magnetic field

Question

From a cylinder of radius r , a cylinder of radius r/2 is removed . Current flowing through the remaining cylinder is I . Then why the magnetuc field inside the cavity is constant

Answer 1

Hi Samagra!

 

This is a classic example of how to use Ampere's law, which relates the total net magnetic field around a closed loop to the current penetrating the area bounded by the loop (in its full result, it is modified into the Ampere-Maxwell law, but we can deal with just Ampere's law here).  One way of expressing this mathematically is that:

 

∫B·dl = μ_oi

 

where

 

B = magnetic field around a loop

dl = infinitesimal length unit vector around loop

i = current passing through the area bounded by the loop

 

The integral on the left above represents line integration around the length of the closed loop, giving the total net magnetic field along the loop.

 

With the cavity off-axis, we lose some of the straightforward symmetric arguments that lead us to the conclusion that B = 0 inside an on-axis cavity.  However, symmetry and previous solutions can still help, as long as the cavity is fully inside the wire (a variation can still help even in other cases).  This may be hard to follow without a diagram, but here goes.

 

Take a point inside the cavity.  We can draw a vector r from the center of the wire to that point.  We can also draw a vector r' from the center of the cavity to that point.  Imagine a circle of radius r' center on the center of the cavity.  If there were still wire inside the cavity, we could conceptually imagine a wire of radius r', centered on the cavity, carrying the same current density (assuming the current is uniform, which we will in the absence of more information).

 

If the wire were still completely solid, we could just figure out the magnetic field at that point a distance r from the center using a nicely symmetric implementation of Ampere's law.  However, with the cavity scooped out, that symmetry is broken.  HOWEVER, we can still consider the magnetic field at that point to be the magnetic field the wire would have there if it were still solid MINUS the contribution from a wire of radius r' centered on the cavity.  

 

For the point inside the cavity discussed above, we can take from previous solutions that the magnetic field at that point in a completely solid wire wraps around the wire (according to the right-hand rule).  Because of this configuration, it will be convenient to use polar coordinates, and describe that field as being in the "theta-hat" direction.  Its magnitude will be the magnitude of a magnetic field at the outer edge of a current-carrying wire, and will be related to the amount of current enclosed by a circle of radius r compared to the current in the entire wire.  By previous results (which we can derive, if needed) this means:

 

B_wire(r) = μ_oI(πr²/πR²)/(2πr) θ-hat = μ_oIr/(2πR²) θ-hat

 

where 

 

r = magnitude of vector r from center of wire to point inside cavity

I(πr^r/πR²) is the fraction of the total current of the original wire enclosed within a circle of radius r, for the given wire with radius R.

 

Likewise, for the magnetic field generated by a conceptual part of the original wire centered on the cavity and having radius r', we have:

 

B_cavity (r') =  μoIr'/(2πR²) θ-hat

 

with r' as the magnitude of vector r'

 

Note that only the current in the part of the now-cavity within a radius r' contributes to the magnetic field at r', since (as with all wires) any current outside a given radius in our conceptual wire (centered on the cavity) contributes nothing to the field at that point r', by the usual symmetry arguments we can use for uniform current-carrying wires in those situations.

 

Now, B_tot at the given point is the vector difference of these two vectors, B_wire(r) and Bcavity(r'), that are in the same polar direction, θ-hat.  This is because, since the cavity now exists, the magnetic field contribution from the part of the wire that used to fill the cavity no longer exists.  So:

 

B_tot = B_wire(r) - B_cavity(r') = μ_oIr/(2πR²) θ-hat - μ_oIr'/(2πR²) θ-hat 

 

This can be rewritten as:

 

B_tot = (μ_oI)/(2πR²)(r θ-hat - r' θ-hat)

 

Now, here is some rewriting to make our understanding of this vector easier.

 

First, in cylindrical polar coordinates,  to get the θ-hat unit vector, you can cross z-hat with r-hat.  So, to get (r θ-hat) as a result, you can take:

 

(z-hat) x( r r-hat) [which equals (magnitude of z-hat)(magnitude of r r-hat)(sine of angle between r-hat and z-hat) θ-hat = (1)(r)sin(90^o) θ-hat = r θ-hat]

 

So we can rewrite B_tot = (μ_oI)/(2πR²)[(z-hat x r r-hat) - (z-hat x r' r-hat)]

 

And since what we are calling (r r-hat) is equivalent to our original vector r, and (r' r-hat) is equivalent to our original r', we have:

 

B_tot = (μ_oI)/(2πR²)[(z-hat x r) - (z-hat x r')]

 

Now, we can take advantage of a particular relationship between r and r'.  Imagine drawing a vector a from the center of the original wire to the center of the cavity.  Once the cavity is placed, this vector is a constant (since the center of the wire and center of the cavity do not change position). Since r and r' point to the same spot -- with r originating at the center of the wire and r' at the center of the cavity -- we can write the vector equation:

 

a + r' = r

 

which can be rearranged to  r' = r - a.  Putting this into the B_tot expression, we get:

 

B_tot = (μ_oI)/(2πR²)[(z-hat x r) - (z-hat x (r-a))]

 

Since the cross product follows the distributive property, this becomes:

 

B_tot = (μ_oI)/(2πR²)[(z-hat x r) - ((z-hat x r) - (z-hat x a))] = (μ_oI)/(2πR²)[(z-hat x r) - (z-hat x r) + (z-hat x a))]

 

The first two terms inside the square brackets cancel out.  Also, a can be rewritten into (a r-hat), where a is the magnitude of the distance from the center of the wire to the center of the cavity.  Then:

 

B_tot = (μ_oI)/(2πR²)(z-hat x a r-hat) = (μ_oIa)/(2πR²) θ-hat

 

Note that this last expression for B_tot does not depend on r or r'.  Hence, it is constant in magnitude and direction (i.e. uniform) for any point inside the cavity (to which r points from the center of the wire and r' from the center of the cavity).  This applies as long as both the wire and the cavity have cylindrical symmetry that helps us write the B field expressions readily using Ampere's law.

 

I hope this helps some, even with no diagrams, but feel free to ask more questions about it.