This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 47. We use Eq. 29-37 where µ is the magnetic dipole moment of the wire loop and B is the magnetic ﬁeld,
as well as Newton’s second law. Since the plane of the loop is parallel to the incline the dipole moment
is normal to the incline. The forces acting on the cylinder are the force of gravity mg , acting downward
from the center of mass, the normal force of the incline N , acting perpendicularly to the incline through
the center of mass, and the force of friction f , acting up the incline at the point of contact. We take the
x axis to be positive down the incline. Then the x component of Newton’s second law for the center of
mg sin θ − f = ma .
For purposes of calculating the torque, we take the axis of the cylinder to be the axis of rotation. The
magnetic ﬁeld produces a torque with magnitude µB sin θ, and the force of friction produces a torque with
magnitude f r, where r is the radius of the cylinder. The ﬁrst tends to produce an angular acceleration in
the counterclockwise direction, and the second tends to produce an angular acceleration in the clockwise
direction. Newton’s second law for rotation about the center of the cylinder, τ = Iα, gives
f r − µB sin θ = Iα .
Since we want the current that holds the cylinder in place, we set a = 0 and α = 0, and use one equation
to eliminate f from the other. The result is mgr = µB . The loop is rectangular with two sides of length
L and two of length 2r, so its area is A = 2rL and the dipole moment is µ = N iA = 2N irL. Thus,
mgr = 2N irLB and
(0.250 kg)(9.8 m/s )
= 2.45 A .
2(10.0)(0.100 m)(0.500 T) ...
View Full Document