Here is the full question
A coil consisting of 100 circular loops with radius 0.6 m carries a 5.0-A current.
(a) Find the magnetic field at a point along the axis of the coil, 0.80 m from the center.
(b) Along the axis, at what distance from the center of the coil is the field magnitude 1/8 as great as it is at the center
Answer:
a) B = 0.000113 T
b) The distance (x) from the center of the coil = 1.038 m
Explanation:
The formula for the magnetic field on the line of axis of the circular loop can be expressed as:
B = [tex]\frac{\mu NIr^2}{2 ( r^2+x^2)^{3/2}}[/tex]
where;
N = number of turns = 100
[tex]\mu =[/tex] [tex]4 \pi * 10^{-7} \ N/m[/tex]
I = current flowing through the coil = 5.0- A
r = radius of the circular loop = 0.6 m
x = distance from the center of the coil
So;
[tex]B = \frac{4 \pi *10^{-7}N/m (100)(5A)(0.6\ m^2)}{2((0.6 \m )^2+(0.8m)^2)^{3/2}}[/tex]
B = 0.000113 T
b)
Also; to determine the distance (x) from the center of the coil; we have the following:
We know that the magnetic field at the center of the coil can be expressed as:
[tex]B = \frac{\mu_o NI}{2 r}[/tex]
Now; given that the field magnitude is 1/8 as great as it is at the center; Then ;
[tex]B' = \frac{1}{8} B[/tex]
∴
[tex]\frac{\mu NIr^2}{2 ( r^2+x^2)^{3/2}}[/tex] = [tex]\frac{1}{8} ( \frac{\mu_o NI}{2 r})[/tex]
[tex]\frac{r^2}{(r^2+x^2)^{3/2}}= (\frac{1}{8}) r[/tex]
[tex]8r^3 = (r^2+x^2)^{3/2}[/tex]
[tex](8r^3)^2 = (r^2+x^2)^3[/tex]
[tex](64r^6) = (r^2+x^2)^3[/tex]
[tex](r^2+x^2) = \sqrt[3]{64r^6}[/tex]
[tex](r^2+x^2) = 4r^2[/tex]
[tex]x^2 = 4r^2- r^2[/tex]
[tex]x^2 = 3 \ r^2[/tex]
[tex]x= \sqrt {3 r^2}[/tex]
[tex]x = 1.73 (r)[/tex]
[tex]x = 1.73 (0.6)[/tex]
[tex]x = 1.038 \ m[/tex]
Therefore, the distance from the center of the coil is = 1.038 m
