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

1.56 m

Step-by-step 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) \\ (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=(\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*10^(-8))=((4\pi*10^(-7))(0.17))/(2\pi r) \\ (2.22*10^(-8))r=((4\pi*10^(-7))(0.17))/(2\pi r)\cdot r \\ (2.22*10^(-8))r=(4\pi*10^(-7)(0.17))/(2\pi) \\ (2.22*10^(-8))r=3.4*10^(-8) \\ ((2.22*10^(-8))r)/(2.22*10^(-8))=(3.4*10^(-8))/(2.22*10^(-8)) \\ r=1.56\text{ m} \end{gathered}`

Therefore, the distance is 1.56 m