Question #a6a34

1 Answer
Oct 5, 2016

Answer:

#B = color(red)( 7/16) (mu I)/(R)#

Explanation:

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/curloo.html

We need only add the components due to the different part circles.

For a full circle of radius R' we have #B = (mu I)/(2R')# as per the attachment.

For BC, which is only a quarter circle of radius #2R#:

#B_(BC) = 1/4 * (mu I)/(2(2R)) = 1/16 (mu I)/(R)#

For DA, we do likewise

#B_(DA) = 3/4 * (mu I)/(2R) = 3/8 (mu I)/(R)#

And adding #implies 7/16 (mu I)/(R)#

NB
We can do it this simply because of the consat radius and because #d vec L times hat r# is the same all along a circle: it is the tangent #d vec L# vector crossed with unit radial #hat r# vector. So we are merely adding the components of the #d vec B #'s for each circle

The direction of the #vec B# field will be into the page as the right hand rule would indicate when applied to this cross product.