Download Quiz 6

October 17, 2008
PHY2054 Discussion-Fall ‘08
Quiz 6 (Chapter 19.1-19.6)
Name:
UFID:
*1. (5pts) An electron enters in a magnetic field with a velocity of 5.00×106 m/s perpendicular to and
into the sheet. The magnetic field has an intensity of 3.00 mT, lies in the sheet and is directed as
shown. Find the direction (you can just draw an arrow) and magnitude of the magnetic force on the
electron.
Since the velocity is perpendicular to the sheet, it is also perpendicular to the
magnetic field. Therefore,
F = evBsin90˚ = 1.6×10-19×5×106×3×10-3×1= 2.4×10-15 N
Since an electron has a negative charge, point your fingers in the direction
opposite the velocity (out of the page) and curl your fingers to the direction of the
field. Then your thumb points in the direction of the force.
**2. (5pts) A wire having a mass per unit length of 0.100 kg/m is placed horizontally on a 30.0˚
incline, where there is a uniform vertical magnetic field. The wire carries a current of 2.00 A, and the
coefficient of the kinetic friction between the wire and the incline is 0.250. What is the magnitude of
the magnetic field if the wire slides up the incline with a constant speed?
When an object is moving at a constant speed, its acceleration is zero and the net force acting on the
object is zero. Since the magnetic field is vertical and the wire is sliding up the incline, the magnetic
force is in the horizontal plane and into the incline. Taking x-axis along the incline and y-axis
perpendicular to the incline, we apply the equilibrium condition to each direction:
ΣFy = 0 ⇒ n-mgcosθ-IBLsinθ = 0 ⇒ n = mgcosθ+IBLsinθ
ΣFx = 0 ⇒ IBLcosθ-mgsinθ-μn = 0 ⇒ IBLcosθ-mgsinθ-μ(mgcosθ+IBLsinθ) = 0
⇒ IBL(cosθ-μsinθ) = mg(sinθ+μcosθ)
⇒ B = (m/L)g(sinθ+μcosθ)/[I(cosθ-μsinθ)]
= 0.1×9.8×(sin30˚+0.25cos30˚)/[2×(cos30˚-0.25sin30˚)] = 0.474 T
Constants
μ0 = 4π×10-7 Tm/A
me = 9.11×10-31 kg
e = 1.6×10-19 C
***3. (5pts) A square coil with a side of 0.200 m consists of 50 closely wrapped turns and has a mass
of 0.400 kg. It carries a current of 3.00 A. The loop is hinged along a horizontal side and is placed in
a uniform magnetic field with a magnitude of 0.0500 T directed vertically down. Find the torque
acting on the coil due to the magnetic field at equilibrium position.
At the equilibrium position, the net torque acting on the loop is zero. Choosing rotational axis along
the hinged side, we have
Στ = 0 ⇒ NIBL²cosθ-mg(L/2)sinθ = 0 ⇒ tanθ = 2NIBL/(mg)
= 2×50×3×0.05×0.2/(0.4×9.8) = 0.765 ⇒ θ = tan-1(0.765) = 37.4˚,
where θ is the angle between the vertical and the plane of the coil.
The torque due to the electric field at this angle is
τm = NIBL²cosθ = 50×3×0.05×0.2²cos37.4˚ = 0.238 Nm
**4. (5pts) Consider the mass spectrometer shown below. The electric field between the plates of the
velocity selector is 900 V/m and the magnetic fields in both the velocity selector and the deflection
chamber is 0.600 T. What is the mass of the singly charged ion if it takes a semicircular path of r =
0.300 mm in the deflection chamber?
The ion travels straight and enters the deflection
chamber only when the net force acting on it is zero.
Thus the velocity of the ion is
evB-eE = 0 ⇒ v = E/B = 900/0.6 = 1500 m/s
Only the magnetic force is exerted on the ion in the
deflection chamber and the ion moves in a circle at a
constant speed. Thus we have
mac = Fc ⇒ m(v²/r) = evB
⇒ m = eBr/v = 1.6×10-19×0.6×0.3×10-3/1500 = 1.92×10-26 kg