Download Problem Sheet 8

PHYS1022 Electricity and Magnetism
Problem Sheet 8 - Workshop
1. The figure shows two long straight wires in the xy
plane and parallel to the x axis. The current in each
wire is 20 A. Find the magnitude of the force per unit
length exerted by one wire on the other.
2. A metal strip 2 cm wide and 0.1 cm thick carries a current
of 20 A in a uniform magnetic field of 2 T, as shown in the
figure. The Hall voltage is measured to be 4.27 μV.
(a) Calculate the drift velocity of the electrons in the strip.
(b) Find the number density of the charge carriers in the strip.
(c) Is point a or point b at the higher potential?
3. A particle with mass m and positive charge q starts from rest
at the origin shown in the figure. There is a uniform electric
field E in the +y-direction and a uniform magnetic field B
directed out of the page. The path of the particle is a cycloid
whose radius of curvature at the top of the cycloid is twice the
y-coordinate at that level.
(a) Explain why the path has this general shape and why it is
repetitive.
2qEy
. (Use energy conservation.)
(b) Prove that the speed at any point is equal to
m
(c) Applying Newton’s second law at the top point and taking as given that the radius
of curvature here equals 2y, prove that the speed at this point is 2E/B.
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Solutions
1. The magnetic field due to a current carrying wire follows from Ampere’s law and a
circular loop distance r away.
The force on the other wire is F = I l x B - everything here is at 90 degrees to each other
2. The current in the perpendicular magnetic field experiences a force of magnitude B q vd
The electrons therefore migrate across the strip setting up a Hall electric field EH = VH/w
(where w is the width) that at equilibrium opposes the magnetic force.
(a)
B q vd = q VH / w
vd = VH / B w = 4.27/2/0.02 = 0.107 mms-1
(b) The current is given by I = n A q vd
n = I/A q vd = 20 / (0.02x0.001) / 1.6x10-19 / 0.107x10-3 = 5.84x1028m-3
(c) the –ve charge electrons move in the opposite direction to I. –e v x B points towards
a where the electrons will accumulate. Thus Vb > Va
3. (a) Initially the electric field accelerates the particle upwards. It is then moving in a
perpendicular B field that generates a force perpendicular to the motion. The radius of
curvature of its circular motion though changes as its speed changes due to E.
(r=mv/qB). At the top point the magnetic field and electric field oppose each other and
this is the point the particle begins to move downwards. When it returns to the bottom it
must be stationnary by energy conservation – the B field does no work since its
orthogonal to the motion. The cycle then repeats.
(b) kinetic energy gained = electrostatic energy lost
½ m v2 = q E y
v=
2qEy
m
(c) q v B - q E = mv2/r = m . (2 q E y/m) / 2 y
(using b)
v = 2 E/B
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