Physics 1310
Name:
Quiz 4 Key
(5 Questions
4 points each)
1. A long straight wire carries a current of 25 A. Calculate the magnitude of the magnetic field at a
point 2.5 cm distant from the wire.
25 A
2.5 cm
B= o I / 2 R = 2 x 10-7 25 / 0.025 T = 2 x 10-4 T
2. A 5.0 cm long segment of wire carries a 20 A current. What is the magnitude of the maximum force
on this wire due to a 2.5 T magnetic field?
20A
5.0 cm
F = ILB = 20 A x 0.05 m * 2.5 T = 2.5 N
3. A particle with mass equal to 6.7x10-27 kg is moving in a uniform magnetic field of magnitude 2.5 T.
The velocity of the particle is perpendicular to the magnetic field. The speed of the particle equals
3.0x106 m/s. The acceleration of the particle equals 3.6x1014 m/s2. Calculate the particle’s charge.
F = ma = qvB  q = ma / vB = 6.7x10-27 x 3.6x1014 m/s2 / (3.0x106 m/s x 2.5 T) = 3.2 x10-19 C
4. Two long straight wires are perpendicular to the plane of the page. The current in the wire on the left
points out of the page. The current in the wire on the right points into the page.
Current is out of the page
.
P
x
Current is into the page
What is the direction of the net magnetic field at point P, the mid-point of a line joining the two wires?
Up the page.
5. As shown below, an electron moves to the right through an area with E and B fields that point at right
angles to each other and to the direction of the electron’s velocity.
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B .
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E
B points out of the page.
E points up.
Electron moves to the right.
Which of the following statements is true for an electron within the area of the combined E and B
fields?
a.
b.
c.
d.
The net force on the electron is always zero for this situation.
The net force on the electron is never zero for this situation.
The net force on the electron is zero when the electron is at rest.
none of the above.
The net force is only zero when v = E/B
Extra-Credit (5 point problem). A particle with
mass equal to 2.0 x 10-26 kg has a negative electric
charge equal to –q. The particle is accelerated from
rest through an electric potential difference. This
electric potential difference equals 6,400 volts.
Because it falls through this electric potential
difference, the particle attains speed v before it enters
a magnetic field, B. The magnitude of this magnetic
field equals 0.200 T. The particle moves in a circular
path in this magnetic field. The radius of the
particle's path in the B-field equals 20.0 cm. The
magnitude of the particle’s charge equals
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
a. 1.6 x 10-19 C
b. 3.2 x 10-19 C
c. 4.8 x 10-19 C
d. 6.4 x 10-19 C
e. 9.6 x 10-19 C
½ mv2 = qV and R = mv / qB  v = qBR/m
so, ½ m(qBR/m)2 = qV  ½ q B2 R2 / m = V  q = 2 m V / (B2 R2)
so, q = 2 x 2.0 x 10-26 kg x 6,400 volts /(0.04 T-m)2 = 1.6 x 10-19 C .