Solutions to Magnetism, Force &
Field
1. At one instant an electron (charge = –1.6  10–19 C) is moving in the xy plane, the
components of its velocity being vx = 5  105 m/s and vy = 3  105 m/s. A magnetic
field of 0.8 T is in the positive x direction. At that instant the magnitude of the
magnetic force on the electron is:
A) 0
B) 2.6  10–14 N
C) 3.8  10–14 N
D) 6.4  10–14 N
E) 1.0  10–13 N
Solutions:
f z   qv y Bx  1.6  1019 C    3  105 m / s    0.8T   3.84  1014 N
Ans: C
3. An ion with a charge of +3.25  10−19 C is in region where a uniform electric
field of 5  104. V/m is perpendicular to a uniform magnetic field of 0.8 T. If its
acceleration is zero then its speed must be:
A) 0
B)
C)
D)
E)
1.6  104 m/s
4.0  105 m/s
6.3  105 m/s
any value but 0
Solutions:
v
E 5  104V / m

 6.25  104 m / s
B
0.8 T
Ans: D
4. An electron is launched with velocity v in a uniform magnetic field B. The
angle between v and B is between 0 and 90o. As a result, the electron follows a
helix, its velocity vector v returning to its initial value in a time interval of:
A) 2 π m / e B
B) 2 π m v / e B
C) 2 π m v sin/ e B
D) 2 π m v cos/ e B
E) none of these
T
Solutions:
2
B

2 m
eB
Ans: A
5. Electrons (mass m, charge –e) are accelerated from rest through a potential
difference V and are then deflected by a magnetic field B that is perpendicular to
their velocity. The radius of the resulting electron trajectory is:
A)

B)
B 2eV m
C)

D)
B 2mV e
2eV m
2mV e


B
B
E) none of these
Solutions:

1 2
mv  eV
2

v
2eV
m
eB v

m r

r
mv m

eB eB
2eV 1

m
B
2mV
e
Ans: C
6.
In a certain mass spectrograph, an ion beam passes through a velocity filter
consisting of mutually perpendicular fields E and B. The beam then enters a region
of another magnetic field B ' perpendicular to the beam. The radius of curvature of
the resulting ion beam is proportional to:
A) E B / B
B) E B / B
C) B B' /E
D) B / E B
E) E / B B
Solutions:
Velocity selector:
r
v


v
E
B
mv
mE

qB qBB
Ans: E
7. A cyclotron operates with a given magnetic field and at a given frequency. If R
denotes the radius of the final orbit, the final particle energy is proportional to:
A) 1/R
B) R
C) R2
D) R3
E) R4
Solutions:
E
1 2 1
mv  mR 2 2
2
2
Ans: C
8. A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with
two equal sides, each 15 cm long. A 0.7 T uniform magnetic field is parallel to the
hypotenuse. The resultant magnetic force on the two sides has a magnitude of:
A) 0
B) 0.21 N
C) 0.30 N
D) 0.41 N
E) 0.51 N
Solutions:
F  I LB.
F on both legs have equal magnitude but opposite directions.
Ans: A
9. You are facing a loop of wire which carries a clockwise current of 3.0 A and
which surrounds an area of 5.8 10−2m2. The magnetic dipole moment of the loop is:
A) 3.0 A  m2, into the page
B) 3.0 A  m2, out of the page
C) 0.17 A  m2, into the page
D) 0.17 A  m2, out of the page
E) 0.17 A  m2, left to right
Solutions:
  IA   3.0 A  0.058m 2   0.17 A  m 2 , into page
Ans: C
10. A loop of current-carrying wire has a magnetic dipole moment of 5  10–4 A 
m2. The moment initially is aligned with a 0.5-T magnetic filed. To rotate the loop so
its dipole moment is perpendicular to the field and hold it in that orientation, you must
do work of:
A) 0
B)
C)
D)
E)
2.5  10–4 J
–2.5  10–4 J
1.0  10–3 J
–1.0  10–3 J
Solutions:
U  μ  B
W    B cos 90   B cos 0   5  104 A  m 2   0.5T  0  1  2.5  104 J
Alternatively,
τ  μB
W    d 
 /2
  B sin  d   B
0
Ans: B
11. The magnitude of the magnetic field at point P, at the center of the semicircle
shown, is given by:
A) 2 0 i / R2
B) 0 i / 2 R
C) 0 i / 4 R
D) 0 i / 2 R
E)
0 i / 4 R
Solutions:
0 I d l  rˆ
4  r 2
No contribution from straight line parts since d l  rˆ  0 .
B
Circular part:
Ans:
B
0 0 i R d
i
 0
2

