tsl185 - the right. The resultant horizontal force is zero....

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Magnetic Force on Semicircular Current (1) Fancy solution : Uniform magnetic field ~ B points out of the plane. Magnetic force on segment ds : dF = IBds = IBRdθ . Integrate dF x = dF sin θ and dF y = dF cos θ along semicircle. F x = IBR Z π 0 sin θdθ = 2 IBR, F y = IBR Z π 0 cos θdθ = 0 . ds dF sin θ θ d θ R dF x y B I tsl185 – p.1/2

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Magnetic Force on Semicircular Current (2) Clever solution : Replace the semicircle by symmetric staircase of tiny wire segments. Half the vertical segments experience a force to the left, the other half a force to
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Unformatted text preview: the right. The resultant horizontal force is zero. • All horizontal segments experience a downward force. The total length is 2 R . The total downward force is 2 IBR . • Making the segments infinitesimally small does not change the result. R x y B I dF tsl186 – p.2/2...
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This note was uploaded on 04/30/2008 for the course PHYS 204 taught by Professor Andrevantonder during the Spring '07 term at Rhode Island.

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tsl185 - the right. The resultant horizontal force is zero....

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