Question

the magnetic field 0.02 m from a wire is 0.1 t. what is the magnitude of the magnetic field 0.01 m from the same wire?
○ 0.01 t
○ 0.05 t
○ 0.1 t
○ 0.2 t

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 \( B=\frac{\mu_0I}{2\pi r} \), where \( \mu_0 \) is the permeability of free space and \( I \) is the current in the wire. For the same wire, \( \mu_0 \) and \( I \) are constant. So we can write the relationship between two magnetic fields \( B_1 \) and \( B_2 \) at distances \( r_1 \) and \( r_2 \) as \( \frac{B_1}{r_1}=\frac{B_2}{r_2} \) (since \( \frac{\mu_0I}{2\pi} \) is a constant, let's call it \( k \), so \( B_1 = \frac{k}{r_1} \) and \( B_2=\frac{k}{r_2} \), then \( B_1r_1=B_2r_2 = k \)).

Step2: Identify the given values

We are given that \( r_1 = 0.02\space m \), \( B_1=0.1\space T \) and \( r_2 = 0.01\space m \). We need to find \( B_2 \).

From the relationship \( B_1r_1=B_2r_2 \), we can solve for \( B_2 \):

\( B_2=\frac{B_1r_1}{r_2} \)

Substitute the given values: \( B_1 = 0.1\space T \), \( r_1=0.02\space m \), \( r_2 = 0.01\space m \) into the formula.

\( B_2=\frac{0.1\space T\times0.02\space m}{0.01\space m} \)

Step3: Calculate the value of \( B_2 \)

First, calculate the numerator: \( 0.1\times0.02 = 0.002 \)

Then divide by the denominator: \( \frac{0.002}{0.01}=0.2\space T \)

Answer:

0.2 T