The Chapter-26 Magnetic Field: (Additional Questions)
1-) An electron that has velocity
r
v = ( 2 106 m/s ) i + ( 3 106 m/s )
j
r
j
moves through the magnetic field B = ( 0.03 T ) i ( 0.15 T ) .
(a) Find the force on the electron.
(b) Repeat your calculation for a proton having the same velocity.
Solution:
(a) The force on the electron is
r
rr
FB = qv B = q vxi + v y Bxi + B y
j
j
(
= q ( vx By v y Bx ) k
)(
)
= ( 1.6 1019 C ) ( 2 106 m/s ) ( 0.15 T ) ( 3 106 m/s ) ( 0.03 T ) k
= ( 6.2 1014 N ) k
r
r
Thus, the magnitude of FB is 6.2 1014 N , and FB points in the positive z direction.
(b) This amount to repeating the above computation with a change in the sign in the change.
r
Thus, FB has the same magnitude but points in the negative z direction.
2-) An electron has a velocity of 1.20 10 4 m/s (in the positive x direction), and an
acceleration of 2.00 1012 m/s2 (in the positive z direction) in uniform electric and magnetic
fields. If the electric field has a magnitude of 20.0 N/C (in the positive z direction), what can
you determine about the magnetic field in the region? What can you not determine?
Solution:
3-) An alpha particle ( q = +2e, m = 4 u ) travels in a circular path of radius 4.5 cm in a
uniform magnetic field with B = 1.2 T. Calculate (a) its speed, (b) its period of revolution, (c)
its kinetic energy in electron-volts, and (d) the potential difference through which it would
have to be accelerated to achieve this energy.
Solution:
(a) For a particle which travels in a circular path in a uniform magnetic field, two forces act
2
on it: Magnetic force is given by FB = qvB , and circular motion yields FC = mv / r . Since
these forces have equal magnitude for equilibrium condition (but have different direction); we
obtain
mv 2
rqB
qvB =
v=
r
m
2
19
rqB ( 4.5 10 m ) ( 2 1.6 10 C ) ( 1.2 T )
v=
=
= 2.6 106 m/s.
27
m
( 4 u ) ( 1.66 10 kg/u )
(b) The period is
T=
2
2 r 2 ( 4.5 10 m )
=
= 1.09 107 s.
6
v
( 2.6 10 m/s )
(c) The kinetic energy of the alpha particle is
27
6
1 2 ( 4 u ) ( 1.66 10 kg/u ) ( 2.6 10 m/s )
K = mv =
= 1.4 105 eV.
19
2
2 ( 1.6 10 J/eV )
2
or
(d)
( 4 u ) ( 1.66 10
1
K = mv 2 =
2
V =
27
kg/u ) ( 2.6 106 m/s )
2
2
= 2.24 1014 J.
K 1.4 105 eV
=
= 7 104 V.
q
2e
or
V =
K
2.24 1014 J
=
= 7 104 V.
q 2 ( 1.6 1019 C )
4-) A proton (charge +e, mass mp), a deuteron (charge +e, mass 2mp), and an alpha particle
(charge +2e, mass 4mp) are accelerated through a common potential difference V. Each of
the particles enters a uniform magnetic field B, with its velocity in a direction perpendicular
to B. The proton moves in a circular path of radius rp. Determine the radii of the circular
orbits for the deuteron, rd, and the alpha particle, r , in terms of rp .
Solution:
5-) charged Singly uranium-238 ions are accelerated through potential difference of 2.00 kV
and enter a uniform magnetic field of 1.20 T directed perpendicular to their velocities.
(a) Determine the radius of their circular path. (b) Repeat for uranium-235 ions. How does the
ratio of these path radii depend on the accelerating voltage and on the magnitude of the
magnetic field?
Solution:
The ratios of the orbit radius for different ions are independent of V and B.
6-) A length L of wire carries a current i . Show that if the wire is formed into a circular coil,
then the maximum torque in a given magnetic field is developed when the coil has one turn
only, and that maximum torque has the magnitude = L2iB / 4 .
Solution:
If N closed loops are formed from the wire of length L, the circumference of each loop is L /N,
the radius of each loop is R = L / 2N, and the area of each loop is A = R2 = (L / 2N)2 =
L2/4N 2. For maximum torque, we orient the plane of the loops parallel to the magnetic field,
so the dipole moment is perpendicular to the field. The magnitude of the torque is then
L2
iL2 B
= NiAB = ( Ni )
B=
.
2
4 N
4 N
To maximize the torque, we take N to have the smallest possible value, 1. Then = iL2 B / 4 .
7-) A circular loop of wire having a radius of 8 cm carries a current of 0.2 A. A vector of unit
r
length and parallel to the dipole moment of the loop is given by 0.6i 0.8 . If the loop is
j
r
located in a uniform magnetic field given by B = (0.25 T)i + (0.3 T)k , find (a) the torque on
the loop (in unit-vector notation) and (b) the magnetic potential energy of the loop.
Solution:
r
The magnetic dipole moment is = (0.6i 0.8 ) , where,
j
= NiA = Ni r 2 = 1(0.2 A) (0.8 m) 2 = 4.02 104 A m 2 .
Here i is the current in the loop, N is the number of turns, A is the area of the loop, and r is
its radius.
(a) The torque is
rrr
= B = (0.6i 0.8 ) (0.25i + 0.3k )
j
= [(0.6)(0.3)(i k ) (0.8)(0.25)( i ) (0.8)(0.3)( k )]
j
j
= ( 0.18 + 0.2k 0.24i ).
j
j
Here i k = , i = k , k = i are used. We also use i i = 0 . Now, we substitute the
jj
value for to obtain
r
= (0.97 104 i 7.2 104 + 8 104 k ) N m.
j
(b) The potential energy of the dipole is given by
rr
U = B = (0.6i 0.8 ) (0.25i + 0.3k )
j
= (0.6)(0.25) = 0.15 = 6 104 J.
Here i i = 1 , i k = 0 , i = 0 , and k = 0 are used.
j
j
8-) A wire is formed into a circle having a diameter of 10.0 cm and is placed in a uniform
magnetic field of 3.00 mT. The wire carries a current of 5.00 A. Find (a) the maximum torque
on the wire and (b) the range of potential energies of the wirefield system for different
orientations of the circle.
Solution:
(a)
(b)
Since
the range of the potential energy is: