Derivation of magnetic field due to a circular loop .

Asked by paul_dmps | 16th Dec, 2009, 07:54: AM

Expert Answer:

Hi dear,

The following figure depicts a circular loop carrying a steady current I. The loop is on y-z plane with the center O at origin and radius R. The x axis is the loop axis. We calculate the magnetic feild at point P on this axis.

Let us take a conducting element dl in the loop. Magnitude dB of the magnetic feild due to dl is

dB = (μο / 4π){ I | dl X r | / r   }

Now r2 = x2 + R2

Direction of dB is perpendicular to the planr formed by dl and r. The net contributtion along x direction is obtained by integrating dBx = dB cosθ over the loop.

cosθ = R/ { x2 + r2} 1/2

dBx = μο I dl / 4π { R / { x2 + r2}3/2  }

Summation of elements dl over the loop yeilds 2π R, the circumference of the loop,Thus the magnetic feild at point P due to entire circular loop is 

B = μο I R2 / [2 X { x2 + r2}3/2 ]

As a special case, the field at the c nenter of the loop . x = 0, hence,

Bo = μοI / 2 R

Hope this helped

Regards

Topperlearning.com

Answered by  | 16th Dec, 2009, 09:49: AM

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