Exam 2 Solutions - Department of Physics

PHY2054 Fall 2014

Exam 2 Solutions

Prof. Paul Avery Prof. Andrey Korytov Oct. 29, 2014

1. A loop of wire carrying a current of 2.0 A is in the shape of a right triangle

with two equal sides, each with length L = 15 cm as shown in the figure. The

B

triangle lies within a 0.7 T uniform magnetic field in the plane of the triangle

and perpendicular to the hypotenuse. The magnetic force on either of the two

sides has a magnitude of:

Answer: 0.15 N ! !!

( ) Solution: The force on each side is F = I ? B L,

F = BLi sin 45? = 0.7 ? 0.15? 2.0 ? 0.707 = 0.15 N.

F = IBL sin

2. Refer to the previous problem. What is the magnitude of the torque on this loop due to the magnetic field (in Nm)?

Answer: 15.8 ? 10-3

( ) Solution: The torque is

! =

?!

?

! B

,

= ?Bsin = (iA)Bsin

( ) = iABsin 90! = 2.0 ? 0.5? 0.152 ? 0.7 = 1.58 ?10-2 Nm.

3. An electrical device with an equivalent resistance of 100 has a maximum instantaneous voltage rating of 60 V. Assume that it is plugged into a standard 120 V electrical outlet in series with a resistance R. What is the approximate minimum value of R needed to keep the instantaneous voltage across the device within its safety limits?

Answer: 183

Solution: The maximum instantaneous voltage from the 120 V outlet is 120 ? 2 = 169.7 V. We want to find the value of R such that the voltage across the device is no more than 60 V.

The device and resistor R are connected in series: RTOT=100+R. The current through the device is I = 169.7/ RTOT. And the voltage across the device is V=100I must satisfy the condition V60

( ) ( ) V. Putting all together, we get 100 169.7 / 100 + R = 60 . Solving for R yields R = 183 .

X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X

4. A 14.0 g conducting rod of length 1.30 m and resistance 8.0 slides freely downward between two vertical conducting rails without friction. The entire apparatus is placed in a 0.43 T uniform magnetic field. Ignore the electrical resistance of the rails. What is the terminal velocity of the rod? (g = 9.8 m/s2) Answer: 3.51 m/s

1

PHY2054 Fall 2014

Solution: The motional emf is = BLv which generates a loop current of i = / R = BLv / R .

At terminal velocity, the magnetic force on this current equals the force due to gravity or mg = BiL = B2L2v / R . Solving yields v = 3.51 m/s.

5. A transformer consists of a 300-turn primary coil and a 2000-turn secondary coil. If the current in the secondary coil is 9.00 A, what is the primary current?

Answer: 60.0 A Solution: Using V2 / V1 = N2 / N1 = 6.67 and i1V1 = i2V2 , with i2 = 9.0 A, we get i1 = 60.0 A.

6. A hair dryer has a power rating of 1200 W when plugged into a standard 110 V AC receptacle. How much current does it draw if it is connected instead to a 110 V DC source?

Answer: 10.9 A

Solution: Power dissipated into hit by a resistor connected to AC of rms emf

rms is P =

2 rms

/

R

,

from where one can find R. The current through a resistor connected to DC source emf 0 is

I =

0 / R . Therefore, I =

0

P

2

= 10.9 A.

rms

7. The number of turns is tripled for an ideal solenoid, and its length is doubled while holding its cross-sectional area constant. If the old inductance is L, what is the new inductance?

Answer: 9L/2

Solution: The inductance is proportional to the cross sectional area, the solenoid length and the square of n = number of turns/length: L = ?0n2A Then, if n 3n / 2 (triple turns over twice the length) and l 2l (length doubled), the new

( ) ( ) inductance will be Lold ? 3 / 2 2 ? 2 = 9 / 2 Lold

2

PHY2054 Fall 2014

8. In the circuit shown, what is the current through the 15 V battery?

12 V +

Answer: 1.18 A upwards

4

Solution: We choose the currents in the branches to be I1 (left

branch upwards), I2 (middle branch upwards) and I3 = I1 + I2

(right branch downwards). Then the loop equations for left-

middle and middle-right (others are possible, but these are convenient) are

-4I1 + 12 - 15 + 8I2 = 0 -8I2 + 15 - 2I3 = 0

Solving for I2 yields I2 = +33 / 28 +1.18 A.

+ 15 V

8

2

9. An ammeter with full-scale deflection for I = 20.0 A has an internal resistance of 31.5 . We need to use this ammeter to measure currents up to 30.0 A. What size resistor should be used to protect the ammeter?

Answer: 63.0

Solution: The "shunt" resistance R is placed in parallel with the ammeter so that part of the current to be measured is shunted around the device to avoid damaging it. We pick R so that at 30 A total current, 10 A goes through the shunt resistor and 20 A through the ammeter itself. The voltage across the ammeter is 20 ? 31.5 = 630 V, which is the same as the voltage across R. Thus to get a current of 10 A through the resistor we require R = 63.0 .

10. An electromagnetic flowmeter is used to measure blood flow rates

during surgery. Blood containing Na+ ions flow to the left through an

B

v

artery with an inner diameter of D = 4.00 mm. The artery is in a

downward magnetic field of B = 0.100 T and develops a Hall voltage VH = 80.0 ?V across its diameter. What is the volumetric flow rate of the blood in (m3/s)?

