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An element Delta l= Delta x hat i is placed at origin and carries a current I=10A. IF Delta x=1cm magnetic field at point P is….. T. |
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Answer» Solution :`|DB| = (mu_0)/(4pi) (Idl sin theta)/(r^2) ` `dl = Delta x = 10^(-2) m , I = 10 A , r = 0.5m = y , mu_0 // 4pi = 10^(-7) (TM)/(A)` ` theta = 90^@ , sin theta = 1` `|dB| = (10^(-7) XX 10 xx 10^(-2))/(25 xx 10^(-2) ) = 4 xx 10^(-8) T` The direction of the field is in the +z-direction. This is so since, `dl xx r = Delta x hati xx y hatj = y Delta x (hati xx hatj) = y Delta x hatk` We REMIND you of the following cyclic property of cross-products `hati xx hatj = hatk , hatj xx hatk = hati , hatk xx hati = hatj` Note that the field is small in magnitude. |
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