Question

Consider a circular conducting wire of radius R carrying a constant current I. The loop lies in the zy-plane and the center of the loop is at origin of the coordinates. The z-axis goes through the center of the loop and it is perpendicular to the plane where the loop lies. The current in the loop flows counterclockwise when viewed from a point on the 2-axis, with 2 > 0 a) From the law of Biot-Savart, show that the magnetic field on the z axis is (10 points) B _IR 2 (2+) Tk Two current loops, identical to the one described in the first part of the problem, are placed parallel to the zy-plane, one with center at (0.0, L/2) and the other with center at (0,0,-L/2), respectively. The current in each of the two loops flows counterclockwise. b) Calculate the total magnetic field along the z axis. (10 points) c) Shows that the derivative of the magnetic field with respect to 2 vanishes at 2 = 0. (10 points) d) Calculate the second derivative of the magnetic field with respect to 10 points) e) What is the value of the ratio L/R, for which the second derivative of the magnetic field at 2 = 0 vanishes? (10 points) The fourrers

Answer 1

a) it has been shown

b)
`B=(\mu_0I)/(2)(\frac{R^2}{(R^2+(z-(L)/(2))^2)^{(3)/(2)}}+\frac{R^2}{(R^2+(z+(L)/(2))^2)^{(3)/(2)}}) \hat k`

c) it has been shown

d)
`(d^2B)/(dz^2)=-\frac{3R^2}{(R^2+(-(L)/(2)+z)^2)^{(5)/(2)}}-\frac{3R^2}{(R^2+((L)/(2)+z)^2)^{(5)/(2)}}+\frac{15R^2(-(L)/(2)+z)^2}{(R^2+(-(L)/(2)+z)^2)^{(7)/(2)}}+\frac{15R^2((L)/(2)+z)^2}{(R^2+((L)/(2)+z)^2)^{(7)/(2)}}`

e)
`L/R = 1`

Step-by-step explanation:

a)
`dB = (\mu_0)/(4\pi)(Idlsin(\theta))/(R^2+z^2)` where
`\theta = 90`°

Then,
`dB=(\mu_0)/(4\pi)(Idl)/(R^2+z^2)`

`dB_a=dBcos(\phi)` where
`cos(\phi)=R/√(R^2+z^2)`

Then,
`dB_a=(\mu_0)/(4\pi)(Idl)/(R^2+z^2)(R)/(√(R^2+z^2))`

So total magnetic field can be found as follows,

`B=\int dB_a=(\mu_0)/(4\pi)(I)/(R^2+z^2)(R)/(√(R^2+z^2))\int\limits^(2\pi R)_0 dl=\frac{\mu_0IR}{4\pi(R^2+z^2)^{(3)/(2)}}2\pi R=(\mu_0I)/(2)\frac{R^2}{(R^2+z^2)^{(3)/(2)}} \hat k`

b) Since the first loop is L/2 distance above center, when we choose arbitrary point at z-axis, the distance will be z - L/2. Also, since the second loop is L/2 below the center, same arbitrary point's distance from the loop will be z + L/2. Moreover, since the position of loops is symmetric, choosing arbitrary point from anywhere will not change the result. So now, let's write the above equation down with corresponding consideration.

Since there are 2 loops, their magnetic fields will be added to each other to find the total magnetic field.

Thus,

`B=(\mu_0I)/(2)(\frac{R^2}{(R^2+(z-(L)/(2))^2)^{(3)/(2)}}+\frac{R^2}{(R^2+(z+(L)/(2))^2)^{(3)/(2)}}) \hat k`

c) The derivative of the equation in part (a) will be as follows,

`(dB)/(dz)=-\frac{3R^2(-(L)/(2)+z)}{(R^2+(-(L)/(2)+z)^2)^{(5)/(2)}}-\frac{3R^2((L)/(2)+z)}{(R^2+((L)/(2)+z)^2)^{(5)/(2)}}`

At
`z=0`, the above equation becomes as follows,

`(dB)/(dz)=\frac{3R^2(L)/(2)}{(R^2+(L^2)/(4))^{(5)/(2)}}-\frac{3R^2(L)/(2)}{(R^2+(L^2)/(4))^{(5)/(2)}}=0`

d) The second derivative of the equation in part (a) will be as follows,

`(d^2B)/(dz^2)=-\frac{3R^2}{(R^2+(-(L)/(2)+z)^2)^{(5)/(2)}}-\frac{3R^2}{(R^2+((L)/(2)+z)^2)^{(5)/(2)}}+\frac{15R^2(-(L)/(2)+z)^2}{(R^2+(-(L)/(2)+z)^2)^{(7)/(2)}}+\frac{15R^2((L)/(2)+z)^2}{(R^2+((L)/(2)+z)^2)^{(7)/(2)}}`

e) At
`z=0`, the above equation becomes as follows,

`-\frac{3R^2}{(R^2+(-(L)/(2))^2)^{(5)/(2)}}-\frac{3R^2}{(R^2+((L)/(2))^2)^{(5)/(2)}}+\frac{15R^2(-(L)/(2))^2}{(R^2+(-(L)/(2))^2)^{(7)/(2)}}+\frac{15R^2((L)/(2))^2}{(R^2+((L)/(2))^2)^{(7)/(2)}}=0`

`-\frac{6R^2}{(R^2+(-(L)/(2))^2)^{(5)/(2)}}+\frac{30R^2(-(L)/(2))^2}{(R^2+(-(L)/(2))^2)^{(7)/(2)}}=0`

`\frac{-6R^2(R^2+(-(L)/(2))^2)+30R^2(-(L)/(2))^2}{(R^2+(-(L)/(2))^2)^{(7)/(2)}}=0`

The denominator cannot be equal to zero.

Hence,

`-6R^2(R^2+(-(L)/(2))^2)+30R^2(-(L)/(2))^2=0\\-12R^4-3R^2L^2+15R^2L^2=0\\-12R^4+12R^2L^2=0\\R^2(L^2-R^2)=0\\R^2(L-R)(L+R)=0\\`

The solutions are
`R=0,R=L,R=-L`. However
`R=0` and
`R=-L` cannot be possible.

So the result is
`R=L`

Thus,
`L/R = 1`