Electric, Gravitational, and Magnetic Fields

SPH4UUnit SummativesJanuary 10, 2014

Part I

Please choose the most correct answer to the following ten questions. (Each question weighs 2 points)

1.) / The diagram shows current I flowing in a circular coil located in a magnetic field. The magnetic force acting on the coil will tend to cause it to
A.) / contract.
B.) / move down the page.
C.) / move up the page.
D.) / expand.
2.) / A solenoid has a length of 0.30 m, a diameter of 0.040 m and 500 windings. The magnetic field at its centre is 0.045 T. What is the current in the windings?
A.) / 2.9 A
B.) / 21 A
C.) / 3.0 A
D.) / 170 A
3.) / A solenoid of length 0.75 m has a radius 0.092 m. A current of 25 A flows through its 4 700 turns. Within the solenoid a 0.10 m long conductor moves at 4.3 m/s perpendicular to the field of the solenoid. What emf is induced between the ends of the conductor?
A.) / 0.085 V
B.) / 0.197 V
C.) / 0.430 V
D.) / 4.80 V
4.) / The electric force on each of two small charged spheres due to the other has a magnitude of F. If the charge of the one sphere is doubled and the distance between centres is tripled, the force on each small charged sphere is:
A.)2F
B.)2F/3
C.)F/3
D.)F/9
E.)2F/9
5.) / The magnitude of the magnetic field a distance r from a long, straight conductor with current I is B. The same magnetic field will have magnitude 2B if the current tripled at a distance of.
A.)2r/3
B.)2r
C.)3r
D.)3r/2
E.)None of these
7.) / 6.) A charged particle travels in a circular path in a magnetic field. What changes to the magnetic field and to the velocity of the particle would both cause the radius of its path to decrease?
A.)Magnetic field change: decrease; velocity change: decrease
B.)Magnetic field change: increase; velocity change: decrease
C.)Magnetic field change: increase; velocity change: increase
D.)Magnetic field change: decrease; velocity change: increase
Which of the following diagrams best shows the magnetic field due to a long straight wire carrying conventional current I as shown?
A.) /
B.) /
C.) /
D.) /
8.) / The diagram below shows an aluminum ring and the current induced in it by the nearby magnet that is free to move along its central axis. The magnet must be
A.) / spinning about its central axis.
B.) / moving to the right.
C.) / moving to the left.
D.) / stationary.
9.) / In which of the following situations would the greatest emf be induced in the coil? All changes occur in the same time interval.
A.) /
B.) /
C.) /
D.) /
10.) / An electric motor rotates at various speeds and the current through the armature changes accordingly. Which pair of conditions occurs when the motor generates the greatest back emf?
A.) / Speed: fastest; Current: largest
B.) / Speed: fastest; Current: smallest
C.) / Speed: slowest; Current: smallest
D.) / Speed: slowest; Current: largest

Part II

Please provide a short answer of no more than 50 words to the following question: (8 points)

Coulomb’s law may be used to calculate the force between charges only under certain conditions. State the conditions and explain why they are imposed.

Ans.:

For Coulomb’s law to be applicable, the charges of the two spheres are to be very small and they are small compared to the distance between them. If these conditions are not met, then the presence of a charged sphere forces the charge of each sphere to redistribute on the surface so that r can no longer be measured from the center. The charge distributions stays nearly uniform because of the repulsive forces between charges on each sphere and the distance is large between the spheres compared to the sizes.

Part III

An electron with a speed of 5.0 x 106 m/s is injected into a parallel plate apparatus, in a vacuum, through a hole in the positive plate. The electron collides with the negative plate at

1.0x 106 m/s .

What is the potential difference between the plates?(10 pts)

(mass of an electron: 9.1 x 10-31 kg, charge of an electron: 1.6 x 10-19 C)

Ans.:

For the free electron,

-Ek = E

½ mvi)2 -vf)2 ) = q V

V = mvi)2 -vf)2 ) / 2q

= 9.1 x 10-31 kg ((5.0 x 106 m/s)2 –( 1.0 x 106 m/s)2)/ (2 x 1.6 x 10-19 C)

= 68 V

The potential difference between plates is 68 V.

A proton, of mass 1.67 x 10-27 kg, moves in a circle in the plane perpendicular to a uniform magnetice field of magnitude 3.6 T. The speed of proton is 5.2 x 106 m/s.

