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An equilateral triangle ofside length 1 is formed from a piece of wire of uniform resistance. The current I is as shown in figure. Find the magnitude of the magnetic field at its center O. |
Answer» Solution : The magnetic FIELD induction at O due to current through PR is `B_(1) = (mu_0)/(4pi) (2I//3)/(r) [ sin 60^@ + sin 60^@]` `= (mu_0)/(4pi)(2I)/(3R)ODOT` (directed outward) The megantic field induction at O due to current through PQR is `B_(2) = 2xx(mu_0(I//3))/(4pir)[sin 60^@ + sin 60^2]` `= (mu_(0)2I)/(2pi3r)OX` (directed inward) `therefore` Resultant magnetic induction at O `rArr B_1 - B_2 = 0` |
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