Download Handout Topic 5 and 10 -11 NEW Selected Problems 3

Handout Topic 5 and 10 -11 NEW Selected Problems 3 SL and HL
SOLUTIONS
1. The element of an electric heater has a resistance R when in operation. What is the resistance of
a second heater that has a power output three times as large at the same operating voltage?
Answer - B
A.
R
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B.
R
3
C.
3R
D.
9R
POWER INV. PROPORTIONAL TO R P. 35 OF PACKET
2. Which one of the following is a correct definition of electric potential difference between two
points? Answer -D
A.
The power to move a small positive charge between the two points.
B.
The work done to move a small positive charge between the two points.
C.
The power per unit charge to move a small positive charge between the two points.
D.
The work done per unit charge to move a small positive charge between the two points.
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3. The diagram below illustrates some equipotential lines between two charged parallel metal
plates.
+
+
+
+
80 V
60 V
0.1 m
40 V
20 V
–
–
The electric field strength between the plates is
A.
6 NC–1.
B.
8 NC–1.
C.
600 NC–1.
D.
800 NC–1.
–
–
Answer -C
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4.
Answer - B
3
5.
Answer -D
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6.
Answer -D
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7.
Answer - A
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Magnetic Force and Current Carrying Wires – Opposites repel, Like attract
8. Answer - A
The electromotive force (emf) of a cell is defined as
A. the power supplied by the cell per unit current from the cell.
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B. the force that the cell provides to drive electrons round a circuit.
C. the energy supplied by the cell per unit current from the cell.
D. the potential difference across the terminals of the cell.
9. This question is about forces on charged particles in electric and magnetic fields.
The diagram shows two parallel plates situated in a vacuum. One plate is at a positive
potential with respect to the other.
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+
Path of positively charged particle
–
A positively charged particle passes into the region between the plates. Initially, the particle
is travelling parallel to the plates.
(a)
On the diagram,
(i)
draw lines to represent the electric field between the plates.
(3)
(ii)
show the path of the charged particle as it passes between, and beyond, the
plates.
(2)
(b)
An electron is accelerated from rest in a vacuum through a potential difference of 750
V.
(i)
Determine the change in electric potential energy of the electron.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(2)
(ii)
Deduce that the final speed of the electron is 1.6 × 107 ms–1.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(2)
(a)
(i)
lines parallel and normal to plates; (ignore any edge effect)
equally spaced;
direction from (+) to (–);
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(ii)
(b)
(i)
(ii)
curved path between plates and no curvature outside;
in downward direction;
change = qV;
= 1.6 ×10–19 × 750
= 1.2 × 10–16 J;
Or 750 eV.
½mv2 = 1.2 × 10–16;
½ × 9.1 × 10–31 × v2 = 1.2 ×10–16 / v2 = 2.64 × 1014
to give v = 1.67 × 107 m s–1;
2
2
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10. This question is about the force between current-carrying wires.
Diagram 1 below shows two long, parallel vertical wires each carrying equal currents in the
same direction. The wires pass through a horizontal sheet of card. Diagram 2 shows a plan
view of the wires looking down onto the card.
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eye
sheet of card
diagram 1
(a)
diagram 2
(i)
Draw on diagram 1 the direction of the force acting on each wire.
(ii)
Draw on diagram 2 the magnetic field pattern due to the currents in the wire.
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11. This question is about forces on charged particles.
A charged particle is situated in a field of force. Deduce the nature of the force-field
(magnetic, electric or gravitational) when the force on the particle
(a)
is along the direction of the field regardless of its charge and velocity.
GRAVITATIONAL
F = mg
if independent of charge therefore must be gravitational.
(b)
is independent of the velocity of the particle but depends on its charge.
ELECTRIC
F=Eq
Magnetic and electric forces depend on charge but only magnetic forces depend on
velocity. There fore ELECTRIC.
(c)
depends on the velocity of the particle and its charge.
MAGNETIC
F = qvB
Magnetic and electric forces depend on charge but only magnetic forces depend on
velocity. Therefore ELECTRIC.
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11 cont. d)
(ii)
Given that the magnetic force is 6.9 × 10–16 N , determine the magnitude of
the magnetic field (B)
F = Bqv = 6.9 × 10–16 N
B=
(6.9  10 16 )
(1.6  10 19  2.7  10 7 )
B = 1.6 × 10–4 T
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