Question

9.00 V is applied to a wire with aresistance of 52.0 2. At what distancefrom the wire is the imagnetic field2.22 x 10-8 T?

Answer 1

Answer:

1.56 m

Explanation:

First, we need to find the current in the wire, so we will use the following equation:

$V=IR$

Where V is the voltage, R is the resistance and I is the current. Replacing the values and solving for I, we get:

$\begin{gathered} 9V=I(52\Omega) \\ \frac{9V}{52\Omega}=I \\ 0.17A=I \end{gathered}$

So, the current is 0.17 A.

Now, the magnetic field generated by a wire with current I at a distance r is equal to:

$B=\frac{\mu I}{2\pi r}$

Where μ is 4π x 10^(-7). So, replacing the values and solving for r, we get:

$\begin{gathered} (2.22\times10^{-8})=\frac{(4\pi\times10^{-7})(0.17)}{2\pi r} \\ (2.22\times10^{-8})r=\frac{(4\pi\times10^{-7})(0.17)}{2\pi r}\cdot r \\ (2.22\times10^{-8})r=\frac{4\pi\times10^{-7}(0.17)}{2\pi} \\ (2.22\times10^{-8})r=3.4\times10^{-8} \\ \frac{(2.22\times10^{-8})r}{2.22\times10^{-8}}=\frac{3.4\times10^{-8}}{2.22\times10^{-8}} \\ r=1.56\text{ m} \end{gathered}$

Therefore, the distance is 1.56 m