Answer: 2.51 ? 10-6

Solution: In the Hall effect, ions are bent in the magnetic field toward the arterial walls until the

( ) force FE = eE = e VH / D due to electric field E generated by the charges on opposite sides of

the diameter exactly balances the magnetic force FB = evB . The force equality condition is

( ) e VH / D = evB . This yields v = VH / BD for the velocity. The volume flow rate F = area ?

( )( ) velocity or F = D2 / 4 VH / BD = DVH / 4B = 2.51 ? 10-6 m3/s.

3

PHY2054 Fall 2014

11. An electromagnet is made by inserting a soft iron core into a solenoid. The solenoid has 1,700 turns, radius 2.4 cm, and length 13.0 cm. When 0.5 A of current flows through the solenoid, the magnetic field inside the iron core has magnitude 0.43 T. What is the relative permeability B of the iron core? Answer: 52.3 Solution: The B field of a solenoid with a ferromagnetic core is B = B?0ni , where B is the permeability of the core material. Here n = 1700 / 0.13 = 13,100 . Using the values shown we obtain B = 52.3.

12. Four long parallel wires pass through the corners of a square with side 14.1 cm as shown in the figure. All four wires carry the same magnitude of current I = 5.7 A in the directions indicated. Find the magnetic field at the center of the square. Answer: 32.3 ?T Solution: Magnetic field lines make circles around lines of current, with their direction going clockwise for wires with currents into the page and counterclockwise for currents going out of the page. At the center of the square they point either at 45? (for the top-left and bottom-right wires) or at -45? (for the top-right and bottom-left wires). By inspection, we see that the y components of the field cancel at the center while the four x components are all identical. If the side length is L, we obtain B = 4 ? 2?0i / 2 L = 32.3 ?T .

13. The armature of an AC generator is a rectangular coil 10.5 cm ? 4.0 cm with 210 turns. It is immersed in a uniform magnetic field of magnitude 1.5 T. If the amplitude of the emf in the coil is 30.0 V, at what angular speed (in rad/sec) is the armature rotating? Answer: 22.7 Solution: Emf induced in a coil of N loops and area A (A = ab ) rotating with angular velocity

in magnetic field B is = NBA sint . Using the values shown for the emf amplitude, B field,

number of turns and coil dimensions yields = 22.7 rad/sec.

14. A dc motor has coils with resistance of 13 and is connected to an emf of 127 V. When the motor operates at full speed, the back emf is 63 V. If the current is 6.1 A with the motor operating at less than full speed, what is the back emf at that time? Answer: 47.7 V

( ) Solution: I = - external back / R . Therefore, back emf is 127 - 13? 6.1 = 47.7 V.

4

PHY2054 Fall 2014

15. A coil has an inductance of 0.31 H and a resistance of 43 . The coil is connected to a 6.4 V battery. After a long time elapses, the current in the coil is no longer changing. What is the energy stored in the coil?

Answer: 3.43 mJ

Solution: After a long time the inductance has no effect and the current is obtained from

i

=

/

R

=

6.4 /

43 =

0.149

A.

The

energy

stored

in

the

coil

is

then

UL

=

1 2

Li2

=

3.43 mJ

.

16. A circuit breaker trips when the rms current exceeds 10.3 A. How many 60 W lightbulbs, connected in parallel, can be run on this circuit without tripping the breaker? (The rms voltage of the circuit is 110 V.)

Answer: 18

Solution: Each lightbulb draws a current of 60/110 = 0.545 A. Since 10.3/0.545 ~ 18.9, we see that no more than 18 lightbulbs can be supported on this circuit.

17. A sample containing carbon (atomic mass 12 u) and oxygen (16 u) is placed in a mass spectrometer. The ions all have the same charge and are accelerated through the same potential difference before entering a uniform magnetic field. For each ion, the spectrometer measures the distance between the entrance point and the place where it strikes the bottom of the region. If the carbon ions strike the detector at a distance of 9.14 cm from the entrance point, at what distance will oxygen ions strike the detector?

Answer: 10.6 cm

Solution: The ions travel in circular paths and land a distance of one diameter from the entrance

point. The centripetal force equation mv2 / r = evB yields for the diameter d = 2r = 2mv / eB .

All ions have the same kinetic energy

1 2

mv2

=

eV

which gives

mv =

2meV . This yields for the

diameter d = 2 2mV / eB2 ~ m , i.e. the distance is proportional to the square root of the mass. The distance for O16 ions can then be determined by scaling the distance for C12 ions:

dO = dC mO / mC = 9.14 16 / 12 = 10.6 cm.

18. The current I in a long wire is going up as shown in the figure and increasing in magnitude. What is the direction of the induced current in

I

the left loop and the right loop? List the direction of the induced current in

the left loop first. (CW = clockwise, CCW = counterclockwise)

Answer: CW, CCW

Solution: Using the right hand rule, we see that the B field is increasing out of the page on the left side and increasing into the page on the right. Lenz' law says that to counteract those fields the induced currents must generate B fields into the page and out of the page, respectively. Thus the currents must flow clockwise (left) and counterclockwise (right).

5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download