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Calculate the magnetic field at the center of a square loop which carries a current of1.5 A , length of each loop is 50 cm . |
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Answer» Solution :Current through the square loop , I = 1.5 A Length of each loop, l = 50 cm = `50 xx 10^(-2)` m According to biot-Savarth law , Magnetic field due to a current carrying straight WIRE B = `(mu_(0)I)/(4 pi a ) ( sin alpha +Sin beta) = (4 pi xx 10^(-7) xx 1.5)/(4 pi xx ((l)/(2)) )(sin 45^(@) + sin 45^(@) )` = `(2 xx 1.5 xx 10^(-7))/(l) ((1)/(sqrt(2)) + (1)/(sqrt(2)) ) = ( 2xx 1.5 xx 10^(-7))/(50 xx 10^(-2)) ((2)/(sqrt(2)) ) ` B = 0.084866 `xx 10^(-5)` T Magnetic field at a point .P.of CENTRE of current carrying square loop `B.= 4 "sides" xx B ` = 4 `xx 0.08487 xx 10^(-5) = 0.33948 xx 10^(-5)` `B. = 3.4 xx 10^(-6) `T 
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