Question

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 1

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 = $\frac{\mu NIr^2}{2 ( r^2+x^2)^{3/2}}$

where;

N = number of turns = 100

$\mu =$ $4 \pi * 10^{-7} \ N/m$

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;

$B = \frac{4 \pi *10^{-7}N/m (100)(5A)(0.6\ m^2)}{2((0.6 \m )^2+(0.8m)^2)^{3/2}}$

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:

$B = \frac{\mu_o NI}{2 r}$

Now; given that the field magnitude is 1/8 as great as it is at the center; Then ;

$B' = \frac{1}{8} B$

∴

$\frac{\mu NIr^2}{2 ( r^2+x^2)^{3/2}}$ = $\frac{1}{8} ( \frac{\mu_o NI}{2 r})$

$\frac{r^2}{(r^2+x^2)^{3/2}}= (\frac{1}{8}) r$

$8r^3 = (r^2+x^2)^{3/2}$

$(8r^3)^2 = (r^2+x^2)^3$

$(64r^6) = (r^2+x^2)^3$

$(r^2+x^2) = \sqrt[3]{64r^6}$

$(r^2+x^2) = 4r^2$

$x^2 = 4r^2- r^2$

$x^2 = 3 \ r^2$

$x= \sqrt {3 r^2}$

$x = 1.73 (r)$

$x = 1.73 (0.6)$

$x = 1.038 \ m$

Therefore, the distance from the center of the coil is  = 1.038 m