P29_040 - 50 cm side, the B x component produces a force i`...

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40. We establish coordinates such that the two sides of the right triangle meet at the origin, and the ` y = 50 cm side runs along the + y axis, while the ` x = 120 cm side runs along the + x axis. The angle made by the hypotenuse (of length 130 cm) is θ =tan 1 (50 / 120) = 22 . 6 , relative to the 120 cm side. If one measures the angle counterclockwise from the + x direction, then the angle for the hypotenuse is 180 22 . 6 = +157 . Since we are only asked to Fnd the magnitudes of the forces, we have the freedom to assume the current is flowing, say, counterclockwise in the triangular loop (as viewed by an observer on the + z axis. We take ~ B to be in the same direction as that of the current flow in the hypotenuse. Then, with B = | ~ B | =0 . 0750 T, B x = B cos θ = 0 . 0692 T and B y = B sin θ =0 . 0288 T . (a) Eq. 29-26 produces zero force when ~ L k ~ B so there is no force exerted on the hypotenuse. On the
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Unformatted text preview: 50 cm side, the B x component produces a force i` y B x k, and there is no contribution from the B y component. Using SI units, the magnitude of the force on the ` y side is therefore (4 . 00 A)(0 . 500 m)(0 . 0692 T) = 0 . 138 N . On the 120 cm side, the B y component produces a force i` x B y k, and there is no contribution from the B x component. Using SI units, the magnitude of the force on the ` x side is also (4 . 00 A)(1 . 20 m)(0 . 0288 T) = 0 . 138 N . (b) The net force is i` y B x k + i` x B y k = 0 , keeping in mind that B x < 0 due to our initial assumptions. If we had instead assumed ~ B went the opposite direction of the current ow in the hypotenuse, then B x > 0 but B y < 0 and a zero net force would still be the result....
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