February 24, 2009
PHY2054 Discussion-Spring ‘09
Quiz 5 (Chapter 19.1-19.6)
Name:
UFID:
*1. (2.5pts) An electron is accelerated through a potential difference of 1500 V from rest and then
enters a region where there is a uniform 2.00 T magnetic field. The magnetic field makes 45.0º to the
direction of the electron’s motion. What is the direction and magnitude of the magnetic force on the
electron?
While the electron is accelerated by an electric field, the mechanical energy of
the electron is conserved. Thus we have
ΔE = 0 ⇒ (1/2)mv2-eΔV = 0
⇒ v = √(2eΔV/m) = √[2×1.6×10-19×1500/(9.11×10-31)] = 2.30×107 m/s
The magnetic field makes 45º to the direction of electron’s motion, thus the
magnitude of the magnetic force is
F = qvBsin45º = 1.6×10-19×2.3×107×2×sin45º = 5.20×10-12 N
To find the direction of magnetic force on a negative charge, point your fingers opposite the motion
of the charge and curl your fingers in the direction of the field, then your thumb points in the
direction of the force. It is out of the page.
**2. (2.5pts) A straight wire of mass 20.0 g and 10.0 cm is suspended from two identical springs,
which are connected across a 30.0 V battery and make a closed circuit. The total resistance of the
circuit is 6.00 Ω. The springs stretch a distance of 5.00 cm before the switch is closed. After the
switch is closed, the horizontal wire moves up by a distance of 3.00 cm due to the magnetic field
directed out of the plane of the sheet. What is the strength of the magnetic field?
Before the switch is closed, the gravity of the wire is cancelled by the
supporting force from two springs. We have
mg-2kxi = 0 ⇒ k = mg/(2xi) = 0.02×9.8/(2×0.05) = 1.96 N/m
After the switch is closed, an upward magnetic force is exerted on the
current-carrying wire. The current in the wire is
I = ΔV/R = 30/6 = 5.00 A
The magnetic force, gravity and spring force must add up to zero for the wire to be in equilibrium.
The field strength is calculated as
mg-2k(xi+Δx)-IBL = 0 ⇒ B = -2kΔx/(IL) = -2×1.96×-0.03/(5×0.1) = 0.235 T
Constants & Formulas
e = 1.6×10-19 C
me = 9.11×10-31 kg
Motion of a charge in a magnetic field:
mp = 1.67×10 -27 kg
Magnetic Force (on moving charge): F = qvBsinθ
R = mv/(qB)
T = 2πm/(qB)
Magnetic Torque: τ = NBIAsinθ
(on current): F = IBlsinθ
**3. (2.5pts) A 0.500-kg square loop with side length of 0.500 m consists of 40 closely wrapped
turns. The loop is hinged along a horizontal side, carries a current of 3.00 A, and is placed in a
vertical magnetic field with a magnitude of 2×10-2 T. Determine the angle that the plane of the coil
makes to the vertical when the loop is in equilibrium.
The loop is in rotational equilibrium, which means the net torque exerted on the loop is zero.
Choosing the rotational axis on the hinged side, we have
NBIAsin(90º+θ)-mg(L/2)sinθ = 0 ⇒ tanθ = 2NBIA/(mgL) = 0.490 ⇒ θ = tan-1(0.490) = 26.1º
***4. (2.5pts) A proton is moving with a speed of 2.00×106 m/s perpendicular to a magnetic field. If
the magnetic field is uniform, the proton follows a circular path. However, if the field is not uniform,
the center of the circle “drifts” with time. Assuming the field consists of two parts, fields with
strength of 1.00 T into the page and 0.800 T into the page, find the drift velocity of the center of the
motion.
The proton follows a semicircular path in the 1-T magnetic field and
then takes another semicircular path with larger radius in the 0.8-T
magnetic field. The time required for the proton to complete these two
semicircular motions is
T = πm/(eB1)+πm/(eB2)
= 3.14×1.67×10-27/(1.6×10-19)×[1/1+1/0.8] = 7.37×10-8 s
During this time interval, the center of the motion moves by a distance of
d =2( r2-r1)=2mv/(eB2)-2mv(eB1) = 2×1.67×10-27×2×106/1.6×10-19×[1/0.8-1/1] = 1.04×10-2 m
The drift velocity is
vd = d/T = 1.04×10-2/(7.37×10-8) = 1.41×105 m/s