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.

Explanation:

Step1: Recall the formula for magnetic field around a current - carrying wire

The magnetic field \( B \) at a distance \( r \) from a long straight current - carrying wire is given by the formula \( B=\frac{\mu_0I}{2\pi r} \), where \( \mu_0 \) is the permeability of free space, \( I \) is the current in the wire, and \( r \) is the distance from the wire.

Assume the initial current is \( I_1 = 2\space A \) and the final current is \( I_2=6\space A \), and the distance \( r \) is constant (\( r = 0.01\space m \)). Let the initial magnetic field be \( B_1=\frac{\mu_0I_1}{2\pi r} \) and the final magnetic field be \( B_2=\frac{\mu_0I_2}{2\pi r} \).

Step2: Find the ratio of the final magnetic field to the initial magnetic field

We calculate the ratio \( \frac{B_2}{B_1} \). Substitute the expressions for \( B_1 \) and \( B_2 \) into the ratio:

\[

$$\begin{align*} \frac{B_2}{B_1}&=\frac{\frac{\mu_0I_2}{2\pi r}}{\frac{\mu_0I_1}{2\pi r}}\\ &=\frac{I_2}{I_1} \end{align*}$$

\]

Since \( I_1 = 2\space A \) and \( I_2 = 6\space A \), then \( \frac{I_2}{I_1}=\frac{6}{2}=3 \). So \( B_2 = 3B_1 \), which means the magnetic field increases to three times its original value.

Answer:

It increases to three times its original value.