Discussion Session for PHY2049
Physics with Calculus 2 - Electromagnetism
Spring 2012
Week # 8
Problem 3
The figure shows a wood cylinder of mass m = 0.25 kgs and length L = 0.1
m, with N = 10 turns of wire wrapped around it longitudinally, so that the
plane of the wire coil contains the long central axis of the cylinder. The
cylinder is released on a plane inclined at an angle θ to the horizontal, with
the plane of the coil parallel to the incline plane. If there is a vertical uniform
magnetic field of magnitude 0.5 T, what is the least current i through the
coil that keeps the cylinder from rolling down the plane?
Solution
The first thing we need to remember comes from physics I: the problem
demands translational and rotational equilibrium. The former implies that
the net force on the cylinder should be equal to zero while the latter means
that the net torque is equal to zero as well. Using the first equation should
give us the limiting value the current must have so that the cylinder remains
still. Let fs be the force of static friction acting on the contact point between
the cylinder and the inclined plane. This is one the forces that makes it
rotate. The other one is the force produced by the electric current interacting
with the external magnetic field. The net force is:
mg sin θ − fs = ma
(1)
The question is now, how do we relate this to the electric current? We see
that the only unknown variable here is fs , therefore the current must be
implicit in the variable fs .
We now move to rotational equilibrium which is ~τnet = 0 where ~τnet is the
1
net torque acting on the cylinder. There are two sources of torque acting
in opposite directions of rotation: the force of friction tries to make the
cylinder to roll down while the currrent creates a torque that tries to make
it go uphill. This in other words means that the torque due to these forces
are opposite in sign. We can write this as
τnet = τf riction − τcurrent
(2)
On one hand we know that τf riction = fs r where r is the radius of the
cylinder (notice that this was not given in the problem, so it better cancels
out in the end). On the other hand, you learned this week that the torque
produces by a loop of current is
~
~τcurrent = µ
~ ×B
~ is the external magnetic field and µ
where B
~ is the magnetic moment defined
as |~
µ| = iAN where i, A and N are the current, the area, and the number of
turns of the loop respectively. The direcction of this µ
~ vector is the direction
perpendicular to the plane of the loop. Since the direction of the external
magnetic field is fixed and pointing up, we have
~ = µB sin θ = iA B sin θ N = i 2rL B sin θN
τcurrent = |~
µ × B|
where we used the fact that the area A of the loop of current is A = 2rL
because the loop is a rectangle of sides L and 2r. We have also used the
~ is
fact that the angle θ that is necessary to compute the cross product µ
~ ×B
precisely the same angle θ that the inclined plane does with the horizontal.
Putting this into (2) we have
f r =i 2rL B sin θN
f =i 2L B sin θN
(3)
where we notice the r canceled out as expected. Since we want the net force
being equal to zero, i.e. the acceleration a must be zero, from (1) we have:
f = mg sin θ. Plugging this into (3) we obtain
i=
mg
0.25 × 9.8
=
= 2.45 A
2N LB
2 × 10 × 0.1 × 0.5
2
(4)