Question

39:06

Answer 1

Answer:

The magnetic field 0.01m from the wire is 0.2T.

Explanation:

The magnetic field $B$ at a distance $R$ due to a wire carrying current $I$ is

$B = \dfrac{\mu_o I}{2\pi R}.$

Now, let us call $B_1$ the magnetic field at $R_1$ and $B_2$ the magnetic field at $R_2$:

$B_1 = \dfrac{\mu_o I}{2\pi R_1},$

$B_2 = \dfrac{\mu_o I}{2\pi R_2}.$

Dividing $B_1$ by $B_2$ we get

$\dfrac{B_1}{B_2} = \dfrac{\dfrac{\mu_o I}{2\pi R_1} }{\dfrac{\mu_o I}{2\pi R_2} }$

$\dfrac{B_1}{B_2} = \dfrac{\mu_o I}{2\pi R_1}* \dfrac{2\pi R_2}{\mu_oI}.$

$\dfrac{B_1}{B_2} = \dfrac{2\pi R_2}{2\pi R_1}$

$\boxed{\dfrac{B_1}{B_2} = \dfrac{R_2}{R_1}}$

Now we put in the numbers

$B_1 = 0.1T,\: \: R_1 = 0.02m$ and $R_2 = 0.01m$ to get:

$\dfrac{0.1T}{B_2} = \dfrac{0.01m}{0.02m}$

$\dfrac{0.1T}{B_2} = 0.5$

solving for $B_2$ we get:

$B_2 = \dfrac{0.1T}{0.5}$

$\boxed{B_2 = 0.2T}$

which is the magnetic field at 0.01 meters from the wire.