AP Physics C 2017

Note    that    since   the velocity    is  perpendicular   to  the magnetic    field,  sinθ    =   sin(90°)    =   1.  Therefore,  E =

vB when the charge  travels in  a   straight    line    thorough    the fields. The electric    force   on  the proton  is

directed    to  the right,  while   the magnetic    force   goes    to  the left.   Therefore,  to  deflect the charge  to  the

left,   either  the electric    field   must    decrease,   the magnetic    field   must    increase,   or  the velocity    of  the

charge  must    be  increased.  Note    that    changing    the charge  will    have    no  effect  because the charge  q

cancels out in  the preceding   equation.

4 . B —A retarding force will be present when there is a change in magnetic flux through the metal loop
of wire:

This    occurs  only    when    the cart    is  entering    the front   edge    and leaving the back    edge    of  the field.  No

change  in  speed   occurs  while   the cart    is  fully   immersed    in  the magnetic    field,  as  there   is  no  change  in

flux.   Therefore,  choice  A   is  incorrect   because it  shows   no  change  in  velocity.   Choices D   and E   are

also    incorrect   because they    show    only    a   single, continuous, retarding   force   slowing the cart    down.

Choice  B   is  the best    of  the remaining   options.    The retarding   force   is

FB =    ILB sinθ    =   ILB

and the current is  given   by

where   the induced voltage is

ε   =   V = BLv

Combining   these   equations,  we  get the retarding   force   on  the cart    and loop:

FB =    B^2     L^2     v

This    means   the magnetic    force   and the acceleration    of  the cart    are directly    related to  the velocity.   The

cart    slows   down    when    it  enters  the magnetic    field   and again   when    it  leaves  the field.  Since   the

velocity    of  the cart    when    exiting the field   will    be  less    than    when    first   entering    the field,  the retarding

force   will    be  less    when    exiting the field.  The resulting   acceleration,   and thus    the slope   of  the v-t

graph,  will    be  less    when    exiting the field.