# Solution - =-F2 + oI1I2L2a b Forces on both vertical edges...

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+ve Solution: Let = , = I1 30A I2 20A = , = , = a 1 cm b 6 cm L 13 cm The magnetic field due to I1 can be found by drawing a circle at a distance r from the wire and applying Ampere's Law: . = B dl μoI1 = B μoI12πr The field goes into the page inside the rectangular loop. (a) Force on top edge of loop = = F1 μoI1I2L2πa Force on bottom edge of loop =
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Unformatted text preview: =-F2 + oI1I2L2a b Forces on both vertical edges will cancel each other. Resultant magnetic force acting on the loop = + = F1 F2 + oI1I2Lb2aa b = . 0 001337143 N (b) Force on wire: The magnetic force on wire due to loop current can be found by applying Newton's 2nd Law: =- . F 0 001337143 N...
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## This note was uploaded on 07/22/2011 for the course ME 3 taught by Professor Prof.ramachandran during the Spring '11 term at Indian Institute of Technology, Kharagpur.

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