PHYSICS 204 CHAPTER 28 SOLUTION
MATERIAL (ALL CORRECT
The Magnetic Field
Q1. A loop of wire carrying a current of 2.0 A is in the shape of a right-angle
triangle with two equal sides, each 15 cm long. A 0.70-T uniform magnetic field is
in the plane of the triangle and is directed as shown in figure 7. The magnetic force
on side 3 is : Ans:0.21 N, into of the page
The Definition of B
Q2. An electron is projected into a uniform magnetic field B = (0.8 k) T. Find the
magnitude of the magnetic force, on the electron when the velocity is:
v = (5.0×105 i + 3.0×105 j) m/sec.
(i, j and k are the unit vectors in the x, y and z directions, respectively).Ans:7.5×10-14
N.
Q3. In figure (4), a loop of wire carrying a current, I, 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 in
the plane of the triangle and is perpendicular to the hypotenuse. The resultant
magnetic force on the two equal sides is: Ans:0.30 N, into the page.
Q4. An electron that has velocity v = 3.2×107 i m/s traveling parallel to a uniform
magnetic field of strength 2.60×10-3 Tesla. The force on the electron is: [i is the unit
, PHYSICS 204 CHAPTER 28 SOLUTION
MATERIAL (ALL CORRECT
vectors in the directions of x]Ans:zero.
Q5. At a point in a uniform magnetic field the acceleration of an electron is 5.0×1014
m/s2 and its speed is 7.0×106 m/s. If the magnitude of the magnetic field is 1.0 mT,
what is the angle between the electrons velocity and the magnetic field?Ans:24
degrees.
Crossed Fields
Q6. In the velocity selector, an electron passes, without deflection, through a region
where there is an electric and a magnetic field perpendicular to each other. If the
electric field is E=4.5×103 V/m and the magnetic field is B=9.0×10-3 T, what is the
kinetic energy of the electron, in electron Volts? ( 1 eV=1.6×10-19 J)Ans:0.71 eV. Q7.
In a region where an electric field E and a magnetic field B are perpendicular to each
other, a beam of charged particles, each of mass m and charge q, passes through this
region. The kinetic energy of a particle passing undeflected normal to these two fields
is:
Ans:m E2 /2B2.
Q8. An electric field of 1.5×103 V/m and a magnetic field of 0.50 T act on a moving
electron to produce no net force. Calculate the minimum speed of the moving
electron.
Ans:3.0×103 m/s.
Q9. A proton with velocity v = (2.0×106) i (m/s) moves horizontally into a region of
space in which there is an electric field E = (- 5000) j (N/C) and a magnetic field B.
Find the smallest magnetic field such that the
MATERIAL (ALL CORRECT
The Magnetic Field
Q1. A loop of wire carrying a current of 2.0 A is in the shape of a right-angle
triangle with two equal sides, each 15 cm long. A 0.70-T uniform magnetic field is
in the plane of the triangle and is directed as shown in figure 7. The magnetic force
on side 3 is : Ans:0.21 N, into of the page
The Definition of B
Q2. An electron is projected into a uniform magnetic field B = (0.8 k) T. Find the
magnitude of the magnetic force, on the electron when the velocity is:
v = (5.0×105 i + 3.0×105 j) m/sec.
(i, j and k are the unit vectors in the x, y and z directions, respectively).Ans:7.5×10-14
N.
Q3. In figure (4), a loop of wire carrying a current, I, 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 in
the plane of the triangle and is perpendicular to the hypotenuse. The resultant
magnetic force on the two equal sides is: Ans:0.30 N, into the page.
Q4. An electron that has velocity v = 3.2×107 i m/s traveling parallel to a uniform
magnetic field of strength 2.60×10-3 Tesla. The force on the electron is: [i is the unit
, PHYSICS 204 CHAPTER 28 SOLUTION
MATERIAL (ALL CORRECT
vectors in the directions of x]Ans:zero.
Q5. At a point in a uniform magnetic field the acceleration of an electron is 5.0×1014
m/s2 and its speed is 7.0×106 m/s. If the magnitude of the magnetic field is 1.0 mT,
what is the angle between the electrons velocity and the magnetic field?Ans:24
degrees.
Crossed Fields
Q6. In the velocity selector, an electron passes, without deflection, through a region
where there is an electric and a magnetic field perpendicular to each other. If the
electric field is E=4.5×103 V/m and the magnetic field is B=9.0×10-3 T, what is the
kinetic energy of the electron, in electron Volts? ( 1 eV=1.6×10-19 J)Ans:0.71 eV. Q7.
In a region where an electric field E and a magnetic field B are perpendicular to each
other, a beam of charged particles, each of mass m and charge q, passes through this
region. The kinetic energy of a particle passing undeflected normal to these two fields
is:
Ans:m E2 /2B2.
Q8. An electric field of 1.5×103 V/m and a magnetic field of 0.50 T act on a moving
electron to produce no net force. Calculate the minimum speed of the moving
electron.
Ans:3.0×103 m/s.
Q9. A proton with velocity v = (2.0×106) i (m/s) moves horizontally into a region of
space in which there is an electric field E = (- 5000) j (N/C) and a magnetic field B.
Find the smallest magnetic field such that the
