Magnetism16719.61First, observe that ()( )(2225.00 cm12.0 cm13.0 cm+= ). Thus, the triangle shown in dashed lines is a right triangle giving 1sinα−12.0 cm67.413.0 cm==°90.02, and 2.6β=°−=°At point P, the field due to wire 1 is ( )( )72410TmA3.00A12.0 T25.0010mπ112IBrµ−−×⋅=×±±±²³´µ¶·±±¸±¸³´µ¶·²¹³´´µ¶·²and it is directed from Ptoward wire 2, or to the left and at 67.4° below the horizontal. The field due to wire 2 has magnitude ( )( )7022220TmA3.00A5.00 T2212.0010mIBr−−=×and at Pis directed away from wire 1 or to the right and at 22.6° below the horizontal. Thus, BB11cos67.44.62 Tx=−°=−sin67.411.1 Ty=−°=−cos22.64.62 Tx=°=+
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.