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A rectangular loop of sides 25 cm and 10 cm carrying current of 15A is placed with its longer side parallel to a long straight conductor 2.0 cm apart carrying a current of 25A. What is the new force on the loop ?

Answer»


Solution :`F_1 = (mu_(0))/(4PI) (2I_1 I_2)/(r_1) xx l = (10^(-7) xx 2 xx 25 xx 15 xx 0.25)/(0.02)=9.38 xx 10^(-4)` N attractive
`F_2 = (mu_(0))/(4pi) (2I_1 I_2)/(r_2) xx l = (10^(-7) xx 2 xx 25 xx 15 xx 0. 25)/(0.12) = 1.56 xx 10^(-4)` N REPULSIVE
Net `F= F_1 = F_2 = 7.82 xx 10^(-4)` N


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