Question

The current in a solenoid is changed from 1.1 A to 4.4 A. By what factor does the magnitude of the magnetic field inside the solenoid increase?

A. 2

B. 4

C. 0.5

D. 0.25

Answer 1

let's suppose that our solenoid has N turns and its length is L, then the field H inside it is given by the subsequent formula:
\[\Large H = {\mu _0}\frac{{NI}}{L},\quad {\mu _0} = 12.56 \times {10^{ - 6}}Henry/meter\]

Answer 2

1.256E-5 ?

Answer 3

yes!
so we can write:
\[\Large \begin{gathered}
{H_1} = {\mu _0}\frac{{N{I_1}}}{L},\quad {H_2} = {\mu _0}\frac{{N{I_2}}}{L} \hfill \\
\hfill \\
\frac{{{H_2}}}{{{H_1}}} = \frac{{{I_2}}}{{{I_1}}} \hfill \\
\end{gathered} \]

Answer 4

ok! what do we plug in?

Answer 5

and therefore:
\[\Large \frac{{{H_2}}}{{{H_1}}} = \frac{{{I_2}}}{{{I_1}}} = \frac{{4.4}}{{1.1}} = ...\]

Answer 6

4! so choice B (4) is our solution? :O

Answer 7

yes!

Answer 8

yay!! thanks!!:D

Answer 9

:)

Answer 10

@Daniellelovee