An electric current passes through a long straight wire. At a distance 5 cm from the wire the magnetic field is `B`. The field at 20 cm from the wire would be

Options

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

Explanation

To solve the problem, we need to find the magnetic field at a distance of 20 cm from a long straight wire carrying an electric current, given that the magnetic field at a distance of 5 cm is \( B \).

### Step-by-Step Solution:

1. **Understanding the Magnetic Field Formula**:
The magnetic field \( B \) around a long straight wire carrying current \( I \) at a distance \( R \) from the wire is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi R} \]
where \( \mu_0 \) is the permeability of free space.

2. **Magnetic Field at 5 cm**:
At a distance of 5 cm (which is 0.05 m) from the wire, we can express the magnetic field as:
\[ B = \frac{\mu_0 I}{2 \pi (0.05)} \]

3. **Magnetic Field at 20 cm**:
Now, we need to find the magnetic field at a distance of 20 cm (which is 0.20 m) from the wire. Using the same formula:
\[ B' = \frac{\mu_0 I}{2 \pi (0.20)} \]

4. **Relating the Two Fields**:
To relate \( B' \) to \( B \), we can set up the ratio:
\[ \frac{B'}{B} = \frac{\frac{\mu_0 I}{2 \pi (0.20)}}{\frac{\mu_0 I}{2 \pi (0.05)}} \]
Simplifying this gives:
\[ \frac{B'}{B} = \frac{0.05}{0.20} = \frac{1}{4} \]

5. **Finding \( B' \)**:
From the above ratio, we can express \( B' \) in terms of \( B \):
\[ B' = B \times \frac{1}{4} = \frac{B}{4} \]

### Conclusion:
Thus, the magnetic field at a distance of 20 cm from the wire is:
\[ B' = \frac{B}{4} \]