Answer:
a) 4.88 cm
b) [tex]4.05194845*10^{-5[/tex] T
Explanation:
The magnetic filed B inside the solenoid can be given as:
[tex]B = \mu_onl[/tex]
where
n = number of turns per meter = 9.98 turns /cm = 998 m⁻¹
[tex]\mu_o[/tex] = [tex]4 \pi*10^{-7} N/A^2[/tex]
l = 24.8 mA = 0.0248 A
[tex]B_{solenoid} = \mu_onl[/tex]
= [tex](4 \pi*10^{-7} N/A^2)(998 \ m^{-1})(0.0248 \ A)[/tex]
= [tex]3.11 * 10^{-5} \ T[/tex] directed toward the axis
Now; the magnetic filed B from the wire is given by the formula;
[tex]B = \frac{\mu_oI}{2 \pi s}[/tex]
where;
s = distance of the wire
Thus; the magnetic field will be directed radially around the wire , therefore perpendicular as well to the solenoid axis
However; for the field to have an angle of 53.9° ; we have:
[tex]tan ( \frac{B_{wire}}{B_{solenoid} }) = tan \ 53.9^0[/tex]
[tex]\frac{B_{wire}}{B_{solenoid} } =1.3713[/tex]
[tex]B_{wire} =1.3713 *3.11*10^{-5}[/tex]
[tex]B_{wire} =4.265*10^{-5} \ T[/tex]
[tex]s = \frac{\mu_ol}{2 \pi B}[/tex]
[tex]s = \frac{4 \pi *10^{-7} N/A)^2*10.4}{2 \pi * 4.265*10^{-5}T}[/tex]
s = 0.0488 m
s = 4.88 cm
b)
What is the magnitude of the magnetic field there?
the magnitude of the field, is just the square root of the sum of the squares of [tex]B_{wire[/tex] and [tex]B_{solenoid[/tex]
[tex]\left(2.722\cdot10^{-5}\ +3.0015\cdot10^{-5}\right)^{2}\ -\ 2\left(2.722\cdot10^{-5}\cdot3.0015\cdot10^{-5}\right)[/tex]
= [tex]1.64182862*10^{-9}[/tex]
The square root of the answer =
[tex]\sqrt{1.64182862*10^{-9}}[/tex]
= [tex]4.05194845*10^{-5[/tex] T