What is radius of the curvature?(12 pts)

Ans. :

q = 1.6 x 10-19 Cv = 5.2 x 106 m/s

m = 1.67 x 10-27 kgB= 3.6 T = 3.6 kg/C.s

FM = Fc

qvB = (mv2)/r

r = (mv)/qB

= (1.67 x 10-27 kg x 5.2 x 106 m/s)/(1.6 x 10-19 C x 3.6 kg/C.s)

= 1.5 x 10-2 m

The radius of the curvature is 1.5 x 10-2 m.

Part I

Please choose the most correct answer to the following ten questions. (Each question weighs 2 points)

1.) / The diagram shows current I flowing in a circular coil located in a magnetic field. The magnetic force acting on the coil will tend to cause it to
A.) / contract.
B.) / move down the page.
C.) / move up the page.
D.) / expand.
2.) / A solenoid has a length of 0.30 m, a diameter of 0.040 m and 500 windings. The magnetic field at its centre is 0.045 T. What is the current in the windings?
A.) / 2.9 A
B.) / 21 A
C.) / 3.0 A
D.) / 170 A
3.) / A solenoid of length 0.75 m has a radius 0.092 m. A current of 25 A flows through its 4 700 turns. Within the solenoid a 0.10 m long conductor moves at 4.3 m/s perpendicular to the field of the solenoid. What emf is induced between the ends of the conductor?
A.) / 0.085 V
B.) / 0.197 V
C.) / 0.430 V
D.) / 4.80 V
4.) / The diagram shows a conductor between a pair of magnets. The current in the conductor flows out of the page. In what direction will the magnetic force act on the conductor?
A.)towards the right
B.)down the page
C.)towards the left
D.)up the page
5.) / The diagram below shows two coils in a magnetic field. An electric current can be induced in the coil oriented with its plane
A.)parallel to a changing magnetic field.
B.)parallel to a constant magnetic field.
C.)perpendicular to a changing magnetic field.
D.)perpendicular to a constant magnetic field.
7.) / Which of the following diagrams best shows the magnetic field due to a long straight wire carrying conventional current I as shown?
A.) /
B.) /
C.) /
D.) /
8.) / The diagram below shows an aluminum ring and the current induced in it by the nearby magnet that is free to move along its central axis. The magnet must be
A.) / spinning about its central axis.
B.) / moving to the right.
C.) / moving to the left.
D.) / stationary.
9.) / In which of the following situations would the greatest emf be induced in the coil? All changes occur in the same time interval.
A.) /
B.) /
C.) /
D.) /
10.) / A proton is travelling at 2.3 x 106 m/s in a circular path in a 0.75 T magnetic field. What is the magnitude of the force of the proton?
A.)1.7 N
B.)1.6 x 10-24 N
C.)2.8 x 10-13 N
D.)2.9 x 10-21 N

Part II

Please provide a short answer of no more than 50 words to the following question: (8 points)

Coulomb’s law may be used to calculate the force between charges only under certain conditions. State the conditions and explain why they are imposed.

Ans.:

Part III

An electron with a speed of 6.1 x 106 m/s is injected into a parallel plate apparatus, in a vacuum, through a hole in the positive plate. The electron collides with the negative plate at
x 106 m/s .

What is the potential difference between the plates?(10 pts)

(mass of an electron: 9.1 x 10-31 kg, charge of an electron: 1.6 x 10-19 C)

Ans.:

For the free electron,

-Ek = E

½ mvi)2 -vf)2 ) = q V

V = mvi)2 -vf)2 ) / 2q

= 9.1 x 10-31 kg ((6.1 x 106 m/s)2 –( 1.9 x 106 m/s)2 )/ (2 x 1.6 x 10-19 C)

= 96 V

The potential difference between plates is 96 V.

A proton, of mass 1.67 x 10-27 kg, moves in a circle in the plane perpendicular to a uniform magnetice field of magnitude 1.8 T. The radius of curvature is 3.0 cm.

What is the speed of proton?(12 pts)

Ans. :

q = 1.6 x 10-19 Cr = 3.0 x 10-2 m

m = 1.67 x 10-27 kgB= 1.8 T = 1.8 kg/C.s

FM = Fc

qvB = (mv2)/r

v = (qBr)/m

= (1.6 x 10-19 C x 1.8 kg/C.s x 3.0 x 10-2 m)/ 1.67 x 10-27 kg

= 5.2 x 106 m/s

The speed of proton is 5.2 x 106 m/s.