Then, by the right-hand rule, a downward component (Bd) of GBwill produce the eastward Fx, and a westward component (Bw) will produce the upward Fy. Specifically, ,.xdywFiLBFiLB==Considering forces along a vertical axis, we find NywFmggiLB=−=−so that ffmgiLBssw==−,.maxµbgIt is on the verge of motion, so we set the horizontal acceleration to zero: ( )0.swFfiLBmgiLB−= ⇒=−The angle of the field components is adjustable, and we can minimize with respect to it. Defining the angle by Bw= Bsinθand Bd= Bcos(which means is being measured from a vertical axis) and writing the above expression in these terms, we obtain ()cossincossinsssmgiLBmgiLBBiLθµµθ=−⇒=+which we differentiate (with respect to ) and set the result equal to zero. This provides a determination of the angle: =°−−tantan..1106031sbgb gConsequently, ( )( )()( )2min0.60 1.0kg 9.8m s0.10T.50A 1.0m cos310.60sin31B°+°(b) As shown above, the angle is ( ) ( )tantan0.6031 .s=°45. (a) The magnetic force must push horizontally on the rod to overcome the force of
This is the end of the preview. Sign up
access the rest of the document.