ch28-p045 - 45. (a) The magnetic force must push...

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Then, by the right-hand rule, a downward component ( B d ) of G B will produce the eastward F x , and a westward component ( B w ) will produce the upward F y . Specifically, ,. xd yw F iLB F iLB = = Considering forces along a vertical axis, we find Ny w Fm g g i L B = −= so that ff m g i L B ss w == , . max µ b g It is on the verge of motion, so we set the horizontal acceleration to zero: ( ) 0. s w Ff i L B m g i L B −= ⇒ = The angle of the field components is adjustable, and we can minimize with respect to it. Defining the angle by B w = B sin θ and B d = B cos (which means is being measured from a vertical axis) and writing the above expression in these terms, we obtain () cos sin cos sin s s s mg iLB mg iLB B iL θµ µθ =− = + which we differentiate (with respect to ) and set the result equal to zero. This provides a determination of the angle: = ° −− tan tan . . 11 060 31 s bg b g Consequently, ( ) ( ) ( ) ( ) 2 min 0.60 1.0kg 9.8m s 0.10T. 50A 1.0m cos31 0.60sin31 B °+ ° (b) As shown above, the angle is ( ) ( ) tan tan 0.60 31 . s = ° 45. (a) The magnetic force must push horizontally on the rod to overcome the force of
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