4  R
4R
(into paper)
E
12. The diagrams show three circuits consisting of concentric circular arcs (either
half or quarter circles of radii r, 2r, and 3r) and radial lengths. The circuits carry the
same current. Rank them according to the magnitudes of the magnetic fields they
produce at C, least to greatest.
A) 1, 2, 3
B) 3, 2, 1
C) 1, 3, 2
D) 2, 3, 1
E) 2, 1, 3
Solutions:
0 I d l  rˆ
I 
 B 0
for arc subtending angle .
2

4
r
4 r
Radial parts don’t contribute.
Upper half circle contributes same field for all cases.
 I 1 
I 4
B1  0   1   0 
4 r  3 
4r 3
B
B2 
0 I  1 
0 I 2

  1   
4 r  3 
4r 3
B3 
0 I  1 1  1  
I 5
   1     0 

4 r  3 2  2  
4r 12
Rank least to greatest :
3,2,1.
Ans: B
13. Which graph correctly gives the magnitude of the magnetic field outside an
infinitely long straight current-carrying wire as a function of the distance r from the
wire?
A) I
B) II
C) III
D) IV
E) V
Solutions:
Ans:
B
0 I
2 r
D
14. Two long straight wires are parallel and carry current in the same direction. The
currents are 8.0 and 12 A and the wires are separated by 0.40 cm. The magnetic field
in tesla at a point midway between the wires is:
A) 0
B) 4.0  10–4
C) 8.0  10–4
D) 0.3  10–4
E) 20  10–4
Solutions:
B
12  8  4  104 T
0 I
 2  107
2 r
0.002
Ans: B
15. Two long straight current-carrying parallel wires cross the x axis and carry
currents I and 3I in the same direction, as shown.
At what value of x is the net
magnetic field zero?
A)
B)
C)
D)
0
1
3
5
E) 7
Solutions:
Parallel currents attract each other  point in between wires.
B
0 I
2 r

point is r from I and 3r from 3I.
Ans: C
16. The diagram shows three equally spaced wires that are perpendicular to the page.
The currents are all equal, two being out of the page and one being into the page.
Rank the wires according to the magnitudes of the magnetic forces on them, from
least to greatest.
A)
B)
C)
D)
1, 2, 3
2, 1 and 3 tie
2 and 3 tie, then 1
1 and 3 tie, then 2
E) 3, 2, 1
Solutions:
f 
f1 
0 I 1 I 2
2 r
0 I 2 
1
0 I 2

1





2 r 
2
4 r
f2 
0 I
 1  1  0
2 r
f3 
0 I 2  1  0 I 2
   1 
2 r  2  4 r
Ans: B
17.
ˆ
The magnetic field at any point is given by B  Ar  k , where r is the
position vector of the point and A is a constant. The net current through a circle of
radius R, in the xy plane and centered at the origin is given by:
A)
B)
C)
D)
E)
 A R2 / 0
2 A R / 0
4 A R3 / 3 0
2 A R2 / 0
A R2 / 2 0
Solutions:
B  A r  kˆ
Ampere’s law:

B is tangential to the circle.
 A R  2 R   0 I

I
2 R 2 A
0
Ans: D
18. A solenoid is 3.0 cm long and has a radius of 0.50 cm. It is wrapped with 500
turns of wire carrying a current of 2.0 A. The magnetic field at the center of the
solenoid is:
A)
B)
C)
D)
E)
9.9  10–8 T
1.3  10–3 T
4.2  10–2 T
16 T
20 T
Solutions:
BL  0 NI
B   4  107 
Ans: C
500  2
 4  102 T
0.03