:
\[ B \propto \frac{1}{r} \]
- Therefore, we can write:
\[ \frac{B_2}{B_1} = \frac{r_1}{r_2} \]

3. **Substitute the Known Values**:
- Substitute \( r_1 = 5 \) cm and \( r_2 = 20 \) cm into the equation:
\[ \frac{B_2}{B} = \frac{5}{20} \]

4. **Simplify the Ratio**:
- Simplifying the right-hand side:
\[ \frac{B_2}{B} = \frac{1}{4} \]

5. **Solve for \( B_2 \)**:
- Rearranging gives us:
\[ B_2 = \frac{1}{4} B \]

### Final Answer:
The magnetic field at a distance of 20 cm from the wire is \( \frac{B}{4} \).