Question

A current in a wire increases from 2 A to 6 A. How will the magnetic field 0.01 m from the wire change?

It increases to four times its original value.
It increases to three times its original value.
It decreases to one-fourth its original value.
It decreases to one-third its original value.

Answer 1

The magnetic field \( B \) around a long straight wire carrying a current \( I \) can be calculated using Ampère's law. The magnetic field 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 (a constant),
- \( I \) is the current through the wire,
- \( r \) is the distance from the wire.

In this case, we have a current that increases from 2 A to 6 A, and we want to see how the magnetic field changes at a distance of 0.01 m from the wire.

1. **Initial magnetic field when \( I = 2 \, A \):**

\[
B_1 = \frac{\mu_0 (2)}{2 \pi (0.01)}
\]

2. **Final magnetic field when \( I = 6 \, A \):**

\[
B_2 = \frac{\mu_0 (6)}{2 \pi (0.01)}
\]

Now, let's find the ratio of \( B_2 \) to \( B_1 \):

\[
\frac{B_2}{B_1} = \frac{\frac{\mu_0 (6)}{2 \pi (0.01)}}{\frac{\mu_0 (2)}{2 \pi (0.01)}} = \frac{6}{2} = 3
\]

So, the final magnetic field \( B_2 \) is three times the initial magnetic field \( B_1 \).

Therefore, the answer is: **It increases to three times its original value.**