Why Does the Taylor Expansion of a Magnetic Field Include a Factor of 1/2?

[Mothrog]

OK, I'm given a magnetic field and told the field is cylindrically symmetric. Obviously, divergence of B must be zero, so taking the divergence in cylindrical coordinates,

(1/r)(B_r) + (dB_r/dr) = -(dB_z/dz)

where B_r is the radial component of the vector and B_z is the z component of the vector.

Then, taking a Taylor expansion about r = 0, z = z_0,

B_r = B_r(0, z_0) + r*(dB_r/dr) + (z - z_0)*(dB_r/dz)

Substituting in the condition from the divergence,

B_r = -r*(dB_z/dz) + z(dB_r/dz)

So, for z = z_0,

B_r = -r*(dB_z/dz)

But, my book says that for small values of r, and z = z_0

B_r = -(r/2)*(dB_z/dz)

Anyone know where that factor of 1/2 might be coming from?

 

[OlderDan]

OK, I'm given a magnetic field and told the field is cylindrically symmetric. Obviously, divergence of B must be zero, so taking the divergence in cylindrical coordinates,

(1/r)(B_r) + (dB_r/dr) = -(dB_z/dz)


where B_r is the radial component of the vector and B_z is the z component of the vector.

Then, taking a Taylor expansion about r = 0, z = z_0,

B_r = B_r(0, z_0) + r*(dB_r/dr) + (z - z_0)*(dB_r/dz)

Substituting in the condition from the divergence,

B_r = -r*(dB_z/dz) + z(dB_r/dz)

So, for z = z_0,

B_r = -r*(dB_z/dz)

But, my book says that for small values of r, and z = z_0

B_r = -(r/2)*(dB_z/dz)

Anyone know where that factor of 1/2 might be coming from?

Doesn't the zero divergence condition demand that B_r(0, z_0) = 0? And I don't see where you get this
B_r = -r*(dB_z/dz) + z(dB_r/dz)
What happened to the z_0?

I see a 1/2 in there if you incorporate this.