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

Step-by-step explanation:

The formula for the magnetic field on the line of axis of the circular loop can be expressed as:

B =
`(\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 = (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 = (\mu_o NI)/(2 r)`

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

`B' = (1)/(8) B`

∴

`(\mu NIr^2)/(2 ( r^2+x^2)^(3/2))` =
`(1)/(8) ( (\mu_o NI)/(2 r))`

`(r^2)/((r^2+x^2)^(3/2))= ((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