Question

a thin ring of radius 5 cm is placed on plane z =1cm so that it's centre is at (0,0,1) if the rings carries 50mA along a h at (0,0,-1) is​

Answer 1

Given:

Radius of ring = 5cm

centre is at (0,0,1)

Current through ring = 50mA

To find:

H at (0 , 0 , -1)

Solution:

$dB = \frac{\mu Idlsin\theta}{4\pi r^{2} }$

θ is angle between dl and r

θ = 90° (here)

$r=\sqrt{R^{2} + z^{2} }$

dl = RdФ

and dBz = dB R/r

$dB_z = \frac{\mu I R^{2} d\phi }{4\pi (R^{2} + z^{2})^{3/2} } \\\\B = B_za_z = \int\limits^{2\pi }_0 {\frac{\mu I R^{2} d\phi }{4\pi (R^{2} + z^{2})^{3/2} }} \, a_z = \frac{\mu I R^{2}}{4\pi (R^{2} + z^{2})^{3/2} }2\pi a_z = \frac{\mu I R^{2}}{2(R^{2} + z^{2})^{3/2} }a_z$

R = 5cm

z = 1 - (-1) = 2cm

H = B / μ

$H = \frac{\frac{\mu I R^{2}}{2(R^{2} + z^{2})^{3/2} }a_z}{\mu} \\\\H = \frac{I R^{2}}{2(R^{2} + z^{2})^{3/2} }a_z$

$H = \frac{50 (5 * 10^{-2} )^{2} }{2*((5 * 10^{-2} )^{2}+(2 * 10^{-2} )^{2})^{3/2} }\\\\H = \frac{50 * 0.0025}{2*(0.0025+0.0004)^{3/2} }$

H = 400.2 az mA/m

Therefore, H at (0 , 0 , -1) is 400.2 az mA/m