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If a straight conductor of length 40cm bent in the form of a square and the current 2A is allowed to pass through square, then find the magnetic induction at the centre of the square loop |
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Answer» Solution :`B_("net")=4B_("side")` `B_("net")=4(mu_(0))/(4pi)XXI/(L//2)(sin45^(@)+sin45^(@))` `=4XX(mu_(0))/(4pi)xxI/(L//2)(sqrt2)=(mu_(0))/(4pi)(8sqrt(21))/L` `=10^(-7)xx8sqrt2xx2xx10=16sqrt2muT`
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