Question

A straight wire carrying a current of 51.471 A lies along the axis of a 7 cm-diameter solenoid. The solenoid is 65.271 cm long and has 271 turns carrying a current of 16.358 A. What is the magnitude, in milliTeslas, of the magnetic field 1.227 cm from the wire?

Answer 1

0.839 milliTeslas

Explanation:

Given:

Current (I) = 51.471 A

Distance (d) = 1.227 cm = 0.01227 m

Current solenoid (Is) = 16.358 A

Distance solenoid (ds) =65.271 cm = 0.65271 m

A sketch of the situation:

We calculate the magnetic field for each case:

`\begin{gathered} B_w=(\mu_0\cdot I)/(2\pi d)=(4\pi\cdot10^(-7)\cdot51.471)/(2\pi\cdot0.01227)=0.00083897\text{ T} \\ \\ B_s=(\mu_0\cdot I_s)/(2\pi\cdot d_s)=(4\pi\cdot10^(-7)\cdot16.358)/(2\pi\cdot0.65271)=0.00000501\text{ T} \end{gathered}`

Therefore, the resulting field due to the wire and the solenoid would be:

`\begin{gathered} B=√((B_w)^2+(B_s)^2) \\ \\ \text{ We replacing:} \\ \\ B=√((0.00083897)^2+(0.00000501)^2) \\ \\ B\cong0.000839\text{ T}=0.839\text{ mT} \end{gathered}`

The correct answer is 0.839 milliTeslas